Evaluating the Series n^2/2^n a Differential WayDate: 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/ |
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