DivGradCurl
Nov2-04, 01:48 PM
I need to obtain the sum of the following series
\sum _{n=1} ^{\infty} \frac{n^2}{2^n}
Well, with the aid of Mathematica, I get the answer, which is 6. What I'm trying to do now is work my way backwards from there. I need to express it through the geometric series
\sum _{n=0} ^{\infty} x^n = \frac{1}{1-x}
In fact, my guess is that I should use
\sum _{n=1} ^{\infty} n x^{n-1} = \frac{1}{\left( x - 1 \right) ^2}
but it can't get it to fit in.
Thank you very much.
\sum _{n=1} ^{\infty} \frac{n^2}{2^n}
Well, with the aid of Mathematica, I get the answer, which is 6. What I'm trying to do now is work my way backwards from there. I need to express it through the geometric series
\sum _{n=0} ^{\infty} x^n = \frac{1}{1-x}
In fact, my guess is that I should use
\sum _{n=1} ^{\infty} n x^{n-1} = \frac{1}{\left( x - 1 \right) ^2}
but it can't get it to fit in.
Thank you very much.