TITLE: Find the limit of $\frac{1}{n+1}-\frac{1}{n+2}+...+\frac{(-1)^{n-1}}{2n}$ as $n\to\infty$
QUESTION [6 upvotes]: Find the limit of $\frac{1}{n+1}-\frac{1}{n+2}+...+\frac{(-1)^{n-1}}{2n}$ as $n\to\infty$.
In the previous part of the question I was asked to prove that as $n\to\infty$ $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}\to log{2}$. I could do this considering the integral of the function $f(x)=\frac{1}{x}$ from $(n+1)$ to $(2n+1)$ and from $(n)$ to $(2n)$ and then using the sandwich rule.
I guess I should use the same method somehow or this previous result.

REPLY [3 votes]: Note that if $n$ is even you can use successive differences to rewrite this sum as 
$$
\frac{1}{(n+1)(n+2)} + \frac{1}{(n+3)(n+4)} + \ldots + \frac{1}{(2n-1)(2n)}
$$
Here there are $n/2$ terms that are each bounded above by $1/n^2$, so the whole sum is bounded between $0$ and $1/2n$.  If $n$ is odd you can find a similar bound.