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: $$\lim_{n \to\infty}\frac{(n+1)^{(-1)^{(n+1)}}\cdot n}{n^{(-1)^n}\cdot(n+1)}$$


2 Answers 2


$a_n = n^{(-1)^n - 1} = \begin{cases} 1 \qquad n \text{ is even} \\ n^{-2} \quad n \text{ is odd} \end{cases}$

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 $1$.

  • 1
    $\begingroup$ Is there another way to prove it than 'easy to see'? $\endgroup$
    – I I
    Commented Sep 13, 2016 at 11:44
  • $\begingroup$ @II I can think of a better way to prove this, but then again, that involves a complicated result. This problem is not easy, it cannot be done with the help of the tests you know so far, because of the alternating term. $\endgroup$ Commented Sep 13, 2016 at 11:51
  • $\begingroup$ If you know what to prove (i.e., have computed the first few terms of the sequence or plotted them), then a simpler thing could be to observe that if $a_n\to L$, then $a_{n+1}-a_n \to L-L=0$, i.e. the difference between consecutive terms must go to zero. But here, you have $a_{2n+1}-a_{2n} = \frac{1}{n^2}-1 \to -1$. Clearly not zero. $\endgroup$
    – Clement C.
    Commented Sep 13, 2016 at 12:06
  • $\begingroup$ @ClementC. Thank you for the clarification. $\endgroup$ Commented Sep 13, 2016 at 12:07

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 $b_n = n^3$

  • $\begingroup$ What if $b_n$ is just bounded? Then limit may not exist but still $\lim_{n \to \infty}a_n b_n=0$? $\endgroup$
    – I I
    Commented Sep 13, 2016 at 11:40
  • 2
    $\begingroup$ Indeed, it is also true with $b$ bounded and $a$ going to $0$. This is why I mentioned there may be other cases, but those are sufficient conditions. $\endgroup$
    – Vincent
    Commented Sep 13, 2016 at 11:41
  • $\begingroup$ I think that you can't just multiply in $\frac{(2k)^{(-1)^{2k}}}{2k} $ ${(-1)}^n=-n$ wolfram alpha gives another result $\endgroup$
    – I I
    Commented Sep 13, 2016 at 11:58

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