Proof that the sum of all series 1/n^m, (n>1,m>1) =1?

[BWV]

Curious about proving that $\sum_{m=2}^\infty \sum_{n=2}^\infty 1/n^m$ = 1

ran this in Matlab and n,m to 2:1000 =0.9990, and n,m 2:10000 =0.9999, so it does appear to converge to 1

 

[mjc123]

What is the sum of 1/n² + 1/n³ + 1/n⁴...?
What is the sum of these sums from n = 2 to ∞?

 

[BWV]

OK, playing with Wolfram Alpha, the series reduces to $\sum_{n=2}^\infty (n(n-1))^{-1}$, which does sum to 1

 

[mjc123]

Playing with Wolfram Alpha? You're not familiar with the result
a(1+r+r²...) = a/(1-r) ?

 

[Hawkeye18]

OK, playing with Wolfram Alpha, the series reduces to $\sum_{n=2}^\infty (n(n-1))^{-1}$, which does sum to 1

The series $\sum_{n=2}^\infty (n(n-1))^{-1} = \sum_{n=2}^\infty \left( \frac{1}{n-1} - \frac1n \right)$ does sum to 1 because $\sum_{n=2}^N \left( \frac{1}{n-1} - \frac1n \right) =1-1/N$

 

[BWV]

Yes, for some reason it took me a while to see it as a geometric series instead of a p-series