$\lim_{n\to\infty} \frac{(-1)^n}{n}=0\,$? What is $\lim_{n\to\infty} \frac{(-1)^n}{n}$?
If this value tends to zero, how can I prove it?
 A: There are several ways to proof this. 
It depends on what you already know.
If you know that $\lim_{n\to\infty} \frac{1}{n}=0=\lim_{n\to\infty}-\frac1n$ then since
$-\frac{1}{n}\leq \frac{(-1)^n}{n}\leq \frac{1}{n}$
It is $\lim_{n\to\infty} \frac{(-1)^n}{n}=0$ by the squeezing theorem. 
Or simply since the sequence $a_n=\frac1n$ converges to $0$ and $(-1)^n$ is bounded (by $\pm 1$) your sequence in question is a null sequence.
Also you can go straight by definition. In that case you can more likly copy the proof of $\lim_{n\to\infty} \frac1n=0$.
A: Remember that if $\sum a_n$ converges, then $\lim_{n\to \infty}a_n=0$. Using the alternating series test method, we know that $\sum_{n=1}^\infty\frac{(-1)^n}{n}$ converges. 
Hence, $\lim_{n\to \infty}\frac{(-1)^n}{n}=0$.
A: I assume that what causes you trouble is the $(-1)^n$ term in the numerator. I also suppose that you already know that $\lim\limits_{n\to \infty}\frac{1}{n}=0$, so I will show you how to compute this limit without using the squeeze theorem or using some overly complicated series test.
Consider the sequence $(x_n)_{n\ge 1}$, $x_n=\frac{(-1)^n}{n}$. We are going to show that the subsequences $(x_{2n})_{n\ge 1}$ and $(x_{2n+1})_{n\ge 1}$ both converge to the same limit, which is $0$.
We have that $\lim\limits_{n\to \infty} x_{2n}=\lim\limits_{n\to \infty}\frac{1}{2n}=0$ and $\lim\limits_{n\to \infty} x_{2n+1}=-\lim_{n\to \infty}\frac{1}{2n+1}=0$.
Since $(x_n)_{n\ge 1}=(x_{2n})_{n\ge 1} \cup (x_{2n+1})_{n\ge 1}$, it follows that $\lim\limits_{n\to \infty} x_{n}=0$.
P.S.: I also assumed that you are not looking for a proof by using the epsilon definition. In the future, please give more information when asking a question.
