Check the convergence of the series : Which test do we apply? Check the convergence of the following series:

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


*$\displaystyle{\sum_{n=1}^{+\infty}\frac{n^{n^2}}{(n+1)^{n^2}}}$
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?

Also, note that the terms of the second sum decay exponentially: Using $\ln(1-x)=-x+O(x^2)$ for $x\to0$, $$\frac{n^{n^2}}{(n+1)^{n^2}}=\left(1-\frac1{n+1}\right)^{n^2}=\exp\left(n^2\ln\left(1-\frac1{n+1}\right)\right)=O(1)\exp\left(-n\right).$$
