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\noindent
\parbox{1.5in}{\bf Math 1120}
\hfill {\Large \bf Series Tests} \hfill
\parbox{1.5in}{\bf \hfill Supplement}

\vspace{0.1in}

\noindent
Since we have a whole bunch of series convergence and divergence tests, it might be nice to put them all in one place. This handout also gives some advice about how to choose the proper test to determine convergence and divergence. First, we recall the following important bit of philosophy:

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\noindent
\fbox{\large\bf Convergence Philosophy:} Finitely many terms (of a sequence or series) cannot effect convergence (or divergence). 

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For what follows, we will assume that our series start at $n=1$ and we will demand that hypotheses hold for all terms. However, because of the above theorem, it is good enough that any of our test hypotheses just hold {\it eventually}. Do keep in mind that when hypotheses are violated by finitely many terms, we can no longer count on results about the sum of our series (like error bounds) to hold.

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Let's begin by reviewing some notation and what we mean by {\it series} and {\it convergence} vs. {\it divergence} of series.

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\noindent
\fbox{\large\bf Series:} Essentially a {\it series} is an infinite sum: $\sum\limits_{k=1}^\infty a_k = a_1+a_2+a_3+\cdots$ 

We call $a_k$ the $k^\text{th}$ {\it term} of the series and $S_n=\sum\limits_{k=1}^n a_k = a_1+a_2+\cdots+a_n$ the $n^\text{th}$ {\it partial sum} of the series. Moreover, the series $\sum\limits_{k=1}^\infty a_k$ {\it converges} if $\lim\limits_{n \to \infty} S_n$ (the limit of the partial sums) converges and {\it diverges} if $\lim\limits_{n \to \infty} S_n$ diverges. Finally, if the series converges, we call the limit of its partial sums, the {\it sum} of the series: $\sum\limits_{k=1}^\infty a_k = \lim\limits_{n \to \infty} S_n = \lim\limits_{n \to \infty} \sum\limits_{k=1}^n a_k$.


\noindent
\fbox{\large\bf Telescoping Series:} If a series has partial sums that collapse down, we refer to it as a {\it telescoping series}.

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Every series is a telescoping series if we use the ``right'' formula. So telescoping refers not so much to the series itself, but to the formula used to define the series. Having a telescoping formula is quite nice\dots{}but rare. In such a case, we can easily tell if the series converges or diverges. Even better, if it does converge, we can find its sum! [Finding sums is very very hard in general.] 

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\noindent
{\bf Quick Examples:} $\displaystyle\sum\limits_{k=1}^\infty \left(\dfrac{1}{k}-\dfrac{1}{k+1}\right)$ is telescoping since $S_n = \left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\cdots+\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)=1-\dfrac{1}{n}$ (the sum collapses down nicely). Notice that $\lim\limits_{n\to\infty} S_n =\lim\limits_{n\to\infty} 1-\dfrac{1}{n} = 1-0=1$. Thus the series converges and, in fact, it sums to $1$.

On the other hand, $\sum\limits_{k=1}^\infty (k^2-(k+1)^2)$ is also telescoping: $S_n=(1^2-2^2)+(2^2-3^2)+\cdots+(n^2-(n+1)^2)=1-(n+1)^2$, but $\lim\limits_{n\to\infty} S_n = \lim\limits_{n\to\infty} 1-(n+1)^2=-\infty$. So even though this series telescopes, it diverges.

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\noindent
\fbox{\large\bf Master Examples:} We have two main classes of examples that we compare series to.

\begin{description}
\item[Geometric Series:] $\sum\limits_{k=0}^\infty ar^k = a+ar+ar^2+\cdots$ is a {\it geometric series} with first term $a$ and ratio $r$. It converges when $|r|<1$ and diverges when $|r| \geq 1$. Moreover, when it converges, it must converge to $\dfrac{a}{1-r}$ (i.e., first term divided by one minus the ratio).

\item[$p$-Series:] A {\it $p$-series}, $\sum\limits_{k=1}^\infty \dfrac{1}{k^p}$, converges if $p>1$ and diverges if $p \leq 1$. The special case $p=1$ yields a divergent series $\sum\limits_{k=1}^\infty \dfrac{1}{k}$ called the {\it harmonic series}. 
\end{description}

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Let's recall one last thing before the tests. 

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\noindent
\fbox{\large\bf Absolute vs. Conditional Convergence:} If $\sum\limits_{n=1}^\infty |a_n|$ converges, we say $\sum\limits_{n=1}^\infty a_n$ {\it converges absolutely}. A very sensible theorem states that if a series converges absolutely, then it must converge too. It is possible for a series to converge but not converge absolutely (the alternating harmonic series is the easiest example). A series that converges but does not converge absolutely is called a {\it conditionally convergent series}.

Anytime we run into a series with both positive and negative terms, we can consider the corresponding series summing absolute values of terms. If that series converges, so does our original series. This is helpful since several tests only work for non-negative (or positive) terms. Unfortunately, if the series of absolute values of terms diverges, we can't say anything about the convergence or divergence of the original series.

We remind ourselves that {\it conditionally convergent series are weird}. The Riemann rearrangment theorem sates that it is possible to permute terms of a conditionally convergent series and make it converge to any desired number or make it diverge to $\pm \infty$. When a series converges absolutely, any permutation of the terms leaves us with an absolutely convergent series that still converges to the same sum as the original series. Absolute convergence is stable. Conditional convergence is fragile.

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Next, we have our various convergence tests. These are listed (more-or-less) in the order we would consider using them.

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\noindent
\fbox{\large\bf Series Convergence Tests:} 

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\begin{description}
\item[$n^\text{th}$ Term Test (for Divergence):] If $\lim\limits_{n \to \infty} a_n \not=0$ (or does not exist), then $\sum\limits_{n=1}^\infty a_n$ diverges.

This is a good first test to try. It doesn't assume anything about the terms, $a_n$, and usually is easy to apply. Unfortunately, we generally have terms limiting to zero (in which case this test says nothing).

\item[Alternating Series Test:] If we have an alternating series: $a_1-a_2+a_3-\cdots$ or $-a_1+a_2-a_3+\cdots$ where $a_n>0$, $a_{n} \geq a_{n+1}$, and $\lim\limits_{n\to \infty} a_n=0$ (i.e., term magnitudes are decreasing to zero), then $\sum\limits_{n=1}^\infty (-1)^n a_n$ $\left(\text{and }\sum\limits_{n=1}^\infty (-1)^{n+1}a_n\right)$ converge. 

While this test can only tell us about convergence of this specific class of alternating series, it is very easy to apply. If we see a series is an alternating series, we should try to apply this test first! This test comes with a bonus fact: When this test applies, the sum of the series lies between any two consecutive partial sums.

\item[Ratio Test:] Let $a_n$ be nonzero and compute $\lim\limits_{n \to \infty} \dfrac{|a_{n+1}|}{|a_n|}=L$ (where $L=\infty$ is allowed). If $L<1$, then $\sum\limits_{n=1}^\infty a_n$ converges absolutely. If $L>1$, then $\sum\limits_{n=1}^\infty a_n$ diverges. If $L=1$ or does not exist, then this test says nothing.

Again, it is nice that we don't need to assume much about our series' terms and it's generally pretty easy to compute the necessary limit. {\it Note:} If one knows about ``limit supremums'' (i.e., $\mathrm{limsup}$), it is possible to eliminate the case of the limit not existing. However, $L=1$ still yields no information.

\item[Root Test:] One can swap out the ratio limit for $\lim\limits_{n \to \infty} \sqrt[n]{|a_n|}=L$ and get the root test (conclusions are the same). 

{\it Ratio/Root Warning:} Ratio/Root tests will give the same conclusions, so if one gives $L=1$, don't bother with the other. These tests work well with series somewhat resembling geometric series. They don't work on series that resemble $p$-series (in such cases you just get $L=1$).

\item[Limit Comparison:] Suppose $a_n>0$ and $b_n>0$ (we need {\it positive} terms). Let $\lim\limits_{n \to \infty} \dfrac{a_n}{b_n}=L$ (where $L=\infty$ is allowed). Then\dots
\begin{itemize}
\item If $0<L<\infty$, either both $\sum\limits_{n=1}^\infty a_n$ and $\sum\limits_{n=1}^\infty b_n$ converge or both diverge. 
\item If $L=0$, then $b_n$'s dominate $a_n$'s (i.e., approximately $\sum\limits_{n=1}^\infty a_n \leq \sum\limits_{n=1}^\infty b_n$). So if $\sum\limits_{n=1}^\infty b_n$ converges, then $\sum\limits_{n=1}^\infty a_n$ converges too. Likewise, if $\sum\limits_{n=1}^\infty a_n$ diverges, then $\sum\limits_{n=1}^\infty b_n$ diverges too. 
\item If $L=\infty$, the opposite occurs: $a_n$'s dominate $b_n$'s (i.e., approximately $\sum\limits_{n=1}^\infty b_n \leq \sum\limits_{n=1}^\infty a_n$). So if $\sum\limits_{n=1}^\infty a_n$ converges, then $\sum\limits_{n=1}^\infty b_n$ converges too. Likewise, if $\sum\limits_{n=1}^\infty b_n$ diverges, then $\sum\limits_{n=1}^\infty a_n$ diverges too. 
\end{itemize}

First, note that we need positive terms. Although if we don't have positive terms we can always check for absolute convergence. However, if we took absolute values and found that our absolute values series diverges, it wouldn't help us determine if the original series converges or not! The other weakness of the limit comparison test is that we need to know of a reasonable series to compare to. Generally, we compare to either geometric series or $p$-series.

\item[Direct Comparison:] Suppose $0 \leq a_n \leq b_n$ (notice that terms must be non-negative). Then $\sum\limits_{n=1}^\infty a_n \leq \sum\limits_{n=1}^\infty b_n$. Therefore, if $\sum\limits_{n=1}^\infty b_n$ converges, then $\sum\limits_{n=1}^\infty a_n$ must converge too. On the other hand, if $\sum\limits_{n=1}^\infty a_n$ diverges, then $\sum\limits_{n=1}^\infty b_n$ must diverge too. 

The same concerns as with the limit comparison test apply here. Typically (but not always) it is easier to compare using a limit. However, there are cases where this test is the better one to use.

\item[Integral Test:] Suppose that for $x \geq 1$, $f(x)$ is continuous, $f(x) \geq 0$ (non-negative), and $f(x)$ is decreasing (generally we check that $f'(x) \leq 0$), then either both $\sum\limits_{n=1}^\infty f(n)$ and $\displaystyle \int_1^\infty f(x)\,dx$ converge or both diverge.

We usually think of this test as a last resort. It takes a fair amount of calculation to check if the test even applies and then even more calculation to apply the test! The upside to this test is that we get a bound on what our series sum might be: $\displaystyle \int_1^\infty f(x)\,dx \leq \sum\limits_{n=1}^\infty f(n) \leq f(1)+ \int_1^\infty f(x)\,dx$. More generally,  $\displaystyle \int_N^\infty f(x)\,dx \leq \sum\limits_{n=N}^\infty f(n) \leq f(N)+ \int_N^\infty f(x)\,dx$.
\end{description}

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