Sum of Series: Find the Solution

[mattmannmf]

Find the sum of the series

E (-1)^(n-1) * (n/(2^{n-1}))

By using the power series 1/ (1+x) = E (-1)^n *x^n,

I am unsure of what to do...

Do I take the derivative with respect to x of both sides?

 

[mattmannmf]

now what i did was this:

1/ (1+x) = E (-1)^n *x^n

now i took the derivative with respect to x:

-(1+x)^-2 = E (-1)^n * n*x^(n-1)

Then i substituted x by 1/2 getting:

-(1+(1/2))^-2 = E (-1)^(n-1) * (n/ 2^(n-1))

Does my math look correct? now my sum should be pretty much what i get from this equation:
-(1+x)^-2
but with the substituted 1/2 for x.. right?

 

[Cyosis]

You will need to differentiate but let's rewrite your original problem to something more suggestive.

You can write your original sum as:

-2 \sum_n (-1)^n n \left(\frac{x}{2}\right)^n

Compare this to the given power series.

 

[Cyosis]

Your maths is almost correct. You make a minus sign error going from step 1 to two. Going from (-1)^n to (-1)^(n-1) requires a multiplication by -1 on both sides.

 

[mattmannmf]

oh ok. so my final answer should be:

(1+(1/2))^-2 = E (-1)^(n-1) * (n/ 2^(n-1))

and to find the sum would be to just do the math for:
(1+(1/2))^-2 = .444

 

[Cyosis]

Yes although I would write 4/9 instead.

 

[mattmannmf]

thank you so much!