summation problem; find closed-form

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June 29th 2007, 04:21 AM
galactus

summation problem; find closed-form

Hello all:

I seen this summation problem and was wondering what insight the learned on this site have:

Given $\frac{2-r}{r(r+1)(r+2)}=\frac{2}{r+2}-\frac{3}{r+1}+\frac{1}{r}$ , show that $\sum_{r=1}^{n}\frac{2-r}{r(r+1)(r+2)}=\frac{2}{n+2}-\frac{1}{n+1}$

I ran this through Maple and got $\frac{2}{n+2}-\frac{1}{n+1}$

but how is that derived?. I tried partial sums but hit a wall. It may be easier than I am making it.
June 29th 2007, 04:35 AM
red_dog

$\displaystyle \sum_{r=1}^n\frac{2-r}{r(r+1)(r+2)}=\sum_{r=1}^n\left(\frac{2}{r+2}-\frac{3}{r+1}+\frac{1}{r}\right)=$
$\displaystyle =\frac{2}{3}-\frac{3}{2}+\frac{1}{1}+$
$\displaystyle +\frac{2}{4}-\frac{3}{3}+\frac{1}{2}+$
$\displaystyle +\frac{2}{5}-\frac{3}{4}+\frac{1}{3}+$
.................................................. ........................
$\displaystyle +\frac{2}{n}-\frac{3}{n-1}+\frac{1}{n-2}+$
$\displaystyle +\frac{2}{n+1}-\frac{3}{n}+\frac{1}{n-1}+$
$\displaystyle +\frac{2}{n+2}-\frac{3}{n+1}+\frac{1}{n}=$
$\displaystyle =\frac{2}{n+1}+\frac{2}{n+2}-\frac{3}{n+1}-\frac{3}{2}+1+\frac{1}{2}=$
$\displaystyle =\frac{2}{n+2}-\frac{1}{n+1}-1+1=$
$\displaystyle =\frac{2}{n+2}-\frac{1}{n+1}$
June 29th 2007, 04:43 AM
galactus

You're a good egg, Red Dog. I was on the same track as you but went off ona tangent. I see.
June 29th 2007, 08:24 AM
MathGuru

i ham a good egg
June 29th 2007, 09:01 AM
topsquark

Quote:

Originally Posted by galactus

$\frac{2-r}{r(r+1)(r+2)}=\frac{2}{r+2}-\frac{3}{r+1}+\frac{1}{r}$ , show that $\sum_{r=1}^{n}\frac{2-r}{r(r+1)(r+2)}=\frac{2}{n+2}-\frac{1}{n+1}$

We have that
$\sum_{r = 1}^n \frac{1}{r + 1} = \sum_{r = 1}^{n} \frac{1}{r} - 1 + \frac{1}{n + 1}$

and
$\sum_{r = 1}^n \frac{1}{r + 2} = \sum_{r = 1}^{n} \frac{1}{r} - 1 - \frac{1}{2} + \frac{1}{n + 1} + \frac{1}{n + 2}$

So....
$\sum_{r=1}^{n}\frac{2-r}{r(r+1)(r+2)}$

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

$= 2 \sum_{r=1}^{n} \frac{1}{r+2} - 3 \sum_{r=1}^{n} \frac{1}{r+1} + \sum_{r=1}^{n} \frac{1}{r}$

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

$= 2 \sum_{r = 1}^{n} \frac{1}{r} - 2 - 1 + \frac{2}{n + 1} + \frac{2}{n + 2} - 3 \sum_{r = 1}^{n} \frac{1}{r} + 3 - \frac{3}{n + 1} + \sum_{r=1}^{n} \frac{1}{r}$

Most of the terms cancel and we are left with
$\sum_{r=1}^{n}\frac{2-r}{r(r+1)(r+2)} = \frac{2}{n + 2} - \frac{1}{n + 1}$

-Dan
June 29th 2007, 10:51 AM
CaptainBlack

Quote:

Originally Posted by galactus

Hello all:

I seen this summation problem and was wondering what insight the learned on this site have:

Given $\frac{2-r}{r(r+1)(r+2)}=\frac{2}{r+2}-\frac{3}{r+1}+\frac{1}{r}$ , show that $\sum_{r=1}^{n}\frac{2-r}{r(r+1)(r+2)}=\frac{2}{n+2}-\frac{1}{n+1}$

I ran this through Maple and got $\frac{2}{n+2}-\frac{1}{n+1}$

but how is that derived?. I tried partial sums but hit a wall. It may be easier than I am making it.

I have a nice method that gives the sum in terms of hypergeometric functions:cool:

RonL
June 29th 2007, 11:06 AM
ThePerfectHacker

Quote:

Originally Posted by CaptainBlank

I have a nice method that gives the sum in terms of hypergeometric functions:cool:

That does sound cool!

My question is where are these "hypergeometric series" used? I keep on hearing and hearing their names but not once I used them.
June 29th 2007, 11:50 AM
galactus

Yes, CB, would you share them with us. I always like learning something new with math. The thing about math, one can always learn something new.
June 29th 2007, 12:28 PM
ThePerfectHacker

Quote:

Originally Posted by galactus

The thing about math, one can always learn something new.

Except for Gauss. I am sure I mentioned this before. One book I had said Gauss was the last mathematician to know everything.

It might seem that is a bad thing, but I would rather know everything (I plan to in a few years) rather than learn new things.
June 29th 2007, 02:55 PM
CaptainBlack

Define the functions:

$ f_{k,n}(x)=\sum_{r=1}^n \frac{x^{r+k}}{r+k} $

Then the sum in question can be written in terms of $f_{0,n}(1)$ , $f_{1,n}(1)$ , and $f_{2,n}(1)$ .

But:

$ f'_{k,n}(x)=\sum_{r=1}^n x^{r+k-1}=\frac{x^k(1-x^n)}{1-x} $
so:

$ f_{k,n}(x)=\int_0^x \frac{\zeta^k(1-\zeta^n)}{1-\zeta} d\zeta $

Which can be written in terms of the hypergeometric functions, but my copy of Abramowitz and Stegun is at work
so I can't check which bits of the formal answer that I have in my note book are valid for the values of $k$ and $x=1$ that we need, so that is my lot for the weekend at least:).

RonL