# $n$-th partial sum and convergence $\sum_{k=1}^{\infty}\frac{1}{k(k+2)}$

Having trouble with finding the $n$-th partial sum, and seeing if it diverges or not of,

$$\sum_{k=1}^{\infty}\frac{1}{k(k+2)}$$

I know that it is a telescoping series, and I can solve $\displaystyle\sum_{k=1}^{\infty}\frac{1}{k(k+1)}$, but the same method doesn't seem to work with this one. Any help would be appreciated.

## 4 Answers

$$\frac{1}{k(k+2)} = \dfrac12\left(\frac{1}{k}-\dfrac{1}{k+1}\right)+\dfrac12\left(\dfrac{1}{k+1}-\frac{1}{k+2}\right)$$

$$\frac{1}{k(k+2)}=\frac{1}{2}\big( \frac{1}{k}-\frac{1}{k+2} \big)$$ this is a telescoping sum $$S_n=\frac{1}{2}+\frac{1}{4}-\frac{1}{2n+2} -\frac{1}{2n+4}$$

$$S_{\infty}=\frac{3}{4}$$ indeed it is converging

Partial sum:

This time instead of cancelling consecutive terms, you would cancel every other term, and you should be left with $\frac12 \left( 1 + \frac12 - \frac1{n+1} - \frac1{n+2} \right)$

Convergence:

$$0 \le \sum_{k=1}^{\infty}\frac{1}{k(k+2)} \le \sum_{k=1}^{\infty}\frac{1}{k(k+1)} = 1 < \infty$$

You don't have to look at the partial sums: it's simpler to use asymptotic equivalence, since it is a series with positive terms: $$\frac1{k(k+2)}\sim_\infty \frac1{k^2},$$ and the latter is a convergent $p$-series.