Series - Tests for Convergence/Divergence

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 $r$ between adjacent terms which are usually written

$\sum _{k=0}^{\infty }a_{0}r^{k}.$

These are convergent if $|r|<1$, and divergent if $|r|\geq 1$. If it is convergent, we can find the sum by the formula

$S={\frac {a_{0}}{1-r}},$

where $a_{0}$ is the first term in the series (if the index starts at $k=2$ or $k=6$, then "$a_{0}$" is actually the first term $a_{2}$ or $a_{6}$, respectively).

p-series. These are series of the form

$\sum _{k=1}^{\infty }{\frac {1}{k^{p}}}.$

If $p>1$, then the series is convergent. On the other hand, if $p\leq 1$, the p-series is divergent.

The Divergence Test

If ${\displaystyle \lim _{k\rightarrow \infty }a_{k}\neq 0,}$ then the series/sum $\sum _{k=0}^{\infty }a_{k}$ diverges.

Note: The opposite result doesn't allow you to conclude a series converges. If ${\displaystyle \lim _{k\rightarrow \infty }a_{k}=0}$ , it merely indicates the series might converge, and you still need to confirm it through another test.

In particular, the sequence $\left\{{\frac {1}{k}}\right\}$ converges to zero, but the sum $\sum _{k=0}^{\infty }{\frac {1}{k}}$ , our harmonic series, diverges.

The Integral Test

Suppose the function $f(x)$ is continuous, positive and decreasing on some interval $[c,\infty )$ with $c\geq 1$, and let $a_{k}=f(k)$. Then the series $\sum _{k=b}^{\infty }a_{k}$ is convergent if and only if $c\geq b$ and

$\int _{c}^{\infty }f(x)\,dx$

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 $k\geq c$ for some $c\geq b$, the function is positive. (Most of the time, $c$ is just my starting index $b$).
• For all $k\geq c$, 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 $c$ instead of $b$ 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 $\sum _{k=1}^{\infty }b_{k}$ is a series with positive terms, and $\sum _{k=1}^{\infty }a_{k}$ 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 $\sum _{k=1}^{\infty }a_{k}$ 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 always 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 $\sum a_{k}$ be a series. Then: \begin{enumerate} \item if ${\displaystyle \lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L<1,}$ the series is absolutely convergent (and therefore convergent),\\

\item if ${\displaystyle \lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L>1}$ or ${\displaystyle \lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L=\infty,}$ the series is divergent,\\

\item if ${\displaystyle \lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L=1,}$ the Ratio Test is inconclusive.\\

\end{enumerate} \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 \[ 0+1+0+\frac{1}{4}+0+\frac{1}{9}+\cdots \]

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 then you could 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.