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Question Number 70030 by Joel122 last updated on 30/Sep/19

$$\mathrm{Find}\mathrm{the}\mathrm{convergence}\mathrm{of}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{\frac{1}{n}+1}{-n^2}$$

Answered by mind is power last updated on 30/Sep/19

$$\mid \sum _{n=1}^{+\mathrm{\infty }}\frac{\frac{1}{n}+1}{-n^2}\mid ⩽\sum _{n=1}^{+\mathrm{\infty }}\frac{2}{n^2}cv$$$$\underset{n=1}{\sum ^{+\mathrm{\infty }}}\frac{\frac{1}{n}+1}{-n^2}=\sum _{n=1}^{+\mathrm{\infty }}\frac{-1}{n^3}-\sum _{n⩾1}\frac{1}{n^2}=-\zeta \left(3\right)-\zeta \left(2\right)$$

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