Series - Tests for Convergence/Divergence - Revision history

Grad: /* The Limit Comparison Test */

2015-04-26T20:13:03Z

The Limit Comparison Test

Grad: /* The Limit Comparison Test */

2015-04-26T04:51:45Z

The Limit Comparison Test

Grad: /* The Limit Comparison Test */

2015-04-26T04:50:48Z

The Limit Comparison Test

Grad: /* The Limit Comparison Test */

2015-04-26T04:50:02Z

The Limit Comparison Test

Grad: /* The Ratio Test */

2015-04-26T04:48:03Z

The Ratio Test

Grad: /* The Ratio Test */

2015-04-26T04:47:32Z

The Ratio Test

Grad: /* The Ratio Test */

2015-04-26T04:46:45Z

The Ratio Test

Grad: /* The Integral Test */

2015-04-26T04:38:40Z

The Integral Test

Grad: /* The Alternating Series Test */

2015-04-26T04:11:11Z

The Alternating Series Test

Grad: /* The Limit Comparison Test */

2015-04-24T05:19:32Z

The Limit Comparison Test

@@ Line 71: / Line 71: @@
 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 <math style="vertical-align: -75%">\lim_{k\rightarrow\infty}\frac{a_{k}}{b_{k}}=c</math> where <math style="vertical-align: 0%">0<c<\infty</math>, then either both series converge, or both series diverge.
+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.
 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.

@@ Line 73: / Line 73: @@
 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: 0%">0<c<\infty</math>, then either both series converge, or both series diverge.
-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: -98%">\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: -98%">\sum_{k=1}^{\infty} b_{k}</math> diverges, then <math style="vertical-align: -98%">\sum_{k=1}^{\infty} a_{k}</math> also diverges.
+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.
 '''<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
@@ Line 82: / Line 82: @@
 :*<math style="vertical-align: -72%">{\displaystyle \lim_{k\rightarrow\infty}\frac{a_{k}}{b_{k}}}=c </math>.
-:* 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: -20%">|r|<1 </math>. Obviously, you would need to state why it is divergent if you're showing it's divergent.
+:* 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)."

@@ Line 70: / Line 70: @@
 == The Limit Comparison Test ==
-Suppose <math style="vertical-align: -98%">\sum_{k=1}^{\infty} a_{k}</math> and <math style="vertical-align: -98%">\sum_{k=1}^{\infty} b_{k}</math>
+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 <math style="vertical-align: -75%">\lim_{k\rightarrow\infty}\frac{a_{k}}{b_{k}}=c</math> where <math style="vertical-align: 0%">0<c<\infty</math>, then either both series converge, or both series diverge.
-Additionally, if <math style="vertical-align: 0%">c=0</math> and <math style="vertical-align:  -98%">\sum_{k=1}^{\infty} b_{k}</math> converges, <math style="vertical-align: -98%">\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: -98%">\sum_{k=1}^{\infty} b_{k}</math> diverges, then <math style="vertical-align: -98%">\sum_{k=1}^{\infty} a_{k}</math> also diverges.
+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: -98%">\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: -98%">\sum_{k=1}^{\infty} b_{k}</math> diverges, then <math style="vertical-align: -98%">\sum_{k=1}^{\infty} a_{k}</math> also diverges.
-'''<u>Note</u>''': First of all, let's mention the fundamental idea here. If some series <math style="vertical-align: -98%">\sum_{k=1}^{\infty} b_{k}</math> converges, then <math style="vertical-align: -98%">\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
+'''<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
 if the one compared to converges/diverges. To use it, you need to state/show:

@@ Line 78: / Line 78: @@
 if the one compared to converges/diverges. To use it, you need to state/show:
-:*<math style="vertical-align: -15%">a_{k} </math> is always positive (really, non-negative).
+:*<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>.

@@ Line 112: / Line 112: @@
 Let <math style="vertical-align: -98%">\sum_{k=1}^{\infty} a_{k}</math> be a series. Then:
-:*If <math style="vertical-align: -82%">\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L<1</math>, 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>, the series is absolutely convergent (and therefore convergent).
-:*If <math style="vertical-align: -82%">\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L>1</math> or <math style="vertical-align: -79%">\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> 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: -82%">\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L=1</math>, the Ratio Test is inconclusive.
+:*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

@@ Line 110: / Line 110: @@
 == 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:
-\item if ${\displaystyle \lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L>1}$
+:*If <math style="vertical-align: -80%">\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L<1</math>, the series is absolutely convergent (and therefore convergent).
-\item if ${\displaystyle \lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L=1,}$
+:*If <math style="vertical-align: -80%">\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L>1</math> or <math style="vertical-align: -77%">\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L=\infty</math>, the series is divergent.
-\end{enumerate}
+:*If <math style="vertical-align: -80%">\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L=1</math>, the Ratio Test is inconclusive.
 == The Root Test ==

@@ Line 33: / Line 33: @@
 == 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 90: / Line 90: @@
 == The Alternating Series Test ==
-If a series $\sum a_{k}$ is
+If a series <math>\sum_{k=1}^{\infty} a_{k}</math> is
-\begin{enumerate}
-\item Alternating in sign, and
+:*Alternating in sign, and
-\item ${\displaystyle \lim_{k\rightarrow0}}|a_{k}|=0,$
-\end{enumerate}
+:*<math>\lim_{k\rightarrow 0}|a_{k}|=0,</math>
 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."
-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.
 == The Ratio Test ==

@@ Line 87: / Line 87: @@
 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.
 == The Alternating Series Test ==