Determine convergence $\sum_{n=1}^\infty\left(\frac{2 n + 2}{2 n + 3}\right)^{n^2}$

Question

Does $$\sum_{n=1}^\infty\left(\frac{2 n + 2}{2 n + 3}\right)^{n^2}$$ converge?

Hi, I was wondering if anyone knows how to solve this problem? I think I can't use root test... because the result is 1 and it is meaningless. Thank you guys very much.

Answer 1

Note that $\lim_{n\to\infty}\sqrt[n]{(\frac{2n+2}{2n+3})^{n^2}}=e^{-\frac{1}{2}}<1$. Then the series converges.

Answer 2

$$\frac{2n+2}{2n+3}=1-\frac1{2n+3}\leqslant1-\frac1{5n}\leqslant\mathrm e^{-1/(5n)}\implies\left(\frac{2n+2}{2n+3}\right)^{n^2}\leqslant\mathrm e^{-n/5}\leqslant(0.9)^n$$

Answer 3

hint:

$$a_n:=\left(\frac{2n+2}{2n+3}\right)^{n^2}=\left(1-\frac1{2n+3}\right)^{n^2}\implies$$

$$ \sqrt[n]{a_n}=\left(1-\frac1{2n+3}\right)^n=\left[\left(1-\frac1{2n+3}\right)^{2n+3}\right]^{1/2}\left(1-\frac1{2n+3}\right)^{-3/2}\xrightarrow[n\to\infty]{}\left(e^{-1}\right)^{1/2}\cdot1=\frac1{\sqrt e}\ldots$$