I was recently helping a college math student with her homework. Her teacher had offered an extra-credit question: Find two alternating series $\sum_{n=1}^\infty (-1)^{n-1}a_n$ such that $a_{n+1} \leq a_n$ for all $n$, but $\lim_{n\to\infty} a_n \neq 0$. One of the provided series should converge, and the other should diverge.

A divergent series was easy to find: $\sum_{n=1}^\infty (-1)^{n-1} \left(1+\frac{1}{n}\right)$. I'm having a much harder time coming up with a convergent series, though. In fact, I suspect there isn't one.

Informally (since it's been many years since I myself studied this topic):

Since $\lim_{n\to\infty}a_n \neq 0$, then it either diverges or converges to some other number. Since the series is positive and monotone nonincreasing, it cannot diverge. Let $L$ be the positive number to which it converges. Then the odd terms of the alternating series converge to $L$ from above, and the even terms converge to $-L$ from below. Each term of the sequence of partial sums then differs from the previous term by at least $2L$, so the series does not converge.

So... Did the teacher offer an impossible problem on purpose, or is there a flaw in my reasoning?

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    $\begingroup$ Your reasoning is correct, the series cannot converge. In fact the limsup and the liminf of the sequence of the partial sums will be at distance $L$ from each other. $\endgroup$ – Did Nov 3 '11 at 19:10
  • $\begingroup$ If $b_n$ is bounded and bounded away from $0$ (e.g. $b_n=(-1)^n$) and $b_na_n\to0$ as $n\to\infty$ then $a_n\to0$ as $n\to\infty$. $\endgroup$ – AD. Nov 3 '11 at 19:11
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    $\begingroup$ I'm betting the student copied the problem incorrectly. Sure, college math teachers make mistakes, but students make many, many more. $\endgroup$ – Gerry Myerson Nov 3 '11 at 23:14

Notice that ($\sum a_n$ converges) $\Longrightarrow$ ($\lim a_n=0$). The contrapositive is sometimes called the test for divergence: ($\lim a_n\ne 0$) $\Longrightarrow$ ($\sum a_n$ diverges).

So indeed, the series you are looking for does not exist.


To answer the question in the title: Yes (just not the way indicated in the body).

There are plenty of examples of nonmonotonic sequences $(a_n)$ of positive numbers such that $\sum\limits_{n=0}^\infty(-1)^na_n$ converges. For example, this will holds whenever $\sum\limits_{n=0}^\infty a_n$ converges. One example would be $a_{2n}=\dfrac{1}{(n+1)^2}$, $a_{2n+1}=\dfrac{1}{(n+1)^3}$.


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