$$\sum_{n=1}^{\infty}\frac{7}{n(n+3)}$$
I know this is convergent but I do not know how to apply "cover-up method" to solve this can I get some help pls?
The answer I got is $77/18$
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$\begingroup$If by cover-up method you mean creative telescoping, well: $$\sum_{n\geq 1}\frac{7}{n(n+3)}=\frac{7}{3}\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+3}\right) = \frac{7}{3}\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}\right)=\color{red}{\frac{77}{18}}.$$ As an alternative: $$ \sum_{n\geq 1}\frac{7}{n(n+3)} = 7\int_{0}^{1}\sum_{n\geq 1}\frac{x^{n+2}}{n}\,dx = 7\int_{0}^{1}-x^2\log(1-x)\,dx $$ and the last integral equals $ 7\int_{0}^{1}-(1-x)^2\log(x)\,dx$ that is simple to compute by IBP (Integration By Parts).
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