Elementary Summation

This can be written as

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

= $\displaystyle \sum_{n = 1}^{\infty} \frac{1}{n^{2} + 5n + 4}$

= $\displaystyle \sum_{ n = 1}^{\infty} \frac{1}{(x +1)(x +4)}$

Seperating

$\displaystyle \sum_{ n = 1}^{\infty} \frac{1}{3}(\frac{1}{x + 1} - \frac{1}{x + 4})$

All terms will cancel out except 3 terms as we put values of n, we will finally get

$\displaystyle \frac{1}{3}( \frac{1}{2} + \frac{1}{3} + \frac{1}{4}) = \boxed{\frac{13}{36}}$

$We\quad will\quad observe\quad that\quad denominators\quad can\quad be\quad \\ expressed\quad as\quad n(n+3)\quad starting\quad from\quad 2\\ It\quad can\quad be\quad done\quad like\quad \frac { 1 }{ n(n+3) } =\frac { a }{ n } +\frac { b }{ n+3 } ........1\quad \\ therefore\quad \frac { 1 }{ n(n+3) } =\frac { a(n+3)+b(n) }{ n(n+3) } \\ cancel\quad denominators\quad \\ 1=\quad a(n+3)+b(n)\\ what\quad now\quad I\quad will\quad do\quad is\quad splitting\quad of\quad factors\\ or\quad partial\quad fractions\\ to\quad make\quad it\quad easy\quad first\quad let\quad a=-3\\ we\quad get\quad b=\frac { -1 }{ 3 } \\ let\quad b=0\quad we\quad get\quad a=\frac { 1 }{ 3 } \\ placing\quad in\quad equation\quad ......1\\ \frac { 1 }{ 3(n) } -\frac { 1 }{ 3(n+3) } \\ take\quad \frac { 1 }{ 3 } \quad as\quad common\quad start\quad putting\quad values\quad \\ from\quad 2\quad and\quad till\quad 10\quad you\quad will\quad observe\quad that\quad \\ all\quad terms\quad except\quad \frac { 1 }{ 2 } ,\frac { 1 }{ 3 } ,\frac { 1 }{ 4 } \quad will\quad be\quad cancelled\quad \\ therefore\quad \frac { 1 }{ 3 } (\frac { 1 }{ 2 } +\frac { 1 }{ 3 } +\frac { 1 }{ 4 } )\Longrightarrow \frac { 13 }{ 36 } \\ therefore\quad m=13\quad n=36\\ so\quad m+n=49\\ Don't\quad u\quad think\quad that\quad It's\quad a\quad very\quad nice\quad and\quad expertized\quad solution$