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How to find the sum of all distinct values of the form $n^{-m}$, where $n,m\in\mathbb{N}$, $n,m>1$?

I can find the value of such sum: $\sum\limits_{n,m>1}n^{-m}$. It is equal to 1: $$\sum\limits_{n,m>1}n^{-m}=\sum\limits_{n=2}^{\infty}\sum\limits_{m=2}^{\infty}n^{-m}=\sum\limits_{n=2}^{\infty}\dfrac{n^{-2}}{1-n^{-1}}=\sum\limits_{n=2}^{\infty}\left(\dfrac{1}{n-1}-\dfrac{1}{n}\right)=1$$ But the sum from the question should be less than it.

Constructor
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  • Where is this problem from? Do you expect this sum to have a closed form expression? – Bruno Joyal Dec 09 '13 at 20:58
  • @Bruno Joyal I saw only simplified version of it (with summation of all values). So I don't know anything about possibility of its solving. – Constructor Dec 09 '13 at 21:07

1 Answers1

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I kind of doubt that this has a nice closed form, though I could easily be wrong. But here's something kind of neat:

Let the sum be $S$. Let's first just consider the terms when the exponent is a prime, since if the exponent is composite, the terms appear for any exponent which is a prime divisor of that number. Of course, the problem then is that we've double counted all the numbers of the form $n^{-pq}$ as both exponent $p$ and exponent $q$, so we need to subtract these off again. This sounds like inclusion-exclusion, and in fact we get:

$$S = \sum_p \sum_{n=2}^\infty n^{-p} - \sum_{p,q} \sum_{n=2}^\infty n^{-pq} + \sum_{p,q,r} \sum_{n=2}^\infty n^{-pqr} + \cdots$$ where all $p,q$ and $p,q,r$, etc are taken to be pairs, triples, etc of distinct primes.

This is just $$S = -\sum_{n=2}^{\infty} \mu(n)(\zeta(n)-1) \approx 0.87446\ldots$$

universalset
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