Find the sum of the following series:
1 + 4/7 + 9/49 + 16/343 + .......... to infinity
I have a more general solution of this kind of problems.
Assume that fn(x)=Sum(k^n*x^k), for k=1,..,Inf
dfn(x)/dx=Sum(k^n*k*x^k1)=Sum(k^(n+1)*x^k1)=
=(1/x)*Sum(k^(n+1)*x^k)=(1/x)*fn+1(x)
> fn+1(x)=x*dfn(x)/dx
We alreday know that f0(x)=Sum(x^k)=s/(1s)
So, f1(x)=s/(1s)^2, and f2(x)=s*(1+s)/(1s)^3
We have 1+4/7+9/49+...=7*Sum(k^2*(1/7)^k)=7*f2(1/7)=
=7*(1/7)*(8/7)/(6/7)^3=49/27
I hope you enjoyed it!!!

Posted by George
on 20060321 19:30:42 