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?

By the Alternating Series Test
$$
\sum_{k=1}^\infty\frac{(-1)^k}k
$$
converges to some limit between $\frac12$ and $1$. Thus,
$$
\lim_{n\to\infty}\frac1{2n}
\le\lim_{n\to\infty}\frac1n\sum_{k=1}^n\frac{(-1)^k}k
\le\lim_{n\to\infty}\frac1n
$$
That is,
$$
\lim_{n\to\infty}\frac1n\sum_{k=1}^n\frac{(-1)^k}k=0
$$

What you have shown in your question is that
$$
\lim_{n\to\infty}\frac1n\sum_{k=1}^n\frac1k=0
$$
which also implies the result about the alternating series.
