$\lim_{n \rightarrow \infty} \big( \frac{1}{n+1}-\frac{1}{n+2}+...+\frac{(-1)^{n-1}}{2n} \big)$ I need to compute
$$\lim_{n \rightarrow \infty} \big( \frac{1}{n+1}-\frac{1}{n+2}+...+\frac{(-1)^{n-1}}{2n} \big).$$
I've previously managed to show that
$$\lim_{n \rightarrow \infty} \Big( \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \Big) \rightarrow \log 2$$
by using the formula 
$$\int_{0}^1 f(x) \mathrm{d}x = \lim_{n\rightarrow \infty} \sum_{i=1}^n f\Big( \frac{i}{n} \Big) \Big( \frac{1}{n}\Big)$$
but unfortunately I don't think the same trick works for what I need here, at least not without some adjustment that I can't think of. Or perhaps there is another way?

$\begin{array}\\
s(n)
&= \big( \frac{1}{n+1}-\frac{1}{n+2}+...+\frac{(-1)^{n-1}}{2n} \big)\\
&= \sum_{k=1}^n\frac{(-1)^{k-1}}{n+k}\\
s(2n)
&= \sum_{k=1}^{2n}\frac{(-1)^{k-1}}{2n+k}\\
&= \sum_{k=1}^{n}\left(\frac{1}{2n-1+k}-\frac{1}{2n+k}\right)\\
&= \sum_{k=1}^{n}\frac{1}{(2n-1+k)(2n+k)}\\
&\le \sum_{k=1}^{n}\frac{1}{(2n)(2n+1)}\\
&= \frac{n}{(2n)(2n+1)}\\
&= \frac{1}{2(2n+1)}\\
&\to 0\\
s(2n+1)
&= \sum_{k=1}^{2n+1}\frac{(-1)^{k-1}}{2n+1+k}\\
&= \sum_{k=2}^{2n+2}\frac{(-1)^{k}}{2n+k}\\
&= \sum_{k=1}^{2n}\frac{(-1)^{k}}{2n+k}-\frac{(-1)^{1}}{2n+1}+\frac{(-1)^{2n+1}}{2n+2n+1}+\frac{(-1)^{2n+2}}{2n+2n+2}\\
&= -s(2n)+\frac{1}{2n+1}-\frac{1}{2n+2n+1}+\frac{1}{2n+2n+2}\\
&= -s(2n)+\frac{8 n + 1}{2 (2 n + 1) (4 n + 1)}\\
|s(2n+1)|
&= |-s(2n)+\frac{8 n + 1}{2 (2 n + 1) (4 n + 1)}|\\
&\le|-s(2n)|+\frac{8 n + 1}{2 (2 n + 1) (4 n + 1)}\\
&\le  \frac{1}{2(2n+1)}+\frac{8 n + 1}{2 (2 n + 1) (4 n + 1)}\\
&\to 0\\
\end{array}
$
