Find `sum_(n=1)^(oo)1/(n^2+5n+6)`

Answer

Let `a_(n)` denote the `n^(th)` terms of the given series. Then `a_n=1/(n^2+5n+6)` By using partial fraction, we get `a_(n)=1/(n+2)-1/(n+3)`.
Let `s_(n)` denote the sum of first `n` terms of the given series. Then
`s_(n)=a_(1+)a_(2+)...+a_(n)=(1/3-1/4)+(1/4-1/5)+(1/5-1/6)+....+(1/(n+2)-1/(n+3)=1/3-1/(n+3))`
But as `n` tends to infinity, `1/(n+3)` tends to zero and hence `1/3-1/(n+3)` tends to `1/3`.In other words `s_(n)` tends to `1/3.`Thus `sum_(n=1)^oo1/(n^2+5n+6)=1/3`.