Find the sum of the following series:

1 + 4/7 + 9/49 + 16/343 + .......... to infinity

We want to find the sum S/7= (1/7 + 2^2/7^2 + 3^2/7^3 + ... + n^2/7^n + ...) Let's generalize and change the 7 to K.

Let's write A= K/(K-1)= 1 + 1/K + 1/K^2 + 1/K^3 + ... + 1/K^n + ... and since we have B= A^2= 1 + 2/K + 3/K^2 + 4/K^3 + ... + (n+1)/K^n + ... and C= A^3= 1 + 3/K + 6/K^2 + 10/K^3 + ... + (n+1)(n+2)/2K^n + ..., then 2C-3B+A equals S/K.

As in this case K=7, S/7= 7/27, so S=49/27.

Posted by Old Original Oskar! on 2006-03-11 22:14:24