http://www.webassign.net/cgi-bin/symimage.cgi?expr=sum_%28n%3D1%29%5Einfinity%20%5C %288%2Fe%2A%2An%20%2B%206%2F%28n%28n%2B1%29%29%5C% 29

I know that this series converges, but how to get its sum?

Break it up into two sums:

8\sum_{n=1}^{\infty}\frac{1}{e^{n}}+6\sum_{n=1}^{\ infty}\frac{1}{n(n+1)}

The second one is a simple telescoping sum. Rewrite it as:

6\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right). Sub in values starting at 1. You will see everything cancels out except for one term and that is the sum.

The first part involving e is a series that can be derived from the e series.
It converges to \sum_{n=1}^{\infty}\frac{1}{e^{n}}=\frac{1}{e-1}. Can you see why.

A better way to obtain the sum for

\sum_{n=1}^{\infty}\frac{1}{e^{n}},

is to note it is a geometric series with first term a_{1}=\frac{1}{e} and ratio r=\frac{1}{e}

Thus, the sum is:

\frac{a_{1}}{1-r}=\frac{\frac{1}{e}}{1-\frac{1}{e}}=\frac{1}{e-1}.