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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.

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  • $\begingroup$ If a series $\sum a_n$ converges then $a_n \to 0$ (why? use Cauchy's convergence test) $\endgroup$
    – Jose Antonio
    Mar 7, 2014 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, 2014 at 9:14

2 Answers 2

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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.

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  • $\begingroup$ Mm, I knew that was true for positive series, but I wasn't sure about alternating series. Thank you! $\endgroup$
    – mowwwalker
    Mar 7, 2014 at 9:02
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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.

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