[seqfan] Re: Sum of expression with harmonic numbers should simplify to Zeta?

[Rob Pratt]

Rewriting the FullSimplify result as

Sum[(1/k^n - 1/(k + 1)^n) HarmonicNumber[k], {k, 1, Infinity}]

and applying summation by parts yields

Sum[(1/k^n) (HarmonicNumber[k] - HarmonicNumber[k-1]), {k, 1, Infinity}]

= Sum[(1/k^n) (1/k), {k, 1, Infinity}]

= Zeta[n+1]

-----Original Message-----
From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of Jean-François Alcover
Sent: Friday, March 29, 2013 9:37 AM
To: Sequence Fanatics Discussion list
Subject: [seqfan] Sum of expression with harmonic numbers should simplify to Zeta?

Dear Seqfans,

Thanks in advance to the (kind) specialist who could explain how can be proved that this expression:

Sum[ HarmonicNumber[k]/k^n - (PolyGamma[k+1] + EulerGamma)/(k+1)^n, {k, 1, Infinity}]

always simplifies to Zeta[n+1].

J.-F. Alcover

P.s.
FullSimplify returns:
Sum[ (-(k*(1+k))^(-n))*(k^n - (1+k)^n)*HarmonicNumber[k], {k, 1, Infinity}]

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