# Does the Leibniz Test say that if the sequence doesn't go to 0 that the series diverges?

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.

• If a series $\sum a_n$ converges then $a_n \to 0$ (why? use Cauchy's convergence test) Commented Mar 7, 2014 at 8:47
• 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 $a_n \to 0$. Commented Mar 7, 2014 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.
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.
$$\frac{(-1)^nn}{\sqrt{n^2+1}}\not\xrightarrow[n\to0]\;0$$ so this series is divergent.