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^k-1)=Sum(k^(n+1)*x^k-1)=

=(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/(1-s)

So, f1(x)=s/(1-s)^2, and f2(x)=s*(1+s)/(1-s)^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!!!