Proof convergence of series using limit of summand 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?
 A: 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.
A: 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.
A: $$ 
\begin{align} 
\,& 0\le\left[\frac{n+\color{red}{(-1)^n}}{2n+10}\right]^{\frac{n}{3}}\le\left[\frac{n+\color{red}{1}}{2n+10}\right]^{\frac{n}{3}} \space\implies \\[4mm] 
\,& 0\le\sum_{n=0}^{\infty}\left[\frac{n+(-1)^n}{2n+10}\right]^{\frac{n}{3}}\le\sum_{n=0}^{\infty}\left[\frac{n+1}{2n+10}\right]^{\frac{n}{3}} \space\rightarrow{\small\text{converges by ratio test or root test}} \\[6mm] 
\,& \small\lim_{n\rightarrow\infty}\left|\frac{a_{\small n+1}}{a_{\small n}}\right|=\lim_{n\rightarrow\infty}\frac{\left[\frac{(n+1)+1}{2(n+1)+10}\right]^{\frac{n+1}{3}}}{\left[\frac{n+1}{2n+10}\right]^{\frac{n}{3}}}=\lim_{n\rightarrow\infty}\sqrt[3]{\frac{1+2/n}{2+12/n}}\,\left[\frac{(n+2)(2n+10)}{(n+1)(2n+12)}\right]^{\frac{n}{3}} =\color{red}{\frac{1}{\sqrt[3]{2}}}\,\lt1
\end{align} 
$$
