# Convergence of alternating series

I was provided the following first terms of a series in a question from my math book:

$-\frac{2}{5}+\frac{4}{6}-\frac{6}{7}+\frac{8}{8}-\frac{10}{9}+...$

I concluded that this is the equivalent of the following sum representation:

$\sum\limits_{n=1}^\infty(-1)^n\frac{2n}{n+4}$

I then proceeded to do the alternating series convergence test by first comparing two following terms:

$b_n=\frac{2n}{n+4}$

$b_{n+1}=\frac{2n+2}{n+5}$

I concluded that $b_{n+1}>b_n$ for at least one $n$ and thus the series is divergent. The limit test is also conclusive:

$\lim_{n->\infty} \frac{2n}{n+4} = \lim_{n->\infty} \frac{2}{1+\frac{4}{n}} = 2$

Since the limit is not equal to $0$, both alternating series tests point out that the series is in fact divergent.

Unfortunately the answer pages in my math book itself say that the series is convergent. Although Wolfram Alpha confirms my thoughts by saying the series is divergent, it shows no proof or method to reach this answer.

Have I done everything right and is the series in fact divergent?

• You are right, the series is definitly divergent, because $\lim_{n\rightarrow\infty} \frac{2n}{n+4} \neq 0$ – user127.0.0.1 Dec 3 '12 at 10:50

You can't say that the series diverges, just because $b_{n+1} > b_n$ for at least one $n$. The alternating series test is not an equivalence, i.e. even if a series fails to satisfy the condtitions, it may still converge.
On the other hand, for your series, it's clear that the terms do not tend to $0$ as $n\to\infty$, so the series is certainly divergent.