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Question Number 82084 by oyemi kemewari last updated on 18/Feb/20

Commented by john santu last updated on 18/Feb/20

$$\frac{1}{2(2^2-1)}+\frac{1}{3(3^2-1)}+\frac{1}{4(4^2-1)}+...$$$$\underset{k=2}{\sum ^{\mathrm{\infty }}}\frac{1}{k(k^2-1)}=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{A}{k}+\frac{B}{k-1}+\frac{C}{k+1}$$$$youcangettheresult$$

Commented by john santu last updated on 18/Feb/20

$$\underset{k=2}{\sum ^{\mathrm{\infty }}}\frac{1}{2(k+1)}+\frac{1}{2(k-1)}-\frac{1}{k}$$

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