# Thread: summation problem; find closed-form

1. ## summation problem; find closed-form

Hello all:

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

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

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

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

2. $\displaystyle \displaystyle \sum_{r=1}^n\frac{2-r}{r(r+1)(r+2)}=\sum_{r=1}^n\left(\frac{2}{r+2}-\frac{3}{r+1}+\frac{1}{r}\right)=$
$\displaystyle \displaystyle =\frac{2}{3}-\frac{3}{2}+\frac{1}{1}+$
$\displaystyle \displaystyle +\frac{2}{4}-\frac{3}{3}+\frac{1}{2}+$
$\displaystyle \displaystyle +\frac{2}{5}-\frac{3}{4}+\frac{1}{3}+$
.................................................. ........................
$\displaystyle \displaystyle +\frac{2}{n}-\frac{3}{n-1}+\frac{1}{n-2}+$
$\displaystyle \displaystyle +\frac{2}{n+1}-\frac{3}{n}+\frac{1}{n-1}+$
$\displaystyle \displaystyle +\frac{2}{n+2}-\frac{3}{n+1}+\frac{1}{n}=$
$\displaystyle \displaystyle =\frac{2}{n+1}+\frac{2}{n+2}-\frac{3}{n+1}-\frac{3}{2}+1+\frac{1}{2}=$
$\displaystyle \displaystyle =\frac{2}{n+2}-\frac{1}{n+1}-1+1=$
$\displaystyle \displaystyle =\frac{2}{n+2}-\frac{1}{n+1}$

3. You're a good egg, Red Dog. I was on the same track as you but went off ona tangent. I see.

4. i ham a good egg

5. Originally Posted by galactus
$\displaystyle \frac{2-r}{r(r+1)(r+2)}=\frac{2}{r+2}-\frac{3}{r+1}+\frac{1}{r}$, show that $\displaystyle \sum_{r=1}^{n}\frac{2-r}{r(r+1)(r+2)}=\frac{2}{n+2}-\frac{1}{n+1}$
We have that
$\displaystyle \sum_{r = 1}^n \frac{1}{r + 1} = \sum_{r = 1}^{n} \frac{1}{r} - 1 + \frac{1}{n + 1}$

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

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

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

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

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

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

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

-Dan

6. Originally Posted by galactus
Hello all:

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

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

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

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

RonL

7. Originally Posted by CaptainBlank
I have a nice method that gives the sum in terms of hypergeometric functions
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.

8. 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.

9. 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.

10. Define the functions:

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

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

But:

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

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

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 $\displaystyle k$ and $\displaystyle x=1$ that we need, so that is my lot for the weekend at least.

RonL