My book explains the Leibniz test by saying:

"Assume a sub n is a positive sequence that converges to 0..."

And goes on to say that that means the alternating series converges. What if the sequence doesn't go to 0? Does the Leibniz Test say that the series diverges in that case?

I'm trying to determine if the following converges or diverges:

$$\sum\limits_{n=0}^\infty \frac{(-1)^n*n}{\sqrt{n^2+1}}$$

I know from the book's answer that it diverges, but I don't know how I was supposed to determine that.

  • $\begingroup$ If a series $\sum a_n$ converges then $a_n \to 0$ (why? use Cauchy's convergence test) $\endgroup$ – Jose Antonio Mar 7 '14 at 8:47
  • $\begingroup$ Denote the $n$-th partial sum as $s_n = \sum_{i=0}^n a_i$, then $s_n \to L$ iff $(s_n)$ is a Cauchy sequence. Given $\varepsilon>0$, for all $n,m\ge n_0(\varepsilon)+1$ we have $|s_n-s_m|< \varepsilon$. If $m=n-1$, so $|s_{n}-s_{n-1}|=|a_{n}|< \varepsilon$. Hence $a_n \to 0$. $\endgroup$ – Jose Antonio Mar 7 '14 at 9:14

If a series $\sum a_n$ converges, then $a_n$ must go to zero. This is not the case in the example you posted, so the series cannot converge.

By the way, the Leibniz test also requires the sequence to be monotonically decreasing.

| cite | improve this answer | |
  • $\begingroup$ Mm, I knew that was true for positive series, but I wasn't sure about alternating series. Thank you! $\endgroup$ – mowwwalker Mar 7 '14 at 9:02

A necessary condition for the convergence of a series $\sum_n a_n$ is that the sequence $(a_n)$ converges to $0$ hence by the contraposition if $(a_n)$ doesn't converge to $0$ then the series diverges.

For the given series we have

$$\frac{(-1)^nn}{\sqrt{n^2+1}}\not\xrightarrow[n\to0]\;0$$ so this series is divergent.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.