# Sum of the series

• Apr 26th 2010, 08:06 PM
Capnredbeard
Sum of the series
Not sure where to go with this:
$\displaystyle \sum{3^n\over(1-2x)^n*n!}$

I tried breaking it up into:
$\displaystyle \sum{1\over(1-2x)^n}$ + $\displaystyle \sum{3^n\over n!}$

Then
$\displaystyle \sum{3^n\over n!}$ would become $\displaystyle e^3-1$

And now I'm stuck.
• Apr 26th 2010, 08:12 PM
dwsmith
Quote:

Originally Posted by Capnredbeard
Not sure where to go with this:
$\displaystyle \sum{3^n\over(1-2x)^n*n!}$

I tried breaking it up into:
$\displaystyle \sum{1\over(1-2x)^n}$ + $\displaystyle \sum{3^n\over n!}$

Then
$\displaystyle \sum{3^n\over n!}$ would become $\displaystyle e^3-1$

And now I'm stuck.

How did you break up that summation?
• Apr 26th 2010, 08:26 PM
Capnredbeard
You're right. I can't break it up like that, but I can do this.
$\displaystyle \sum{{{3\over1-2x}^n}*{1\over n!}}$

And $\displaystyle {1\over n!} = e$
• Apr 26th 2010, 08:32 PM
dwsmith
Is there any indexing for this series?
• Apr 26th 2010, 08:33 PM
Failure
Quote:

Originally Posted by Capnredbeard
Not sure where to go with this:
$\displaystyle \sum{3^n\over(1-2x)^n*n!}$

If the summation is indexed by n then you have simply

$\displaystyle \sum_{n=0}^\infty \frac{3^n}{(1-2x)^n n!}=\sum_{n=0}^\infty \frac{1}{n!}\left(\frac{3}{1-2x}\right)^n=\mathrm{e}^{\frac{3}{1-2x}}$
• Apr 26th 2010, 08:34 PM
Capnredbeard
n=0 to infinity