You have a mistake in the last line, it is easy to prove that ##\lim_{n \to +\infty}(1-\frac{1}{n+2})^{n+2}=e^{-1}=\lim_{n \to +\infty}(1-\frac{1}{n+2})^{n}## so the final limit is 1.
Instead upper bound the series by the hyperharmonic series (p-series for p=2 )which is a well know result that it converges, hence by the comparison criterion this series converges too.
#3
member 587159
Comparing with the convergent series
$$\sum_{n=1}^\infty \frac{1}{n^2} =\frac{\pi^2}{6}$$ yields the result.
Alternatively, use the Cauchy condensation criterium or the integral test.
#4
Michael_0039
59
7
Thanks for your answer. I wanted to use D'alembert criterion to see if a solution can be found.