# Thread: Sum of the series

1. ## 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.

2. 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?

3. 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 $\displaystyle {1\over n!} = e$

4. Is there any indexing for this series?

5. 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

$\displaystyle \sum_{n=0}^\infty \frac{3^n}{(1-2x)^n n!}=\sum_{n=0}^\infty \frac{1}{n!}\left(\frac{3}{1-2x}\right)^n=\mathrm{e}^{\frac{3}{1-2x}}$

6. n=0 to infinity