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Identities on the zeta function and harmonic numbers

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Leonard Euler gave a general identity involving the harmonic number H_n and the Riemann zeta function \zeta(n) for natural numbers n.

\displaystyle{2\sum_{n=1}^{\infty}\frac{H_n}{n^m} = (m+2)\zeta(m+1)-\sum_{k=1}^{m-2}\zeta(m-k)\zeta(k+1)}.

A special case of this is the identity

\displaystyle{\sum_{n=1}^{\infty}\frac{H_n}{n^2} = 2\zeta(3)}.

I first saw the above identity in Alex Youcis’s blog Abstract Nonsense and in course of further investigation, I was able to find several identities involving the Riemann zeta function and the harmonic numbers. While it is practically impossible to go through the entire mathematical literature to see if a formula is new or a rediscovery, in this post we shall see a few identities which are not given in the above links. Also if I encounter an identity elsewhere at any point of time, I have taken care to delete it from this post.

For n \ge 1 the following relations hold.

Entry 1.

\displaystyle{\sum_{r=1}^{\infty}H_r \Big\{\frac{1}{r^{2n}} - \frac{1}{(r+1)^{2n}}\Big\}=\zeta(2n+1)}.

Entry 2.

\displaystyle{\sum_{r=1}^{\infty}H_r \Big\{\frac{1}{r^{4n-1}} - \frac{1}{(r+1)^{4n+1}}\Big\}=\zeta(4n)}.

Entry 3.

\displaystyle{\sum_{r=1}^{\infty}H_r \Big\{\frac{1}{r^{4n+1}} - \frac{1}{(r+1)^{4n+1}}\Big\}=\zeta(4n+2)}.

Entry 4.

\displaystyle{\sum_{r=1}^{\infty}\frac{H_r}{(r+1)^{2n}} = n\zeta(2n+1) - \sum_{k=2}^{n}\zeta(k)\zeta(2n+1-k)}.

Entry 5.

\displaystyle{\sum_{r=1}^{\infty}\frac{H_r}{(r+1)^{4n-1}} = \frac{4n-3}{4}\zeta(4n) - \sum_{k=1}^{n-1}\zeta(2k+1)\zeta(4n-2k-1)}.

Entry 6.

\displaystyle{\sum_{r=1}^{\infty}\frac{H_r}{(r+1)^{4n+1}}= \frac{4n-1}{4}\zeta(4n+2)-\frac{\zeta(2n+1)^2}{2}}

\displaystyle{-\sum_{k=1}^{n-1}\zeta(2k+1)\zeta(4n-2k+1)}.

Entry 7.

\displaystyle{\sum_{r=1}^{\infty}\frac{H_r}{(r+2)^{2n}} = n\zeta(2n+1) - 2n - \sum_{k=2}^{n}\zeta(k)\zeta(2n+1-k) + \sum_{k=2}^{2n}\zeta(k)}.

 

 

Entry a. If |x| < 1 then

\displaystyle{\sum_{n=1}^{\infty}\zeta(4n)x^{4n} = \frac{1}{2}-\frac{\pi x}{4}(\cot(\pi x) +\coth(\pi x))}.

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