Find $\lim_{x\to\infty}\frac{1}{x}\sum_{n\leq x}\frac{(-1)^n}{n}$ What is the limit of
$$\lim_{x\to\infty}\frac{1}{x}\sum_{n\leq x}\frac{(-1)^n}{n}?$$
I know that 
$ \sum_{n \leq x } \frac{(-1)^n}{n}= \sum_{n \leq x }\frac{(-1)^n}{x}+\int_{1}^{x} \sum_{n \leq t } \frac{(-1)^n dt}{t^2}
\leq 1+\int_{1}^{x}  \frac{ dt}{t}=1+\log(x) $.
would that mean that this limit converges?

Since the series $\sum_n \frac{(-1)^n}{n}$ converges, the limit is $0$.
