Simplifying a sum to calculate its limit

Question

I am currently learning about limits and have hit a small roadblock.

Given

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

I want to calculate the limit on my own (which results in $\frac{3}{4}$). The book in which this is explained (and solved) calculates the limit correctly and step by step. What it does not show is how to simplify the sum from the above to the following:

$$\frac{n(3n+5)}{4(n+1)(n+2)}$$

I appreciate any help!

Answer 1

HINT

Note that

$$\sum_{n=1}^\infty \frac{1}{n(n+2)}=\frac12\sum_{n=1}^\infty \left(\frac{1}{n}- \frac{1}{n+2}\right)=\frac12\left(1\color{red}{-\frac13}+\frac12\color{red}{-\frac14+\frac13-\frac15+\frac14-\frac16+...}\right)$$

Answer 2

Hint: $${1\over n(n+2)} = {1\over 2}{2\over n(n+2)} = {1\over 2}{(n+2)-n\over n(n+2)} = {1\over 2}({1\over n} -{1\over n+2}) $$