Check out some similar questions!

western50 · Dec 10, 2010, 08:37 PM

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

galactus · Dec 11, 2010, 04:24 AM

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.

galactus · Dec 12, 2010, 02:22 AM

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:

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