Evaluating the Series n^2/2^n a Differential Way

Date: 11/05/2010 at 11:56:38
From: Lathish
Subject: Value of Summation of  Series to infinity

How do you go about finding a solution to the following series?

   sigma(n^2/2^n, n = 0, ... infinity)

I know the series converges, but I am not able to find its limit.

Date: 11/05/2010 at 14:21:53
From: Doctor Ali
Subject: Re: Value of Summation of  Series to infinity

Hi Lathish!

Thanks for writing to Dr. Math.

Actually, this question has several solutions. Let me explain my favorite
one.

Let's assume that

     |x| < 1

Then, we'll have

     INF
     ---               1
     )   x^n      = -------
     ---             1 - x
     n=0

Does that make sense?

Now let's take the derivative of both sides with respect to x. We'll have

     INF
     ---    n-1         1
     )   n x      = -----------
     ---             (1 - x)^2
     n=0

We can multiply the sides by x to get rid of (n - 1) in the power of x.
Then we'll have

     INF
     ---    n           x
     )   n x      = -----------
     ---             (1 - x)^2
     n=0

Here again, we can take a derivative with respect to x and get

     INF
     ---  2  n-1      x + 1
     )   n  x     = -----------
     ---             (1 - x)^3
     n=0

If we multiply the sides by x a second time, out comes your original summation:

     INF
     ---  2  n       x (x + 1)
     )   n  x     = -----------
     ---             (1 - x)^3
     n=0

Now, all you need to do is to substitute 1/2 for x to arrive at the answer.

As you know, you can continue doing this to find the following summation
in general:

     INF
     ---  k  n
     )   n  x   = ...
     ---
     n=0

Does that make sense?

Please write back if you still have any difficulties.

- Doctor Ali, The Math Forum
  http://mathforum.org/dr.math/