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Question Number 116480 by Dwaipayan Shikari last updated on 04/Oct/20

$$\frac{1}{8}+\frac{1}{18}+\frac{1}{30}+\frac{1}{44}+\frac{1}{60}+\frac{1}{78}+\frac{1}{98}+\frac{1}{120}+........$$

Answered by Olaf last updated on 04/Oct/20

$$u_n=\frac{1}{n^2+7n}=\frac{1}{n(n+7)}=\frac{1}{7}(\frac{1}{n}-\frac{1}{n+7})$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{u}_n=\frac{1}{7}(\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n+7})$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{u}_n=\frac{1}{7}(\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-\underset{n=8}{\sum ^{\mathrm{\infty }}}\frac{1}{n})$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{u}_n=\frac{1}{7}\underset{n=1}{\sum ^7}\frac{1}{n}=\frac{1}{7}\times \frac{363}{140}=\frac{363}{980}$$$$$$

Commented by Dwaipayan Shikari last updated on 04/Oct/20

$$\mathrm{Great}\mathrm{sir}!$$

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