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

Difficulty level: Moderate task
Solution require uncommon idea
Cs translation
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