# Telescoping Sum: 3/(n(n+3))

1. Sep 27, 2009

### EstimatedEyes

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

Determine whether the series is convergent or divergent by expressing $$s_{n}$$ as a telescoping sum. If it is convergent, find its sum.
$$\sum\frac{3}{n(n+3)}$$

2. Relevant equations

3. The attempt at a solution

Partial Fraction Decomposition: $$\frac{1}{n} - \frac{1}{n+3}$$

Partial Sum: $$s_{n}= (\frac{1}{1} - \frac{1}{4}) + (\frac{1}{2} - \frac{1}{5}) + (\frac{1}{3} - \frac{1}{6}) + (\frac{1}{4} - \frac{1}{7}) + ... + (\frac{1}{n} - \frac{1}{n+3})$$

From the above partial sum, I deduced that the negative term of the nth term is canceled out by the positive term in the n+4th term. However, from there, I am not able to come up with a general formula for the sum of the series. Any help would be appreciated; thanks!

2. Sep 27, 2009

### Dick

So which terms DON'T cancel? I count six of them for large n.

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