Finding converging majorant for $\sum_{n=1}^\infty\left(\frac{n^2}{n^2+1}\right)^{n^3}$

Question

I could use some tips/help finding a converging majorant for $$\sum_{n=1}^\infty \left(\frac{n^2}{n^2+1}\right)^{n^3}.$$ Thank you.

Answer 1

Note that

$${\left(\frac{n^2}{n^2+1}\right)^{n^3}}=e^{-n^3\log(1+\frac1{n^2})}\sim\frac1{e^n}$$

Answer 2

Why not directly to use the $\;n\,-$th root test?:

$$\lim_{n\to\infty}\sqrt[n]{\left(\frac{n^2}{1+n^2}\right)^{n^3}}=\lim_{n\to\infty}\left(\frac{n^2}{1+n^2}\right)^{n^2}=\lim_{n\to\infty}\frac1{\left(1+\frac1{n^2}\right)^{n^2}}=\frac1e<1$$

Answer 3

For a convergent majorant, note that, by the binomial theorem, for $n\geq 1$, $$\left(1+\frac{1}{n^2}\right)^{n^2}=\sum_{k=0}^{n^2}\binom{n^2}{k}\frac{1}{n^{2k}}\geq \sum_{k=0}^{1}\binom{n^2}{k}\frac{1}{n^{2k}}=1+1=2.$$ Then $$\left(\frac{n^2}{n^2+1}\right)^{n^3}=\left(\frac{1}{\left(1+\frac{1}{n^2}\right)^{n^2}}\right)^{n}\leq \left(\frac{1}{2}\right)^{n}.$$