<s>[INST] Check the convergence of the series : Which test do we apply? Check the convergence of the following series:

*

*$\displaystyle{\sum_{n=1}^{+\infty}\frac{n^2+1}{n^4+n}}$


*$\displaystyle{\sum_{n=1}^{+\infty}\frac{n^{n^2}}{(n+1)^{n^2}}}$
For the first series do we use the comparison test? But with which sequence do we compare $\frac{n^2+1}{n^4+n}$ ? Do we maybe do the following? $$\frac{n^2+1}{n^4+n}<\frac{n^2+n^2}{n^4+n}<\frac{n^2+n^2}{n^4}=\frac{2n^2}{n^4}=\frac{2}{n^2}$$
As for the second series I thought that we could use the ratio test but I am not sure if that works because we get $$\frac{a_{n+1}}{a_n}=\frac{\frac{(n+1)^{(n+1)^2}}{(n+2)^{(n+1)^2}}}{\frac{n^{n^2}}{(n+1)^{n^2}}}=\frac{(n+1)^{(n+1)^2}(n+1)^{n^2}}{(n+2)^{(n+1)^2}n^{n^2}}=\frac{(n+1)^{(n+1)^2+n^2}}{(n+2)^{(n+1)^2}n^{n^2}}$$ So is it better to apply an other test here?
 [/INST] For the second, it is better to use the root test.
$$\lim_{\infty}a_n^\frac 1n=\lim_{\infty}(\frac{n}{n+1})^n$$
think of
$$\lim_{\infty}(\frac{n+1}{n})^n=e$$
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