Originally Posted by
abender $\displaystyle \sum^\infty_{n=1} \frac{40n}{(n+3)(n+4)(n+5)}=40\cdot \sum^\infty_{n=1} \frac{n}{(n+3)(n+4)(n+5)}$
Let $\displaystyle \sum a_n, \sum b_n$ be two positive sequences such that $\displaystyle a_n\leq b_n$ for all $\displaystyle n$.
If $\displaystyle \sum b_n$ converges, so does $\displaystyle \sum a_n$. If $\displaystyle \sum a_n$ diverges, so does $\displaystyle \sum b_n$.
$\displaystyle \sum^\infty_{n=1}\frac{n}{(n+3)(n+4)(n+5)} \leq \sum^\infty_{n=1} \frac{n}{n^3} = \sum^\infty_{n=1} \frac{1}{n^2}$
We now have a series of the form $\displaystyle \sum^\infty_{n=1}\frac{1}{n^p}$. Can you take it from here?
-Andy