# Thread: Sum of a series

1. ## Sum of a series

Find the sum of the series:

$\sum^{\infty}_{n=0} (-1)^n \frac{2^{n+1}x^{6n}}{n!}$

$= \sum^{\infty}_{n=0} \frac{(-2^nx^6)^n}{n!}$

$= e^{-2^nx^6}$

Where am I wrong? Do I need an index shift because of the $2^{n+1}$ ?

2. Originally Posted by mollymcf2009
Find the sum of the series:

$\sum^{\infty}_{n=0} (-1)^n \frac{2^{n+1}x^{6n}}{n!}$

$= \sum^{\infty}_{n=0} \frac{(-2^nx^6)^n}{n!}$

$= e^{-2^nx^6}$

Where am I wrong? Do I need an index shift because of the $2^{n+1}$ ?

$2 \sum^{\infty}_{n=0} (-1)^n \frac{2^n x^{6n}}{n!}$

3. Also you need

(-2*x^6)^n

not (-2^n x^6)^n

you get 2e^(-2x^6)

4. Originally Posted by skeeter
$2 \sum^{\infty}_{n=0} (-1)^n \frac{2^n x^{6n}}{n!}$
Ok, here is what I did, but I'm not sure it's right:

$2 \sum^{\infty}_{n=0} \frac{(-2x^6)^n}{n!}$

When do I multiply that 2 back in, do I just put it in front of the e or does it get multiplied back into the argument before I take the e?

$e^{-4x^6}$ or is it $2e^{-2x^6}$

So sum is:

$\frac{1}{4x^6}$ or $\frac{1}{x^6}$

Also, am I supposed to have an x in my answer? That doesn't seem like an "answer".

Thanks!!!!

5. No the sum of the series is period.

Since sum(x^n/n!) = e^x and you are substituting (-2x^6) for x

The sum is a Taylor Series so it better have an x in it.

What it can't have is n which is what you're summing over.

not sure where you're getting 1/x^6 or 1/(4*x^6) from but they have nothing to do with the series.

6. the sum of the original series is $2e^{-2x^6}$

where/how/why are you getting $\frac{1}{4x^6}$ or $\frac{1}{x^6}$ ?

is there some other info you haven't posted?

7. Originally Posted by Calculus26
No the sum of the series is period.

Since sum(x^n/n!) = e^x and you are substituting (-2x^6) for x

The sum is a Taylor Series so it better have an x in it.

What it can't have is n which is what you're summing over.

not sure where you're getting 1/x^6 or 1/(4*x^6) from but they have nothing to do with the series.
Ugggg...totally see what I did...lol!

My brain is doing:

$e^{-ln(4x^6)}$

Now see where I'm getting my crazy answer?

Lordy, lordy....it's gonna be a long night...

Thanks y'all!

8. You're alright-- I thought as much but coudn't figure out why you were taking logarithms .

Stream of consciousness often finds its way into my calculations--thats why I never became an engineer-- at least with math things don't blow up