<s>[INST] Problem with summation by method of difference Question: What would be the result of: $$\sum_{k=1}^{n}\frac{1}{n(n+2)}$$
My Approach:
Let $T_n$ denote the $n^{th}$ term of the given series. Then we have
$$T_1=\frac12 \left(\frac11-\frac13\right)$$
$$T_2=\frac12 \left(\frac12-\frac14\right)$$
$$T_3=\frac12 \left(\frac13-\frac15\right)$$
And so on up till
$$T_n=\frac12 \left(\frac1n-\frac1{n+2}\right)$$
I can see that the series telescopes and the terms start to cut each other after an interval of one. My only problem is, how do I find the terms that remain in the end?
 [/INST] For all $k \geq 1$ you have
$$
\frac{1}{k(k+2)} = \frac{1}{2} \left( \frac{1}{k} - \frac{1}{k+2} \right)
$$
so
$$
\begin{aligned}
\sum_{k=1}^n \frac{1}{k(k+2)} &= \frac{1}{2}\sum_{k=1}^n \left( \frac{1}{k} - \frac{1}{k+2} \right) \\
&= \frac{1}{2}\left( \sum_{k=1}^n \frac{1}{k} - \sum_{k=1}^n \frac{1}{k+2} \right) \\
&= \frac{1}{2}\left( \sum_{k=1}^n \frac{1}{k} - \sum_{k=3}^{n+2} \frac{1}{k} \right) \\
&= \frac{1}{2}\left( 1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2}\right) \\
\end{aligned}
$$
</s>