$\lim_{n \rightarrow \infty} \big( \frac{1}{n+1}-\frac{1}{n+2}+...+\frac{(-1)^{n-1}}{2n} \big)$ 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?

Since:
$$\sum_{k=1}^{n}\frac{(-1)^k}{k} \rightarrow \sum_{k=1}^{\infty}\frac{(-1)^k}{k}=\log 2$$
And also:
$$\sum_{k=1}^{2n}\frac{(-1)^k}{k} \rightarrow \sum_{k=1}^{\infty}\frac{(-1)^k}{k}=\log 2$$
Substracting one from the other we obtain:
$$\sum_{k=n+1}^{2n}\frac{(-1)^k}{k} \rightarrow \log 2 - \log 2 = 0$$
Which is either your sum, or its negative and thus the limit in question is zero.
