In the problem below, I got the partial fraction decomposition but don't know how to get the answer:
$$\sum^\infty_{k=1}{\frac{1}{k(k+2)}}$$
I found the answer, using an online calculator, to be $\frac{3}{4}$.
When I did the partial fraction decomposition I got:
$$\sum^\infty_{k=1}\frac{1}{2k}-\frac{1}{2(k+2)}$$
Then I got the sum of
$$\left(\frac{1}{2}\right)+\dotsi-\frac{1}{2(n+2)}$$
Why does this equal $\frac{3}{4}$ and not $\frac{1}{2}$?
Best Answer
For $n \geq 3,$ the partial sums $$\begin{align*}\sum_{k=1}^n \left(\frac{1}{2k} - \frac{1}{2(k+2)}\right) &= \sum_{k=1}^n \frac{1}{2k} - \sum_{k=1}^n \frac{1}{2(k+2)} \\ &= \sum_{k=1}^n \frac{1}{2k} - \sum_{k=3}^{n+2} \frac{1}{2k} \\ &= \left(\frac{1}{2(1)} + \frac{1}{2(2)} + \sum_{k=3}^n \frac{1}{2k}\right) - \left(\sum_{k=3}^n \frac{1}{2k} + \frac{1}{2(n+1)} + \frac{1}{2(n+2)}\right) \\ &= \frac{1}{2(1)} + \frac{1}{2(2)} - \frac{1}{2(n+1)} - \frac{1}{2(n+2)} \\ &= \frac{3}{4} - \frac{1}{2(n+1)} - \frac{1}{2(n+2)}\end{align*}$$ have limit $\frac{3}{4}.$