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 Post subject: Fun with summations (1)
PostPosted: Thu, 25 Oct 2007 17:53:54 UTC 
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Evaluate the following:

$\sum_{n=1}^{\infty}{(-1)^{\lfloor 2^nx\rfloor}\over 2^n} for 0<x<1 and \lfloor y\rfloor is the greatest integer function.

$\sum_{n=1}^{\infty}{s(n)\over n(n+1)} where s(n) is the number of 1s in the binary expansion of n.

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PostPosted: Sun, 4 Jan 2009 16:12:30 UTC 
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I recognise the first one at sight...
Spoiler:
The (-1)^(...) suggests the Haar wavelets... and once you get that it isn't long to get 1-2x as the function with the appropriate "Fourier" coefficients. Finally, right-continuity together with the values at diadic rationals force 1-2x as the required sum.


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PostPosted: Tue, 6 Jan 2009 15:11:37 UTC 
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And for the second one:
Spoiler:
Convert to double sum by splitting up s(n) into places where there are 1, then swap the order of sum (since all summands are nonnegative), so the contribution from the 2^k-place 1s evaluates to (1/2^k)log(2), from 1/(n(n+1))=1/n-1/(n+1) and the well known alternating series 1-(1/2)+(1/3)-(1/4)+...=log(2). So the result is 2*log(2).


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