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In class today I was looking at the sum $1 +\frac{1}{2}+\frac{1}{3}+\frac{1}{4}...$ and with a bit of fiddling, managed to come up with the following: $$\sum_{n=2}^\infty \left(\sum_{m=2}^\infty \frac{1}{n^m} \right) = 1$$ I managed to show this in 2 different ways: Method 1 and Method 2, mostly the same method, yes.

But I was wondering if this is at all meaningful, has it been used in anything at all? Does it mean anything apart from it's something that's pretty cool?

Simon Goodwin
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  • Are you asking for its meaning in physical sense, like how to interpret the result geometrically? – BigbearZzz Aug 30 '16 at 08:07
  • I'm interested in anything regarding it, I'm just generally interested in any interpretation since I happened to stumble across it. – Simon Goodwin Aug 30 '16 at 08:16
  • That result is correct: it is just using standard results in geometric series and telescopic ones. What I don't see clear is what has the harmonic series to do with all this... – DonAntonio Aug 30 '16 at 08:20
  • @DonAntonio The only thing it has to do with the harmonic series is how I came up with it, since I noticed when I expanded the harmonic series using the infinite sum for fractions such as 1 = 1/2 + 1/4 + 1/8... 1/2 = 1/3 + 1/9 + 1/27 ... and 1/3 = 1/4 + 1/16 + 1/64 ... I could expand the series into a new series that had itself less 1 which when cancelled gave me the series above i.e when you take all the expansions and remove the first terms you get what must equal 1. This is what I posted above. – Simon Goodwin Aug 30 '16 at 22:27

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It is one of the several identities involving the Riemann Zeta Function $$\sum_{n=2}^{\infty}(\zeta(n)-1)=1.$$ Here there is another one $$\sum_{n=2}^{\infty}(-1)^n\cdot (\zeta(n)-1)=\frac{1}{2}.$$ You can try to prove it by considering $$\sum_{n=2,4,6,\dots}(\zeta(n)-1)=\frac{3}{4}$$ and $$\sum_{n=3,5,7\dots}(\zeta(n)-1)=\frac{1}{4}.$$

Robert Z
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  • Oh, huh, that's cool, thanks a lot. I think I've only heard of the Riemann Zeta function in numberphile videos and in an old New Zealand Scholarship Calculus paper I did, I'll look into it – Simon Goodwin Aug 30 '16 at 08:18
  • @Simon Goodwin You're welcome. You can try to prove also the identity with $(-1)^n$ by following the hint above. – Robert Z Aug 30 '16 at 08:23