A calculus problem by Ronak Agarwal

Interestingly moulded simple problem if you know basel series,

Well first split them up as partial fractions to get

$\frac { 1 }{ { n }^{ 2 }(n+1) } =\frac { 1 }{ { n }^{ 2 } } -\frac { 1 }{ n } +\frac { 1 }{ n+1 }$

Now we know the sum of the first term is

$\frac { 1 }{ { n }^{ 2 } } =\frac { \pi ^{ 2 } }{ 6 }$

(This is a standard result and the proof is really interesting and derived using factor theorem on the expansion of sin(x) , check it out http://en.wikipedia.org/wiki/Basel_problem )

Now the second and third terms are individually divergent but they cancel out each other leaving '-1' as

$\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ 1+n } } =\sum _{ m=2 }^{ \infty }{ \frac { 1 }{ m } } =\sum _{ m=1 }^{ \infty }{ \frac { 1 }{ m } } -1\quad =\quad \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ n } } -1\\$

from this we get

$\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ 1+n } } -\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ n } } =\quad -1\\$

so we have the sum as

$\frac { \pi ^{ 2 } }{ 6 } -1\quad =\quad 0.64493$