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$\sum_{1}^{\infty } \frac{2n+3}{n(n+1)(n+2)(n+3)}$

How to find partial sum, sum and prove convergence by definition? Thanks a lot.

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closed as off-topic by user223391, NCh, mlc, Dragonemperor42, David K May 19 '17 at 21:26

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$$\sum_{n=1}^{\infty } \frac{2n+3}{n(n+1)(n+2)(n+3)}$$

Resolve it into partial fractions,

$$=\frac{1}{2}\sum_{n=1}^{\infty }[\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}]$$

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Hint: use that $$\frac{2n+3}{n(n+1)(n+2)(n+3)}=-1/2\, \left( n+1 \right) ^{-1}-1/2\, \left( n+2 \right) ^{-1}+1/2\, \left( n+3 \right) ^{-1}+1/2\,{n}^{-1} $$

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Hint: $\;\;\cfrac{2n+3}{n(n+1)(n+2)(n+3)}=\cfrac{1}{n(n+2)} - \cfrac{1}{(n+1)(n+3)}$

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