Re: Sum With Harmonic #s and Mobius Funct: Puzzle
- From: "Leroy Quet" <qqquet@xxxxxxxxxxxxxx>
- Date: 17 Oct 2005 11:36:50 -0700
Leroy Quet wrote:
> Here is a math puzzle with an unexpectedly simple answer, given the
> relative strangeness of the infinite sum.
>
> First, a couple of definitions, then the problem.
>
> H(k) = sum{j=1 to k}1/j, the kth harmonic number.
> H(x) can be defined for non-integer x as:
> H(x) = Euler's constant + the derivative of ln(Gamma(x+1)).
>
> mu(k) is the Moebius(Mobius) function, defined by
> sum{k>=1} mu(k)/k^r = 1/zeta(r).
>
>
> The problem:
>
> Find a closed form for
>
> --- ---
> \ \ 2
> / / (H(k +x/m) - H(k)) mu(m)/m,
> --- ---
> m>=1 k>=0
>
> for those x's where the double sum converges.
>
>
> The sum in linear-mode:
>
> sum{m>=1} sum{k>=0} (H(k +x/m) - H(k))^2 mu(m)/m
>
>
> I will give my solution (which may be wrong) in a few days if no one
> answers before then.
>
>
> I do not know for which x's the double sum converges. (It could be for
> all x where |x|<1, or it could be for all x not = to a negative
> integer. Or it could be something else.)
> If someone wants to reply with the range of x's where the sum
> converges, even if they have not found the closed form of the sum, that
> would be a good thing.
>
> thanks,
> Leroy Quet
Where things converge, I get:
sum{m>=1} sum{k>=0} (H(k +x/m) - H(k))^2 mu(m)/m
= (3*x^2 + 2*x^3)/(1+x)^2.
Very rough overview of proof.
The above result can be found from the result:
sum{k>=0} sum{j>=1} (H(k +x/j) - H(k))^2 * a(j) =
sum{k>=1} (sum{j>=1} a(j)/j^(k+1)) (-x)^(k+1) (k+2) zeta(k+2),
for {a(k)}'s and x's where the sum converges (absolutely, just to be
safe).
This result, in turn, can be found from the following result,
which I have written about recently both on sci.math and rec.puzzles:
For n>=1,
(n+2)*zeta(n+2) =
sum{k>=1} sum{j=1 to n} zeta(j+1,k) zeta(n+2-j,k),
where zeta(y,x) = sum{j>=0} 1/(j+x)^y, the Hurwitz zeta function.
(And zeta(n) = zeta(n,1), of course.)
thanks,
Leroy Quet
.
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