# Determine whether the series \sum_{n=1}^{\infty} \frac{9}{e^n} + \frac{6}{n(n+1)} is convergent...

## Question:

Determine whether the series {eq}\sum_{n=1}^{\infty} \frac{9}{e^n} + \frac{6}{n(n+1)} {/eq} is convergent or divergent. ? If it is convergent, find its sum.

## Sum of Convergent Series:

This series is the sum of two convergent series and therefore convergent. It sum is equal to the sum of the limits of the two series. The first is geometric series and the second is a telescoping series.

For the first piece, we have a geometric series. Remember the geometric series:

{eq}\displaystyle \sum_{n=1}^{\infty} r^n = \frac{r}{1- r } \quad \mbox{ for} \, |r|<1. {/eq}

So {eq}\displaystyle \sum_{n=1}^{\infty} \frac{9}{e^n} = 9 \sum_{n=1}^{\infty} (\frac{1}{e})^n = 9 \frac{1/e}{1- 1/e} = \frac{9}{e-1}. {/eq}

The second piece is a telescoping series. Using partial fraction decomposition, we have:

{eq}\displaystyle \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} . {/eq}

Find the N-th partial sum:

{eq}\displaystyle s_N= \sum_{n=1}^{N} \frac{1}{n(n+1)} = \sum_{n=1}^{N} \left[ \frac{1}{n} - \frac{1}{n+1} \right] = \frac{1}{1} - \frac{1}{N+1} {/eq}

This is the result of all the cancellations.

Taking the limit of {eq}\displaystyle s_N, {/eq} an N goes to infinity, we find:

{eq}\displaystyle \sum_{n=1}^{\infty}\frac{1}{n(n+1)} = \lim_{N\rightarrow \infty} \sum_{n=1}^{N} \frac{1}{n(n+1)} = \lim_{N\rightarrow \infty} (1 - \frac{1}{N+1}) =1. {/eq}

Therefore,

{eq}\displaystyle \sum_{n=1}^{\infty} \frac{9}{e^n} + \frac{6}{n(n+1)} = \sum_{n=1}^{\infty} \frac{9}{e^n} + 6 \sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \frac{9}{e-1} +6 = \frac{6 e +3}{e-1} {/eq}

is convergent.