Evaluate a sum which almost looks telescoping but not quite:$\sum_{k=2}^n \frac{1}{k(k+2)}$

Question

Suppose I need to evaluate the following sum:

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

With partial fraction decomposition, I can get it into the following form:

$$\sum_{k=2}^n \left[\frac{1}{2k}-\frac{1}{2(k+2)}\right]$$

This almost looks telescoping, but not quite... so at this point I am unsure of how to proceed. How can I evaluate the sum from here?

Answer 1

\begin{align*} \dfrac{1}{k(k+2)}&=\dfrac{1}{2}\left(\dfrac{1}{k}-\dfrac{1}{k+2}\right)\\ &=\dfrac{1}{2}\left(\left(\dfrac{1}{k}-\dfrac{1}{k+1}\right)+\left(\dfrac{1}{k+1}-\dfrac{1}{k+2}\right)\right), \end{align*} splitting the sum and doing telescope twice.

Answer 2

$\require{cancel}$ HINT

Look at the first few terms: $$\frac14\cancel{-\frac18}+\frac16\cancel{-\frac1{10}}\cancel{+\frac18}-\frac1{12}+\cancel{\frac1{10}}-\cdots$$ Do you notice anything particular?

Answer 3

One approach you could use is to manually check which terms cancel out. Notice that:

$$ \sum^{n}_{k=2} \dfrac{1}{k} = \dfrac{1}{2} + \dfrac{1}{3} + \left[\dfrac{1}{5} + ... + \dfrac{1}{n}\right]$$

And,

$$ \sum^{n}_{k=2} \dfrac{1}{k+2} = \left[\dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7} + ... + \dfrac{1}{n}\right] + \dfrac{1}{n+1} + \dfrac{1}{n+2} $$

The terms in the square brackets get cancelled out on subtraction.

Answer 4

Hint :

$1/2(\sum_{k=2}^{n}\dfrac{1}{k} -\sum_{k=2}^{n}\dfrac{1}{k+2} ):$.

$(1/2)\sum_{k=2}^{n}\dfrac{1}{k}=$

$(1/2)(\dfrac{1}{2} + \dfrac{1}{3} +.......\dfrac{1}{n})$ ;

$(1/2)\sum_{k=2}^{n}\dfrac{1}{k+2}=$

$(1/2)(\dfrac{1}{4}+...\dfrac{1}{n} +\dfrac{1}{n+1} + \dfrac{1}{n+2} ).$