TITLE: Is there an alternating series that satisfies only one of the conditions of the Alternating Series Test that nonetheless converges?
QUESTION [11 upvotes]: 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?

REPLY [2 votes]: 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 $a_{2n}=\dfrac{1}{(n+1)^2}$, $a_{2n+1}=\dfrac{1}{(n+1)^3}$.