Question

Evaluate:

$\displaystyle\sf{\sum_{n=1}^{\infty}\dfrac{1}{n^2+5n+6}}$​

Answer 1

$\large\underline{\sf{Solution-}}$

$\displaystyle\rm :\longmapsto\:\sum_{n=1}^{\infty}\dfrac{1}{ {n}^{2} + 5n + 6}$

$\displaystyle \: \rm \: = \:\sum_{n=1}^{\infty}\dfrac{1}{ {n}^{2} + 3n + 2n+ 6}$

$\displaystyle \: \rm \: = \:\sum_{n=1}^{\infty}\dfrac{1}{n(n + 3) +2(n+ 3)}$

$\displaystyle \: \rm \: = \:\sum_{n=1}^{\infty}\dfrac{1}{(n + 3)(n+ 2)}$

$\displaystyle\: \rm \: = \:\sum_{n=1}^{\infty}\dfrac{3 - 2}{(n + 3)(n+ 2)}$

$\displaystyle\: \rm \: = \:\sum_{n=1}^{\infty}\dfrac{3 - 2 + n - n}{(n + 3)(n+ 2)}$

$\displaystyle\: \rm \: = \:\sum_{n=1}^{\infty}\dfrac{(n + 3) - (n + 2)}{(n + 3)(n+ 2)}$

$\displaystyle\: \rm \: = \:\sum_{n=1}^{\infty}\bigg(\dfrac{(n + 3)}{(n + 3)(n + 2)} - \dfrac{(n + 2)}{(n + 3)(n + 2)} \bigg)$

$\displaystyle\: \rm \: = \:\sum_{n=1}^{\infty}\bigg(\dfrac{1}{n + 2} - \dfrac{1}{n + 3} \bigg)$

$\: \rm \: = \bigg(\dfrac{1}{3} - \dfrac{1}{4}\bigg) + \bigg(\dfrac{1}{4} - \dfrac{1}{5} \bigg) + \bigg(\dfrac{1}{5} - \dfrac{1}{6} \bigg) + - - - -$

$\: \rm \: = \:\dfrac{1}{3}$

Hence,

$\purple{ \displaystyle\bf :\longmapsto\:\sum_{n=1}^{\infty}\dfrac{1}{ {n}^{2} + 5n + 6} = \dfrac{1}{3} }$

Additional Information :-

$\rm :\longmapsto\: log(1 + x) = \sum_{n=1}^{\infty}\dfrac{ {( - 1)}^{n - 1} {x}^{n} }{n}$

$\rm :\longmapsto\: - log(1 - x) = \sum_{n=1}^{\infty}\dfrac{{x}^{n} }{n}$

$\rm :\longmapsto\: log(2) = 1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + - - - -$