Series - Tests for Convergence/Divergence

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

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

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 start 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.

The Integral Test

Suppose the function $f(x)$ is continuous, positive and decreasing on some interval $[b,\infty)$ with $b\geq1$, and let $a_{k}=f(k).$ Then the series $\sum_{k=b}^{\infty}a_{k}$ is convergent if and only if for some $c\geq b,$ \[ \int_{c}^{\infty}f(x)\, dx \]

is convergent (not infinite). 

\emph{\uline{Notes}}\emph{:} This test, like many of them has a few requirements. In order to use it on a test, you need to state/show: \begin{itemize} \item For all $k\geq c$ for some $c\geq b,$ the function is positive. (Most of the time, $c$ is just my starting index \emph{b}.) \item For all $k\geq c,$ the function is decreasing. \item The integral is convergent (or divergent, if you're proving divergence). \end{itemize} \textbf{\uline{Then}},\textbf{ }you can say, ``By the Integral Test, the series is convergent (or divergent).

I wrote this with \emph{c} instead of \emph{b} for a lower bound to indicate \emph{\uline{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 $\sum a_{k}$ and $\sum b_{k}$ are series \uline{with positive terms}. Then \begin{enumerate} \item If $a_{k}\leq b_{k}$ for all $k$ and $\sum b_{k}$ is convergent, then $\sum a_{k}$ is convergent. \item If $a_{k}\geq b_{k}$ for all $k$ and $\sum b_{k}$ is divergent, then $\sum a_{k}$ is divergent. \end{enumerate} \emph{\uline{Notes}}\emph{: }Requirements for this test include showing (or at least stating): \begin{itemize} \item For all $k\geq c$ for some $c$ bigger than our starting index, $a_{k}$ is positive. (Most of the time, $c$ is just the starting index.) \item For all $k\geq c$, $a_{k}\leq b_{k}$ for convergence, or $a_{k}\geq b_{k}$ for divergence. \item \textbf{This is important - }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 Comparison Test, the series is convergent (or divergent).

The Limit Comparison Test

Suppose $\sum a_{k}$ and $\sum b_{k}$ 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 $\sum b_{k}$ 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 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 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.

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The Alternating Series Test

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. You only need to do two things: \begin{enumerate} \item Mention the series is alternating (even though it's usually obvious). \item Show the limit converges to zero. \end{enumerate} \textbf{\uline{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, \textbf{\uline{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.