Convergence of series $\sum (\frac{n^4}{n^4+2})^{n^5-3}$

Question

Does the series $\sum_{n=1}^{\infty} (\frac{n^4}{n^4+2})^{n^5-3}$ converge?

I started by using the root test:

$$\sqrt[n]{|a_n|}=\left(\frac{n^4}{n^4+2}\right)^{\frac{n^5-3}{n}}=\left(1+\frac{2}{n^4}\right)^{-\frac{n^5-3}{n}}$$

I don't understand the convergence very well so I'm not sure what to do next? Can I make substitution $n^4=u$ and try to make it converge into $e$ or is that not possible?

Answer 1

Note that: $$\lim_{n\to\infty} \left(1+\frac{2}{n^4}\right)^{-\frac{n^5-3}{n}}=\\ \lim_{n\to\infty} \left(1+\frac{2}{n^4}\right)^{-n^4+\frac{3}{n}}=\\ \lim_{n\to\infty} \left(1+\frac{2}{n^4}\right)^{-n^4}\cdot \lim_{n\to\infty} \left(1+\frac{2}{n^4}\right)^{\frac{3}{n}}=\\ \left(\lim_{n\to\infty} \left(1+\frac{2}{n^4}\right)^{\frac{n^4}{2}}\right)^{-2}\cdot 1=e^{-2}<1.$$ So, based on the root test the original series converges.

Answer 2

Yes, it converges. Note that$$\lim_{n\to\infty}\left(1+\frac2{n^4}\right)^{n^4}=e^2.$$Therefore,$$\lim_{n\to\infty}\left(\frac{n^4}{n^4+2}\right)^{n^4}=e^{-2}$$and, since $0<e^{-2}<1$, the series $\sum_{n=1}^\infty(e^{-2})^n$ converges. Therefore, the series$$\sum_{n=1}^\infty\left(\frac{n^4}{n^4+2}\right)^{n^5}$$converges and it is easy to deduce from this that your series converges too.

Answer 3

As an alternative to root test we have

$$\left(\frac{n^4}{n^4+2}\right)^{n^5-3}=e^{(n^5-3)\log \left(\frac{n^4}{n^4+2}\right)}=e^{-(n^5-3)\log \left(1+\frac{2}{n^4}\right)}=$$$$=e^{-(n^5-3) \left(\frac{2}{n^4}+O(n^{-8})\right)}=e^{-2n+O(n^{-3}))}\sim\frac1{e^{2n}}\le \frac1{n^2}$$

therfore the given series converges by limit comparison test with $\sum \frac1{e^{2n}}$ which converges by comparison test with $\sum \frac1{n^2}$.