Testo nascosto, fai click qui per vederlo
Since \( (1) = (3)\) (after a simplification and a change of variable), we focus on \( (2) \) and \( (3)\).
\( (2) . \) We have that \[ \sum_{n = 0}^\infty \frac{(n+1)^2}{n!} = \sum_{n = 0}^\infty \frac{n^2}{n!} + \frac{2n}{n!} + \frac{1}{n!}. \]Since everything here is absolutely convergent we can harmlessly permute the summation order: \[ \sum_{n = 0}^\infty \frac{n}{n!} = \sum_{n = 1}^\infty \frac{1}{(n-1)!} = e \] and \[\sum_{n = 0}^\infty \frac{n^2}{n!} = \sum_{n = 1}^\infty \frac{n}{(n-1)!} = \sum_{n=0}^\infty \frac{n+1}{n!} = 2e. \] In conclusion the sum of \( (2) \) equals \(5 e \).
\( (3). \) With similar computations it turns out that the sum of \( (3) \) equals \( 15 e\).
I would like to suggest a problem, too:
Using \( \sum_{n=1}^\infty \frac{1}{n^2}= \frac{\pi^2}{6} \), compute \[ \sum_{n=1}^\infty \frac{1}{(n+2)^2} \left( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n \cdot (n+1)}\right). \]