# Prove the following identity

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.

-
The identity you're trying to prove should have $m$'s in it. – Christopher A. Wong Jan 15 '13 at 23:10
Oops. Sorry!! My bad. – user45099 Jan 15 '13 at 23:11
The summation is from $n = 1$ to $n = m$, so the right-hand side surely has $m$'s, not $n$'s. – Christopher A. Wong Jan 15 '13 at 23:13
@ChristopherA.Wong Thank You. – user45099 Jan 15 '13 at 23:14

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*}

-

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)

-
Thanks. Looks shorter. – user45099 Jan 15 '13 at 23:30
@user1709828: welcome! – user 1618033 Jan 15 '13 at 23:32
I was just going to post that $$\frac1{n(n+1)(n+2)}=\frac12\left(\frac1{n(n+1)}-\frac1{(n+1)(n+2)}\right)$$ Too late... – robjohn Jan 16 '13 at 0:00
@robjohn: hehe. :-) – user 1618033 Jan 16 '13 at 0:01
It forced me to write a more general answer :-) – robjohn Jan 16 '13 at 0:43

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 $$\sum_{n=1}^m\frac1{n(n+1)(n+2)}=\frac12\left(\frac12-\frac1{(m+1)(m+2)}\right)$$

-
This almost qualifies as a swatting flies with a sledgehammer answer. – Rick Decker Jan 16 '13 at 2:04
@RickDecker: It's simply a generalization of Chris's Sister's answer. As such, it does pack a bit more power, but it is not that complicated. – robjohn Jan 16 '13 at 2:26
@robjohn: nice. A generalization is welcome! (+1) – user 1618033 Jan 16 '13 at 8:46

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...

-
I don't get it: what is "totally redundant"? – DonAntonio Jan 15 '13 at 23:26
The partial fraction computation was not necessary. It's already done in question so not needed and Brian M. Scott already had the remaining portion. Never mind!. It is self contained and good. Thank You. – user45099 Jan 15 '13 at 23:29
You realize that nobody here can't look what other people do while writing an answer, so it may happen with very basic questions that there are two or more answers that appeared within a few minutes from each other and very similar...you're not a newbie in this site, you should know this. – DonAntonio Jan 16 '13 at 2:54