General formula for this sum $\sum_{i=3}^n \frac{1}{(i-2) i (i+2)} = ?$

Question

I have been trying to get the general formula for this sum but I am stuck. I tried to do a partial fraction but couldn't see any pattern.

$$\sum_{i=3}^n \frac{1}{(i-2) i (i+2)} = ?$$

I need to find the general formula, than to prove it using induction but I can't figure out the formula.

Answer 1

Hint: you can rewrite it as:

$$\frac{1}{4}\sum_{i=3}^n\left(\frac{1}{2}\frac{1}{i-2}-\frac{1}{i}+\frac{1}{2}\frac{1}{i+2}\right)$$

It's a telescopic sum, most elements will cancel each other.

Answer 2

as @Andronicus showed, it decomposes to three fraction. Now every farction like $\frac{1}{m}$ appears three times, two of them by multiple of $+\frac{1}{2}$ and one time by multiple of $-1$. So they cancel each other, therefore only some terms of first and last remain: \begin{equation} \begin{split} 4\sum_{i=3}^n & = \sum_{i=3}^n (\frac{1}{2}\frac{1}{i-2}-\frac{1}{i}+\frac{1}{2}\frac{1}{i+2}) = \\ & = (\frac{1}{2}\frac{1}{3-2}+\frac{1}{2}\frac{1}{4-2}+\frac{1}{2}\frac{1}{5-2}+\frac{1}{2}\frac{1}{6-2}) \\ & - (\frac{1}{3}+\frac{1}{4}) \\ & - (\frac{1}{n}+\frac{1}{n-1}) \\ & = (\frac{1}{2}\frac{1}{n+2}+\frac{1}{2}\frac{1}{(n-1)+2}+\frac{1}{2}\frac{1}{(n-2)+2}+\frac{1}{2}\frac{1}{(n-3)+2}) \\ \end{split} \end{equation} The above formula is true only for any $n \geqslant 6$, because two middle lines suppose to have no mutual terms.