Sum of a series
Task number: 2903
Using \(\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}n=\ln 2 \) and \(\displaystyle \sum_{n=1}^{\infty}\frac{a^n}{n!}=e^a \) determine the sum of the series
Variant 1
\(\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)} \)
Variant 2
\(\displaystyle \sum_{n=1}^{\infty}\frac{n^2}{n!} \)
Variant 3
\(\displaystyle \sum_{n=1}^{\infty}\frac{2^n(n+1)}{n!} \) a