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The following series is known to diverge because the limit does not exists. I am not sure why this limit does not exist and looking just for some explanation and detail of calculating this limit.

$$\sum \frac{n(-1)^{n+1}}{\sqrt{n^2+1}}$$

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We have $$\lim_{n \to \infty} \left\vert \frac{n(-1)^{n+1}}{\sqrt{n^2+1}}\right\vert = \lim_{n \to \infty} \left\vert \frac{n}{\sqrt{n^2+1}}\right\vert = \lim_{n \to \infty} \left\vert \frac{1}{\sqrt{1 + \frac{1}{n^2}}}\right\vert = 1$$ Hence the sequence $$\frac{n(-1)^{n+1}}{\sqrt{n^2+1}}$$ does not converge to $0$ which is necessary for any series to converge.

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    I thought the divergent test was just for $\sum a_n$ not $\sum |a_n|$ – 2017-02-11
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    @fr14 If we consider $\sum a_n$ it is necessary that $a_n \to 0$. But $$a_n \to 0 \qquad \Leftrightarrow \qquad |a_n| \to 0$$ since $|\cdot|$ is a norm. – 2017-02-11
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Notice that

$$\lim_{n\to\infty}\frac n{\sqrt{n^2+1}}=\lim_{n\to\infty}\frac1{\sqrt{1+\frac1{n^2}}}=1$$

And if $\lim_{n\to\infty}|a_n|\ne0$, then the series fails the term test.

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    I was $2$ seconds faster :P still +1 – 2017-02-11
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    @Simply Beautiful Art why did you not consider the alternating part? – 2017-02-11
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    @fr14 It's not necessary for the term test. It's like saying$$1-1+1-1+\dots$$might converge, even though it fails the term test when we ignore the alternating part. – 2017-02-11
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    okay so the nth term test is always $\sum |a_n | \ne 0$ then it diverges? – 2017-02-11
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    @fr14 $\lim|a_n|\ne0$ implies divergence. – 2017-02-11
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    yes sorry that's what i meant, okay thank you very much i thought it was just $lima_n \ne 0$ not $lim|a_n| \ne 0$ that's where i got a little mixed up – 2017-02-11
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    @fr14 (well, they kind of end up being the same thing)$$\lim a_n\ne0\implies|\lim a_n|\ne0\implies\lim|a_n|\ne0?$$ – 2017-02-11