Diverges or converges? Find sequence of a limit I have a problem, I don't know how to transform the ratio test to get a proper result. Also, if $\lim_{n \to \infty}a_nb_n=\lim_{n \to \infty}a_n  \cdot\lim_{n \to \infty}b_n$and$a_n$ converges to $0$, then why doesn't $\lim_{n \to \infty}a_nb_n=0$ for every $b_n$? I think there's a proof of that and I can't derive it myself. Anyway here is the main problem:
$$\lim_{n \to \infty}\frac{n^{{(-1)}^n}}{n}$$
I was experimenting with limits and wolfram alpha cannot compute the answer for that limit.
From ratio test we get:
$$\lim_{n \to\infty}\frac{(n+1)^{(-1)^{(n+1)}}\cdot n}{n^{(-1)^n}\cdot(n+1)}$$
 A: $a_n = n^{(-1)^n - 1} = \begin{cases} 1 \qquad n \text{ is even} \\ n^{-2} \quad n \text{ is odd} \end{cases}$
The point is , whenever you are doing the ratio test for this question, you will fall into trouble because the sequence has an alternating term, like $(-1)^n$, which is nagging and won't disappear. Hence, this route is the best one to use:
There is a result (easy to prove):

Suppose that $a_n$ is a convergent sequence. Then, every subsequence of $a_n$ also converges to the same limit as $a_n$.

Now, it is easy to see that $a_n$ does not converge at all, because one subsequence of odd terms $\{n^{-2}\}$converges to $0$, and another of even terms $\{ 1\}$ to $1$.
A: Concerning your limit, note that $\frac{(2k)^{(-1)^{2k}}}{2k} = 1$ and that $\frac{(2k+1)^{(-1)^{2k+1}}}{2k+1} = \frac{1}{(2k+1)^2}$, so your limit doesn't exist, which explains why wolphram alpha doesn't find it.
Concerning your proposition, $\lim\limits_{n \rightarrow + \infty} a_nb_n = \lim\limits_{n \rightarrow + \infty} a_n \times \lim\limits_{n \rightarrow + \infty} b_n$ is not true all the time. If these sufficient (but not necessary) conditions are met :


*

*all limits in the proposition exist

*all limits in the proposition are finite
Then you can say it. It may be also true with different conditions not included in that case, but not all the time. Consider


*

*$a_n=(-1)^n$ and $b_n=(-1)^n$

*$a_n = \frac{(-1)^n}{n}$ and $b_n=n$

*$a_n=\frac{1}{n^2}$ and $b_n = n^3$
