Sum of a Series: Finding the Convergence and Limit of a Series

[zeion]

Homework Statement

Find the sum of the series.

$$\sum_{k=0}^\infty \frac{1}{(k+1)(k+3)}$$

Homework Equations

The Attempt at a Solution

$$= \frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \frac{1}{3\cdot5} + ... + \frac{1}{(n+1)\cdot(n+3)}$$

$$= \frac{1}{2} [(1-\frac{1}{3}) + (\frac{1}{2} - \frac{1}{4}) + (\frac{1}{3} - \frac{1}{5}) + ... + (\frac{1}{(n+1)} - \frac{1}{(n+3)})$$

$$= \frac{1}{2}[ 1 + (\frac{1}{2} + \frac{1}{3} + \frac{1}{n+1}) - (\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + ... + \frac{1}{n+3}) ]$$

So here
$$(\frac{1}{2} + \frac{1}{3} + \frac{1}{n+1}) \to 1$$

$$(\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + ... + \frac{1}{n+3}) \to 0$$

Then the whole thing sums to 1?

 

[JSuarez]

Ok, things went very bad from the second to the third line in your argument; why don't you try to cancel the terms with opposite signs, instead of grouping them?