# Homework Help: Showing the sum of this telescoping series

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1. Feb 2, 2016

### Euler2718

1. The problem statement, all variables and given/known data

Determine whether each of the following series is convergent or divergent. If the series is convergent, find its sum

$$\sum_{i=1}^{\infty} \frac{6}{9i^{2}+6i-8}$$

2. Relevant equations

Partial fraction decomposition

$$\frac{1}{3i-2} - \frac{1}{3i+4}$$

3. The attempt at a solution

The divergence test is inconclusive, so I wrote as partial fractions and started analysing the nth sum:

$$S_{n} = \left( 1-\frac{1}{7} \right) + \left( \frac{1}{4} - \frac{1}{10} \right) + \left( \frac{1}{7} - \frac{1}{13} \right) + \left( \frac{1}{10} - \frac{1}{16} \right) + \dots$$

1 and 1/4 are the only terms that do not cancel, but how do I show this in the nth case? I'm having trouble writing it generally.

2. Feb 2, 2016

### Staff: Mentor

Include the general term in your expansion:
$S_{n} = \left( 1-\frac{1}{7} \right) + \left( \frac{1}{4} - \frac{1}{10} \right) + \left( \frac{1}{7} - \frac{1}{13} \right) + \left( \frac{1}{10} - \frac{1}{16} \right) + \dots + \left( \frac{1}{3n-2} - \frac{1}{3n+4} \right) + \dots$
If you add in the term before and the one after the last term I wrote above, you should see how the telescoping happens.

3. Feb 2, 2016

### Ray Vickson

Show that for every term of the form $1/(2n)$ there will be another term $-1/(2n)$ (corresponding to just two possible values of $i$) and for every term $1/(2n+1)$ there is a cancelling term $-1/(2n+1)$---again, corresponding to exactly two values of $i$.

4. Feb 2, 2016

### Euler2718

Thanks guys, I got it now.