
Stanton D. answered 07/28/20
Tutor to Pique Your Sciences Interest
Hi Lily S.,
The usual way of showing this is by ratioing the series to itself (using self-similarity), and demonstrating that the sum can be so calculated. So:
let P = sum (n=1 to infinity) of 1/n^k
then k*P = 1 + sum(n=1 to infinity) of 1/n^k (the whole series has been "shifted" by one term)
subtract: P(k-1) = 1
P = 1/(k-1) which exists and is finite for k>1
Right?
-- Cheers, -- Mr. d.