Help in finding the sum of the series!

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June 8th 2008, 07:07 PM
Vedicmaths

Help in finding the sum of the series!

Find the sum of: -

A) (-2/3)^k when K goes from 1 to infinity...I got (3/5)!

B) 2^k / [3^(2k+1) when k goes from 0 to infinity...I got (9/7)!

c) 1 / k(k+1)...when K goes from 1 to infinity...I got (1)

d) 1 / k(k+2)...when K goes from 1 to infinity...I am not sure how to split this term in two different term.

I have got the solution for starting three, but I am not sure if I have done it right or not. I used the geometric series formula (1/1-r) in order to solve them.
Please help me making it sure!

Thanks!
June 8th 2008, 07:24 PM
TheEmptySet

Quote:

Originally Posted by Vedicmaths

Find the sum of: -

A) (-2/3)^k when K goes from 1 to infinity...I got (3/5)!

B) 2^k / [3^(2k+1) when k goes from 0 to infinity...I got (9/7)!

c) 1 / k(k+1)...when K goes from 1 to infinity...I got (1)

d) 1 / k(k+2)...when K goes from 1 to infinity...I am not sure how to split this term in two different term.

I have got the solution for starting three, but I am not sure if I have done it right or not. I used the geometric series formula (1/1-r) in order to solve them.
Please help me making it sure!

Thanks!

For d) you can use partial fraction decomp with this form

$\frac{1}{k(k+2)}=\frac{A}{k}+\frac{B}{K+2}$

Or note that

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

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

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

writing out a few terms of each of the sums gives

$\frac{1}{2} \left( 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...\right)$
$-\frac{1}{2}\left(\frac{1}{3}+\frac{1}{4}+...\right )$

by telescoping we get

$\frac{1}{2}\left( 1+\frac{1}{2}\right)=\frac{3}{4}$
June 9th 2008, 03:30 AM
Moo

Hi !

Quote:

B) 2^k / [3^(2k+1) when k goes from 0 to infinity...I got (9/7)!

Not quite :D

$\frac{2^k}{3^{2k+1}}=\frac 13 \cdot \frac{2^k}{3^{2k}}=\frac 13 \cdot \frac{2^k}{9^k}$

Just to note that you forgot $\frac 13$ (Wink)