for which $\alpha>1$ this series converges

Question

I have to understand for which $\alpha>1$ the following series converges: $$ \sum_{n=1}^{\infty}\left[1-\left(1-\frac{1}{n^\alpha}\right)^{2n}\right] $$ My question is: can I approximate $(1-1/n^\alpha)^{2n}\approx e^{-2/n^{\alpha-1}}$? In this way I find that $$ 1-\left(1-\frac{1}{n^\alpha}\right)^{2n}\sim_{asympt.}1-e^{-\frac{2}{n^{\alpha-1}}}\sim_{asympt.}\frac{1}{n^{\alpha-1}} $$ and I can deduce that the series converges for $\alpha>2$. Sorry for the triviality of my question but seems to me I'm missing something.

Answer 1

Yes that's absolutely fine to guess the solution indeed

$$\left(1-\frac{1}{n^\alpha}\right)^{2n}=e^{2n\log \left(1-\frac{1}{n^\alpha}\right)}=e^{-\frac2{n^{\alpha-1}}+O\left(\frac1{n^{2\alpha-1}}\right)}=1-\frac2{n^{\alpha-1}}+O\left(\frac1{n^{2\alpha-1}}\right)$$

then

$$1-\left(1-\frac{1}{n^\alpha}\right)^{2n}=\frac2{n^{\alpha-1}}+O\left(\frac1{n^{2\alpha-1}}\right)$$

and we formalize noting that the latter converges for any $\alpha>1$ by limit comparison test with $\sum \frac1{n^{\alpha-1}}$.