Does it $\sum_{n=1}^{\infty} ({1}- \frac{1}{n^4})^{ n^5}$ converge?

Question

$$\sum_{n=1}^\infty\left(1-\frac1{n^4}\right)^{n^5}$$

I assume that I should rewrite it in some way like$$\sum_{n=1}^{\infty} ({1}+ \frac{1}{n})^{ n} \rightarrow e$$ but I still unfortunately don't know how to do it in this case.

Answer 1

Well, we know $\left(1-\frac1{n^4}\right)^{n^4}\to e^{-1}$ as $n\to\infty$, so intuitively, the terms tend to $e^{-n}$.

To do this more rigorous; note that $\left(1-\frac1{n^4}\right)^{n^4}$ approaches $e$ from below; that is, $\left(1-\frac1{n^4}\right)^{n^4}<e$ for all $n$, so that your sum is in fact less than our comparison:

$$\sum_{n=1}^\infty\left(1-\frac1{n^4}\right)^{n^5}<\sum_{n=1}^\infty\frac{1}{e^n}$$

And since the latter converges (and both are bounded below by $0$, and are strictly increasing), yours must converge.

Answer 2

Using root test: $$\lim_\limits{n\to\infty} \sqrt[n]{a_n} =\lim_\limits{n\to\infty} \left(1-\frac{1}{n^4}\right)^{n^4} =\frac{1}{e}<1.$$ So, it converges.