The convergence of the series $\sum (-1)^n \frac{n}{n+1}$ and the value of its sum This sum seems convergent, but how to find its precise  value? 
$$\sum\limits_{n=1}^{\infty}{(-1)^{n+1} \frac{n}{n+1}} = \frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}+...=-0.3068... $$ Any help would be much appreciated. 
 A: The limit of the terms of the sum is not zero, thus it does not converge.
This is a well known theorem in calculus. If you are unfamiliar with it, you can nevertheless understand it intuitively: if the terms don't eventually all become very close to zero, then the sum is either getting arbitrarily large (if all the terms are the same sign) or the sum is jumping around (if the terms are alternating, which they are in this case).
A: It is true that since the partial sums do not approach a single limit, the series does not converge. The partial sums instead hover around two numbers, accumulating at one after adding every even term, and accumulating at the other after adding every odd term, and thus alternating between the two. Lets find out what these two accumulation points are.
Taking the even partial sum,
$$\lim_{m\rightarrow \infty} \sum_{n=1}^{2m} (-1)^{n+1} \frac{n}{n+1}$$
$$=\lim_{m\rightarrow \infty} \sum_{n=1}^{m} (-1)^{2n} \frac{2n-1}{2n} + (-1)^{2n+1} \frac{2n}{2n+1}$$
$$=\lim_{m\rightarrow \infty} \sum_{n=1}^{m} \frac{2n-1}{2n} - \frac{2n}{2n+1}$$
$$=\lim_{m\rightarrow \infty} \sum_{n=1}^{m} \frac{(2n-1)(2n+1)-(2n)^2}{(2n)(2n+1)}$$
$$=\lim_{m\rightarrow \infty} \sum_{n=1}^{m} \frac{-1}{(2n)(2n+1)}$$
$$=\lim_{m\rightarrow \infty} \sum_{n=1}^{m} -\frac{1}{2n}+\frac{1}{2n+1}$$
$$=\lim_{m\rightarrow \infty} \sum_{n=2}^{2m} \frac{(-1)^{n-1}}{n}$$
$$=\lim_{m\rightarrow \infty} -1 + \sum_{n=1}^{2m} \frac{(-1)^{n-1}}{n}$$
$$=-1 + \ln 2$$
At first I split up the sum between even and odd terms, and then was able to add them back together later. That last sum is the alternating harmonic series, which is conditionally convergent, a consequence that is implied by the others' answers. This constant $\ln 2 -1$ is the number that you think the series converges to. But because the absolute value of the terms tends towards $1$, the series actually alternates between $\ln 2$ and $\ln2 -1$ when you consider the odd partial sums. So while the series does not converge, manipulation show that a different form of it could conditionally converge. That's why you always have to compute the partial sums as well.
A: Hint: This series can't converge, since |$a_n| \rightarrow 0$ is not true, because $\frac{n}{n+1}=1$, if $n \rightarrow \infty$(This is a neccessary term).
A: Be a little more careful. You can see that the limit of each term goes to $1$. With the alternator thrown in there, you essentially have the sum $$1-1+1-1\ldots$$ for large enough $n$. For a sum to be convergent you need $$\lim_{n \to \infty}a_n = 0$$ 
