# just wanna add another one

• Mar 29th 2010, 08:39 PM
wopashui
just wanna add another one
$\sum\limits_{k = 1}^\infty$ $1/[2k(k+1)]$

$\sum\limits_{k = 1}^\infty$ $3^{k-1}/4^{3k+1}$

find the sum of the series

I got the last one is 61/ (12*64), but doubt the answer, just want to confirm it
• Mar 29th 2010, 08:43 PM
mr fantastic
Quote:

Originally Posted by wopashui
$\sum\limits_{k = 1}^\infty$ $1/[2k(k+1)]$

find the sum of the series

It's a telescoping series. Express $\frac{1}{k (k+1)}$ in partial fraction form.

(You should have learned something from your thread here and compared it to the first question you asked: http://www.mathhelpforum.com/math-he...um-series.html)
• Mar 30th 2010, 03:31 AM
HallsofIvy
Quote:

Originally Posted by wopashui
$\sum\limits_{k = 1}^\infty$ $1/[2k(k+1)]$

$\sum\limits_{k = 1}^\infty$ $3^{k-1}/4^{3k+1}$

find the sum of the series

I got the last one is 61/ (12*64), but doubt the answer, just want to confirm it

$\frac{3^{k-1}}{4^{3k+1}}= \frac{3^k(3^{-1}}{4^{3k}(4)}$ $= \frac{1}{12}\frac{3^k}{(4^3)^k}= \frac{1}{12}\left(\frac{3}{64}\right)^k$ so the second is a geometric series with first term $\frac{1}{12}$ and common ratio $\frac{3}{64}$.

I think you have a fraction upside down.