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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\displaystyle \lim _{k\rightarrow \infty }a_{k}\neq 0,}} then the series/sum Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle a_{k}=f(k)} . Then the series $\sum _{k=b}^{\infty }a_{k}$ is convergent if and only if $c\geq b$ and

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum _{k=1}^{\infty }b_{k}} is a series with positive terms, and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum _{k=1}^{\infty }a_{k}} is a series with eventually positive terms. Then

• If for all $k\geq c$ for some $c$ greater than or equal to our starting index, and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum _{k=1}^{\infty }b_{k}} is convergent, then Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum _{k=1}^{\infty }a_{k}} is convergent.

• If Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle a_{k}\geq b_{k}} for all $k$ and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum _{k=1}^{\infty }b_{k}} is divergent, then Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum _{k=1}^{\infty }a_{k}} is divergent.

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

• For all $k\geq c$ for some $c$ greater than or equal to our starting index, $a_{k}$ is positive. (Most of the time, $c$ is just the starting index.)

• For all $k\geq c$, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle a_{k}\leq b_{k}} for convergence, or Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle a_{k}\geq b_{k}} for divergence.
• (This is important) State why Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum _{k=1}^{\infty }a_{k}} and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum _{k=1}^{\infty }b_{k}} are series with positive terms. If Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{k\rightarrow \infty }{\frac {a_{k}}{b_{k}}}=c} where Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 0<c<\infty } , then either both series converge, or both series diverge.

Additionally, if $c=0$ and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum _{k=1}^{\infty }b_{k}} converges, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum _{k=1}^{\infty }a_{k}} also converges. Similarly, if Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle c=\infty }   and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum _{k=1}^{\infty }b_{k}} diverges, then Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum _{k=1}^{\infty }cb_{k}} converges where Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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).

• Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle k>c} for a particular $c$ ) meeting the requirements.

The Alternating Series Test

If a series Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum _{k=1}^{\infty }a_{k}} is

• Alternating in sign, and

• Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{k\rightarrow \infty }\left|{\frac {a_{k+1}}{a_{k}}}\right|=L<1} , the series is absolutely convergent (and therefore convergent).

• If Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{k\rightarrow \infty }\left|{\frac {a_{k+1}}{a_{k}}}\right|=L>1} or Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{k\rightarrow \infty }\left|{\frac {a_{k+1}}{a_{k}}}\right|=L=\infty } , the series is divergent.

• If Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 1+{\frac {1}{4}}+{\frac {1}{9}}+\cdots ,}

and you could then apply the Ratio Test.

The Root Test

Let Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \displaystyle \sum _{k=0}^{\infty }a_{k}} be a series. Then:

• If Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\displaystyle \lim _{k\rightarrow \infty }{\sqrt[{k}]{|a_{k}|}}}=L=\infty ,}

the series is divergent.

• If Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\displaystyle \lim _{k\rightarrow \infty }{\sqrt[{k}]{|a_{k}|}}}=L=1} , the Root Test is inconclusive.

Contributions to this page were made by John Simanyi