# [SOLVED] Infinite Sum

• Apr 8th 2010, 07:40 PM
Haven
[SOLVED] Infinite Sum
Prove that:

$\sum_{i=4}^{\infty} \frac{2}{i(i-1)(i-2)(i-3)} = \frac{1}{3}$

I think I'm supposed to use the fact that any decreasing sequence $a_1 \dots a_n$ can be written as:
$a_n-a_1 = (a_n - a_{n+1}) + \dots (a_3 - a_2) + (a_2 - a_1)$
• Apr 8th 2010, 08:25 PM
Failure
Quote:

Originally Posted by Haven
Prove that:

$\sum_{i=4}^{\infty} \frac{2}{i(i-1)(i-2)(i-3)} = \frac{1}{3}$

I think I'm supposed to use the fact that any decreasing sequence $a_1 \dots a_n$ can be written as:
$a_n-a_1 = (a_n - a_{n+1}) + \dots (a_3 - a_2) + (a_2 - a_1)$

Well, something like that, only a little more complicated I think. To get started, you can use a "partial fraction decomposition" like this:

$\sum_{i=4}^\infty \frac{2}{i(i-1)(i-2)(i-3)}=\sum_{i=4}^\infty\left({\color{blue}\frac{1}{3 }\cdot\frac{1}{i-3}-\frac{1}{3}\cdot\frac{1}{i}}+{\color{red}\frac{1}{ i-1}-\frac{1}{i-2}}\right)$
Now determine the sum of the blue and red parts separately.
• Apr 8th 2010, 08:40 PM
Haven
EDIT: nevermind i figured it out
• Apr 8th 2010, 09:08 PM
Drexel28
Quote:

Originally Posted by Haven
Prove that:

$\sum_{i=4}^{\infty} \frac{2}{i(i-1)(i-2)(i-3)} = \frac{1}{3}$

I think I'm supposed to use the fact that any decreasing sequence $a_1 \dots a_n$ can be written as:
$a_n-a_1 = (a_n - a_{n+1}) + \dots (a_3 - a_2) + (a_2 - a_1)$

There is a much more general way to solve these kind. I.e. $\sum_{n=0}^{\infty}\frac{1}{(n+1)\cdots(n+\ell)}=\ frac{1}{\ell(\ell-1)(\ell-1)!}$. Ask to see a proof if you're interested.