Prove the following identity

Question

I am having some trouble proving following identity without use of induction, with which it is trivial.

$$\sum_{n=1}^{m}\frac{1}{n(n+1)(n+2)}=\frac{1}{4}-\frac{1}{2(m+1)(m+2)}$$

I did expand the expression: $$\sum_{n=1}^{m}\left( \frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2(n+2)} \right)$$

I have no idea how to proceed further.

Answer 1

HINT:

$$\begin{align*} \sum_{n=1}^m\left(\frac1{2n}-\frac1{n+1}+\frac1{2(n+2)}\right)&=\frac12\sum_{n=1}^m\left(\left(\frac1n-\frac1{n+1}\right)-\left(\frac1{n+1}-\frac1{n+2}\right)\right)\\\\ &=\frac12\left(\sum_{n=1}^m\left(\frac1n-\frac1{n+1}\right)-\sum_{n=1}^m\left(\frac1{n+1}-\frac1{n+2}\right)\right)\\\\ &=\frac12\left(\sum_{n=1}^m\left(\frac1n-\frac1{n+1}\right)-\sum_{n=2}^{m+1}\left(\frac1n-\frac1{n+1}\right)\right) \end{align*}$$

Answer 2

Hint: by partial fraction

$$\frac{1}{n(n+1)(n+2)}=\frac{A}{n}+\frac{B}{n+1}+\frac{C}{n+2}\Longrightarrow$$

$$1=A(n+1)(n+2)+Bn(n+2)+Cn(n+1)$$

Choosing smartly values for $\,n\,$ above , we get

$$1=2A\Longrightarrow A=\frac{1}{2}\;\;,\;\;1=-B\Longrightarrow B=-1\;\;,\;\;1=2C\Longrightarrow C=\frac{1}{2}$$

so

$$\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\left(\frac{1}{n}-\frac{2}{n+1}+\frac{1}{n+2}\right)$$

So summing up

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

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

Now just check that there's a lot of cancellation up there...

Answer 3

A straightforward way. Write the sum as

$$\sum_{n=1}^{m}\left(\frac{1}{2n(n+1)}-\frac{1}{2(n+1)(n+2)}\right)=\frac{1}{4}-\frac{1}{2(m+1)(m+2)}$$ Q.E.D. (the sum telescopes)

Answer 4

This is a special case $(k=3)$ of the following $$ \begin{align} \sum_{n=k}^m\frac1{\binom{n}{k}} &=\frac{k}{k-1}\sum_{n=k}^m\left(\frac1{\binom{n-1}{k-1}}-\frac1{\binom{n\vphantom{1}}{k-1}}\right)\\ &=\frac{k}{k-1}\left(1-\frac1{\binom{m}{k-1}}\right) \end{align} $$ Noting that $n(n+1)(n+2)\dots(n+k-1)=k!\binom{n+k-1}{k}$ yields $$ \sum_{n=1}^m\frac1{k!\binom{n+k-1}{k}}=\frac1{k-1}\left(\frac1{(k-1)!}-\frac1{(k-1)!\binom{(m+1)+(k-2)}{k-1}}\right) $$ that is, $$ \sum_{n=1}^m\underbrace{\frac1{n(n+1)\dots(n+k-1)}}_{k\text{ factors}}=\frac1{k-1}\left(\frac1{(k-1)!}-\underbrace{\frac1{(m+1)(m+2)\dots(m+k-1)}}_{k-1\text{ factors}}\right) $$ Substituting $k=3$ gives $$ \sum_{n=1}^m\frac1{n(n+1)(n+2)}=\frac12\left(\frac12-\frac1{(m+1)(m+2)}\right) $$