Difference between revisions of "Series - Tests for Convergence/Divergence"

Latest revision as of 12:13, 26 April 2015

This page is meant to provide guidelines for actually applying series convergence tests. Although no examples are given here, the requirements for each test are provided.

Important Series

There are two series that are important to know for a variety of reasons. In particular, they are useful for comparison tests.

Geometric series. These are series with a common ratio Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r} between adjacent terms which are usually written

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=0}^{\infty}a_{0}r^{k}.}

These are convergent if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |r|<1} , and divergent if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |r|\geq1} . If it is convergent, we can find the sum by the formula

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S=\frac{a_{0}}{1-r},}

where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_{0}} is the first term in the series (if the index starts at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k=2} or Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k=6} , then "Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_{0}} " is actually the first term Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_{2}} or Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_{6}} , respectively).

p-series. These are series of the form

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^{\infty}\frac{1}{k^{p}}.}

If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p>1} , then the series is convergent. On the other hand, if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p\leq1} , the p-series is divergent.

The Divergence Test

If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\displaystyle \lim_{k\rightarrow\infty}a_{k}\neq0,}} then the series/sum Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=0}^{\infty}a_{k}} diverges.

Note: The opposite result doesn't allow you to conclude a series converges. If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\displaystyle \lim_{k\rightarrow\infty}a_{k}=0}} , it merely indicates the series might converge, and you still need to confirm it through another test.

In particular, the sequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\{ \frac{1}{k}\right\} } converges to zero, but the sum Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=0}^{\infty}\frac{1}{k}} , our harmonic series, diverges.

The Integral Test

Suppose the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)} is continuous, positive and decreasing on some interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [c,\infty)} with Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c\geq1} , and let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_{k}=f(k)} . Then the series Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=b}^{\infty}a_{k}} is convergent if and only if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c\geq b} and

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{c}^{\infty}f(x)\, dx}

is convergent (not infinite).

Note: This test, like many of them, has a few specific requirements. In order to use it on a test, you need to state/show:

For all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k\geq c} for some Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c\geq b} , the function is positive. (Most of the time, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c} is just my starting index Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b} ).
For all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k\geq c} , the function is decreasing.
The integral is convergent (or divergent, if you're proving divergence).

Then, you can say, "By the Integral Test, the series is convergent (or divergent)."

I wrote this with Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c} instead of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b} for a lower bound to indicate you only need to show the series and function are "eventually" decreasing, positive, etc. In other words, we don't care what happens at the beginning (or head) of a series - only at the end (or tail).

The Comparison Test

Suppose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^{\infty} b_{k}} is a series with positive terms, and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^{\infty} a_{k}} is a series with eventually positive terms. Then

If for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k\geq c} for some Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c} greater than or equal to our starting index, and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^{\infty} b_{k}} is convergent, then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^{\infty} a_{k}} is convergent.

If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_{k}\geq b_{k}} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^{\infty} b_{k}} is divergent, then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^{\infty} a_{k}} is divergent.

Note: Requirements for this test include showing (or at least stating):

For all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k\geq c} for some Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c} greater than or equal to our starting index, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_{k}} is positive. (Most of the time, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c} is just the starting index.)

For all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k\geq c} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_{k}\leq b_{k}} for convergence, or Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_{k}\geq b_{k}} for divergence.
(This is important) State why $\sum _{k=1}^{\infty }b_{k}$ is convergent, such as a p-series with $p>1$ , or a geometric series with $|r|<1$ . Obviously, you would need to state why it is divergent if you're showing it's divergent.

Then, you can say, "By the Comparison Test, the series is convergent (or divergent)."

The Limit Comparison Test

Suppose $\sum _{k=1}^{\infty }a_{k}$ and $\sum _{k=1}^{\infty }b_{k}$ are series with positive terms. If $\lim _{k\rightarrow \infty }{\frac {a_{k}}{b_{k}}}=c$ where $0<c<\infty$ , then either both series converge, or both series diverge.

Additionally, if $c=0$ and $\sum _{k=1}^{\infty }b_{k}$ converges, $\sum _{k=1}^{\infty }a_{k}$ also converges. Similarly, if $c=\infty$ and $\sum _{k=1}^{\infty }b_{k}$ diverges, then $\sum _{k=1}^{\infty }a_{k}$ also diverges.

Note: First of all, let's mention the fundamental idea here. If some series $\sum _{k=1}^{\infty }b_{k}$ converges, then $\sum _{k=1}^{\infty }cb_{k}$ converges where $c\neq \pm \infty$ is a constant. This test shows that one series eventually is just like the other one multiplied by a constant, and for that reason it will also converge/diverge if the one compared to converges/diverges. To use it, you need to state/show:

$a_{k}$ is eventually positive (really, non-negative).

${\displaystyle \lim _{k\rightarrow \infty }{\frac {a_{k}}{b_{k}}}}=c$ .

State why $\sum _{k=1}^{\infty }b_{k}$ is convergent, such as a p-series with $p>1$ , or a geometric series with $|r|<1$ . Obviously, you would need to state why it is divergent if you're showing it's divergent.

Then, you can say, "By the Limit Comparison Test, the series is convergent (or divergent)."

Like the Comparison Test and the Integral Test, it's fine if the first terms are kind of "wrong" - negative, for example - as long as they eventually wind up (for $k>c$ for a particular $c$ ) meeting the requirements.

The Alternating Series Test

If a series $\sum _{k=1}^{\infty }a_{k}$ is

Alternating in sign, and

$\lim _{k\rightarrow 0}|a_{k}|=0,$

then the series is convergent.

Note: This is a fairly straightfoward test. You only need to do two things:

Mention the series is alternating (even though it's usually obvious).

Show the limit converges to zero.

Then, you can say, "By the Alternating Series Test, the series is convergent."

As an additional detail, if it fails to converge to zero, then you would say it diverges by the Divergence Test, not the Alternating Series Test.

The Ratio Test

Let $\sum _{k=1}^{\infty }a_{k}$ be a series. Then:

If $\lim _{k\rightarrow \infty }\left|{\frac {a_{k+1}}{a_{k}}}\right|=L<1$ , the series is absolutely convergent (and therefore convergent).

If $\lim _{k\rightarrow \infty }\left|{\frac {a_{k+1}}{a_{k}}}\right|=L>1$ or $\lim _{k\rightarrow \infty }\left|{\frac {a_{k+1}}{a_{k}}}\right|=L=\infty$ , the series is divergent.

If $\lim _{k\rightarrow \infty }\left|{\frac {a_{k+1}}{a_{k}}}\right|=L=1$ , the Ratio Test is inconclusive.

Note: Both this and the Root Test have the least requirements. The Ratio Test does require that such a limit exists, so a series like

$0+1+0+{\frac {1}{4}}+0+{\frac {1}{9}}+\cdots$

could not be assessed as written with the Ratio Test, as division by zero is undefined. You might have to argue it's the same sum as

$1+{\frac {1}{4}}+{\frac {1}{9}}+\cdots ,$

and you could then apply the Ratio Test.

The Root Test

Let $\displaystyle \sum _{k=0}^{\infty }a_{k}$ be a series. Then:

If ${\displaystyle \lim _{k\rightarrow \infty }{\sqrt[{k}]{|a_{k}|}}}=L<1,$ the series is absolutely convergent (and therefore convergent).

If ${\displaystyle \lim _{k\rightarrow \infty }{\sqrt[{k}]{|a_{k}|}}}=L>1$ or ${\displaystyle \lim _{k\rightarrow \infty }{\sqrt[{k}]{|a_{k}|}}}=L=\infty ,$

the series is divergent.

If ${\displaystyle \lim _{k\rightarrow \infty }{\sqrt[{k}]{|a_{k}|}}}=L=1$ , the Root Test is inconclusive.

@@ Line 24: / Line 24: @@
 == The Divergence Test ==
-If <math style="vertical-align: -65%">{\displaystyle \lim_{k\rightarrow\infty}a_{k}\neq0,}</math>
+If <math style="vertical-align: -65%">{\displaystyle \lim_{k\rightarrow\infty}a_{k}\neq0,}</math> then the series/sum  <math style="vertical-align: -98%">\sum_{k=0}^{\infty}a_{k}</math> diverges.
-then the series/sum
-::<math>\sum_{k=0}^{\infty}a_{k}</math>
-diverges.
 '''<u>Note</u>:''' The opposite result <u>''doesn't''</u> allow you to conclude a series converges. If <math style="vertical-align: -60%">{\displaystyle \lim_{k\rightarrow\infty}a_{k}=0}</math>&thinsp;, it merely indicates the series <u>''might''</u> converge, and you still need to confirm it through another test.
-In particular, the sequence <math style="vertical-align: -18%">\left\{ \frac{1}{k}\right\}  </math> converges to zero, but the sum <math style="vertical-align: -42%">\sum_{k=0}^{\infty}\frac{1}{k}</math>&thinsp;, our harmonic series, diverges.
+In particular, the sequence <math style="vertical-align: -38%">\left\{ \frac{1}{k}\right\}  </math> converges to zero, but the sum <math style="vertical-align: -65%">\sum_{k=0}^{\infty}\frac{1}{k}</math>&thinsp;, our harmonic series, diverges.
 == The Integral Test ==
-Suppose the function <math style="vertical-align: -20%">f(x)</math> is continuous, positive and decreasing on some interval <math style="vertical-align: -22%">[b,\infty)</math> with <math style="vertical-align: -13%">b\geq1</math>,
+Suppose the function <math style="vertical-align: -20%">f(x)</math> is continuous, positive and decreasing on some interval <math style="vertical-align: -22%">[c,\infty)</math> with <math style="vertical-align: -13%">c\geq1</math>,
-and let <math style="vertical-align: -21%">a_{k}=f(k)</math>. Then the series <math style="vertical-align: -87%">\sum_{k=b}^{\infty}a_{k}</math> is convergent if and only if for some <math style="vertical-align: -13%">c\geq b</math>,
+and let <math style="vertical-align: -21%">a_{k}=f(k)</math>. Then the series <math style="vertical-align: -87%">\sum_{k=b}^{\infty}a_{k}</math> is convergent if and only if  <math style="vertical-align: -13%">c\geq b</math> and
 ::<math>\int_{c}^{\infty}f(x)\, dx</math>
@@ Line 45: / Line 40: @@
 is convergent (not infinite).
-'''<u>Note</u>:''' This test, like many of them has a few requirements. In order to use it on a test, you need to state/show:
+'''<u>Note</u>:''' This test, like many of them, has a few specific requirements. In order to use it on a test, you need to state/show:
 :* For all <math style="vertical-align: -15%">k\geq c</math> for some <math style="vertical-align: -15%">c\geq b</math>, the function is positive. (Most of the time, <math style="vertical-align: 0%">c</math> is just my starting index <math style="vertical-align: 0%">b</math>).
@@ Line 61: / Line 56: @@
 :* If for all <math style="vertical-align: -15%">k\geq c</math> for some <math style="vertical-align: 0%">c</math> greater than  or equal to our starting index, and <math style="vertical-align: -98%">\sum_{k=1}^{\infty} b_{k}</math> is convergent, then <math style="vertical-align: -98%">\sum_{k=1}^{\infty} a_{k}</math> is convergent.
-:* If <math style="vertical-align: -18%">a_{k}\geq b_{k}</math> for all <math style="vertical-align: 0%">k</math> and <math style="vertical-align: -98%">\sum_{k=1}^{\infty} b_{k}</math> is divergent, then <math style="vertical-align: -98%">\sum_{k=1}^{\infty} a_{k}</math> is divergent.
+:* If <math style="vertical-align: -15%">a_{k}\geq b_{k}</math> for all <math style="vertical-align: 0%">k</math> and <math style="vertical-align: -98%">\sum_{k=1}^{\infty} b_{k}</math> is divergent, then <math style="vertical-align: -98%">\sum_{k=1}^{\infty} a_{k}</math> is divergent.
 '''<u>Note</u>:''' Requirements for this test include showing (or at least stating):
@@ Line 68: / Line 64: @@
 :* For all <math style="vertical-align: -15%">k\geq c</math>, <math style="vertical-align: -15%">a_{k}\leq b_{k}</math> for convergence, or <math style="vertical-align: -15%">a_{k}\geq b_{k}</math> for divergence.
-:* ''(This is important)'' State why <math style="vertical-align: -98%">\sum_{k=1}^{\infty} b_k</math> is convergent, such as a ''p''-series with <math style="vertical-align: -18%">p>1</math>, or a geometric series with <math style="vertical-align: -24%">|r|<1</math>. Obviously, you would need to state why it is divergent if you're showing it's divergent.
+:* ''(This is important)'' State why <math style="vertical-align: -98%">\sum_{k=1}^{\infty} b_k</math> is convergent, such as a ''p''-series with <math style="vertical-align: -20%">p>1</math>, or a geometric series with <math style="vertical-align: -24%">|r|<1</math>. Obviously, you would need to state why it is divergent if you're showing it's divergent.
 ''Then'', you can say, "By the Comparison Test, the series is convergent (or divergent)."
@@ Line 74: / Line 70: @@
 == The Limit Comparison Test ==
-Suppose $\sum a_{k}$ and $\sum b_{k}$
+Suppose <math style="vertical-align: -100%">\sum_{k=1}^{\infty} a_{k}</math> and <math style="vertical-align: -100%">\sum_{k=1}^{\infty} b_{k}</math>
-are series with positive terms. If
+are series with positive terms. If <math style="vertical-align: -75%">\lim_{k\rightarrow\infty}\frac{a_{k}}{b_{k}}=c</math> where <math style="vertical-align: -5%">0<c<\infty</math>, then either both series converge, or both series diverge.
-\[
-\lim_{k\rightarrow\infty}\frac{a_{k}}{b_{k}}=c
-\]
- where $0<c<\infty,$ then either both series converge, or both series
-diverge. Additionally, if $c=0$ and $\sum b_{k}$ converges, $\sum a_{k}$
-also converges. Similarly, if $c=\infty$ and $\sum b_{k}$ diverges,
-then $\sum a_{k}$ also diverges.
-\emph{\uline{Notes}}\emph{: }First of all, let's mention the idea
+Additionally, if <math style="vertical-align: 0%">c=0</math> and <math style="vertical-align:  -100%">\sum_{k=1}^{\infty} b_{k}</math> converges, <math style="vertical-align: -100%">\sum_{k=1}^{\infty} a_{k}</math> also converges. Similarly, if <math style="vertical-align: -5%">c=\infty</math>&thinsp; and <math style="vertical-align: -100%">\sum_{k=1}^{\infty} b_{k}</math> diverges, then <math style="vertical-align: -100%">\sum_{k=1}^{\infty} a_{k}</math> also diverges.
-here. If some series $\sum b_{k}$ converges, then
-\[
-\sum cb_{k}
-\]
- converges where $c\neq\pm\infty$ is a constant. This test shows
-that one series \emph{\uline{eventually}} is just like the other
-one multiplied by a constant, and for that reason it will also converge/diverge
-if the one compared converges/diverges. To use it, you need to state/show:
-\begin{itemize}
-\item $a_{k}$ is always positive (really, non-negative).
-\item ${\displaystyle \lim_{k\rightarrow\infty}\frac{a_{k}}{b_{k}}}=c$.
-\item State why $\sum b_{k}$ is convergent, such as a $p$-series with
-$p>1$, or a geometric series with $|r|<1.$ Obviously, you would
-need to state why it is divergent if you're showing it's divergent.
-\end{itemize}
-\textbf{\uline{Then}}, you can say, ``By the Limit Comparison
-Test, the series is convergent (or divergent).''
-Like the Comparison Test and the integral test, it's fine if the first
+'''<u>Note</u>''': First of all, let's mention the fundamental idea here. If some series <math style="vertical-align: -100%">\sum_{k=1}^{\infty} b_{k}</math> converges, then <math style="vertical-align: -100%">\sum_{k=1}^{\infty} cb_{k}</math> converges where <math style="vertical-align: -22%">c\neq\pm\infty </math> is a constant. This test shows that one series <u>''eventually''</u> is just like the other one multiplied by a constant, and for that reason it will also converge/diverge
-terms are kind of ``wrong'' - negative, for example - as long as
+if the one compared to converges/diverges. To use it, you need to state/show:
-they eventually wind up (for $k>c$ for a particular $c$) meeting
+:*<math style="vertical-align: -15%">a_{k} </math> is eventually positive (really, non-negative).
+:*<math style="vertical-align: -72%">{\displaystyle \lim_{k\rightarrow\infty}\frac{a_{k}}{b_{k}}}=c </math>.
+:* State why <math style="vertical-align: -100%">\sum_{k=1}^{\infty} b_{k} </math> is convergent, such as a ''p''-series with <math style="vertical-align: -20%">p>1 </math>, or a geometric series with <math style="vertical-align: -20%">|r|<1 </math>. Obviously, you would need to state why it is divergent if you're showing it's divergent.
+''Then'', you can say, "By the Limit Comparison Test, the series is convergent (or divergent)."
+Like the Comparison Test and the Integral Test, it's fine if the first terms are kind of "wrong" - negative, for example - as long as they eventually wind up (for <math style="vertical-align: 0%">k>c</math> for a particular <math style="vertical-align: 0%">c</math>&thinsp;) meeting
 the requirements.
-\hrulefill
+== The Alternating Series Test ==
+If a series <math>\sum_{k=1}^{\infty} a_{k}</math> is
+:*Alternating in sign, and
-== The Alternating Series Test ==
+:*<math>\lim_{k\rightarrow 0}|a_{k}|=0,</math>
-If a series $\sum a_{k}$ is
-\begin{enumerate}
-\item Alternating in sign, and
-\item ${\displaystyle \lim_{k\rightarrow0}}|a_{k}|=0,$
-\end{enumerate}
 then the series is convergent.
-\emph{\uline{Notes}}\emph{: }This is a fairly straightfoward test.
+'''<u>Note</u>:''' This is a fairly straightfoward test. You only need to do two things:
-You only need to do two things:
-\begin{enumerate}
+:*Mention the series is alternating (even though it's usually obvious).
-\item Mention the series is alternating (even though it's usually obvious).
-\item Show the limit converges to zero.
+:*Show the limit converges to zero.
-\end{enumerate}
-\textbf{\uline{Then}}, you can say, ``By the Alternating Series
+''Then'', you can say, "By the Alternating Series Test, the series is convergent."
-Test, the series is convergent.''
-As an additional detail, if it fails to converge to zero, then you
+As an additional detail, if it fails to converge to zero, then you would say it diverges by the Divergence Test, ''<u>not</u>'' the Alternating Series Test.
-would say it diverges by the Divergence Test, \textbf{\uline{NOT}}
-the Alternating Series Test.
 == The Ratio Test ==
-Let $\sum a_{k}$ be a series. Then:
+Let <math style="vertical-align: -98%">\sum_{k=1}^{\infty} a_{k}</math> be a series. Then:
-\begin{enumerate}
-\item if ${\displaystyle \lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L<1,}$
+:*If <math style="vertical-align: -84%">\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L<1</math>, the series is absolutely convergent (and therefore convergent).
-the series is absolutely convergent (and therefore convergent),\\
+:*If <math style="vertical-align: -84%">\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L>1</math> or <math style="vertical-align: -83%">\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L=\infty</math>, the series is divergent.
+:*If <math style="vertical-align: -84%">\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L=1</math>, the Ratio Test is inconclusive.
+'''<u>Note</u>:''' Both this and the Root Test have the least requirements. The Ratio Test ''<u>does</u>'' require that such a limit exists, so a series like
-\item if ${\displaystyle \lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L>1}$
+<math>0+1+0+\frac{1}{4}+0+\frac{1}{9}+\cdots</math>
-or ${\displaystyle \lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L=\infty,}$
-the series is divergent,\\
-\item if ${\displaystyle \lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L=1,}$
+could not be assessed as written with the Ratio Test, as division by zero is undefined. You might have to argue it's the same sum as
-the Ratio Test is inconclusive.\\
-\end{enumerate}
+<math>1+\frac{1}{4}+\frac{1}{9}+\cdots,</math>
-\emph{\uline{Notes}}\emph{: }Both this and the Root Test have the
-least requirements. The Ratio Test \emph{\uline{does}} require
-that such a limit exists, so a series like
-\[
-+1+0+\frac{1}{4}+0+\frac{1}{9}+\cdots
-\]
- could not be assessed as written with the Ratio Test, as division
-by zero is undefined. You might have to argue it's the same sum as
-\[
-+\frac{1}{4}+\frac{1}{9}+\cdots,
-\]
- and then you could apply the Ratio Test.
+and you could then apply the Ratio Test.
 == The Root Test ==