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I have the following series: $\sum^\infty_{n=0}(\frac{n+(-1)^n}{2n+10})^{\frac{n}{3}}$ and I have to check wether it converges or not. Which convergence test should I take? I think the limit of the summand is zero so the series should converge but how do I prove that?

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    Probably root test because of the $n/3$. – 2017-02-05
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    "I think the limit of the summand is zero so the series should converge" **No**, this is *not* enough. That the summand goes to zero is *necessary*, but *not sufficient* -- as the series $\sum_n \frac{1}{n}$ clearly shows. – 2017-02-05

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Hint:$$\frac{n+(-1)^n}{2n+10}\leq\frac{n+5}{2n+10}=\frac{1}{2}.$$ Now use the comparison test for positive series and apply the geometric series.

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As Ahmed S. Attaalla suggest note that $$\sqrt[n]{\left|\frac{n+\left(-1\right)^{n}}{2n+1}\right|^{n/3}}=\left|\frac{n+\left(-1\right)^{n}}{2n+1}\right|^{1/3}\rightarrow\sqrt[3]{\frac{1}{2}}$$ hence the series converges by the root test.

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    For clarity, adding "by the [root test](https://en.wikipedia.org/wiki/Root_test)" at the end would help the readers unfamiliar with it (and who haven't read the comments). – 2017-02-05
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    @ClementC. Done! – 2017-02-05
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$$ \begin{align} \,& 0\le\left[\frac{n+\color{red}{(-1)^n}}{2n+10}\right]^{\frac{n}{3}}\le\left[\frac{n+\color{red}{1}}{2n+10}\right]^{\frac{n}{3}} \space\implies \\[4mm] \,& 0\le\sum_{n=0}^{\infty}\left[\frac{n+(-1)^n}{2n+10}\right]^{\frac{n}{3}}\le\sum_{n=0}^{\infty}\left[\frac{n+1}{2n+10}\right]^{\frac{n}{3}} \space\rightarrow{\small\text{converges by ratio test or root test}} \\[6mm] \,& \small\lim_{n\rightarrow\infty}\left|\frac{a_{\small n+1}}{a_{\small n}}\right|=\lim_{n\rightarrow\infty}\frac{\left[\frac{(n+1)+1}{2(n+1)+10}\right]^{\frac{n+1}{3}}}{\left[\frac{n+1}{2n+10}\right]^{\frac{n}{3}}}=\lim_{n\rightarrow\infty}\sqrt[3]{\frac{1+2/n}{2+12/n}}\,\left[\frac{(n+2)(2n+10)}{(n+1)(2n+12)}\right]^{\frac{n}{3}} =\color{red}{\frac{1}{\sqrt[3]{2}}}\,\lt1 \end{align} $$