- View SourcePS: The proof is elementary

sum_{n>1,m>1} 1/n^m = sum_{n>1} 1/(n*(n-1))

= sum_{n>1} (1/(n-1) - 1/n) = 1/(2-1) = 1

Numerical check:

? \p60

realprecision = 67 significant digits (60 digits displayed)

? print(suminf(n=2,zeta(n)-1))

1.00000000000000000000000000000000000000000000000000000000000