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

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  • #1
BWV
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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
 

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  • #2
mjc123
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What is the sum of 1/n2 + 1/n3 + 1/n4...?
What is the sum of these sums from n = 2 to ∞?
 
  • #3
BWV
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OK, playing with Wolfram Alpha, the series reduces to ##\sum_{n=2}^\infty (n(n-1))^{-1}##, which does sum to 1
 
  • #4
mjc123
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Playing with Wolfram Alpha? You're not familiar with the result
a(1+r+r2...) = a/(1-r) ?
 
  • #5
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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##
 
  • #6
BWV
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Yes, for some reason it took me a while to see it as a geometric series instead of a p-series
 

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