Number Theory – Generalized Harmonic Numbers and Riemann Zeta Function

co.combinatoricsnt.number-theoryriemann-zeta-functionsequences-and-series

  • The $n$-th harmonic number is defined as
    $$
    H_{n}=\sum\limits_{k=1}^{n}\frac{1}{k},
    $$
    and the generalized harmonic numbers are defined by
    $$
    H_{n}^{(m)}=\sum\limits_{k=1}^{n}\frac{1}{k^m}.
    $$
  • It is known result that
    $$
    \sum\limits_{n=1}^{\infty}\frac{H_{n}}{n^3}=\frac{5}{4}\zeta(4).
    $$
    Here $\zeta(n)$ denotes Riemann zeta function.

I conjecture that
$$
\sum\limits_{n=1}^{\infty}\frac{H_{n}^{(2)}}{n^2}=\frac{7}{4}\zeta(4).
$$

Is there a way to prove it?

Best Answer

This is not a proof.

$$S_p=\sum\limits_{n=1}^{p}\frac{H_{n}^{(2)}}{n^2}$$ $$S_p=\frac{\left(H_p^{(2)}\right){}^2}{2}-\frac{H_p^{(4)}}{2}-\frac{\psi ^{(3)}(p+1)}{6}+\frac{\pi ^4}{90}$$ Expanded as series $$S_p=\frac{7 \pi ^4}{360}-\frac{\pi ^2}{6 p}+\frac{6+\pi ^2}{12 p^2}+O\left(\frac{1}{p^3}\right)$$

Now, if you consider $$T_k=\sum\limits_{n=1}^{\infty}\frac{H_{n}^{(2k)}}{n^{2k}}$$ you generate

$$T_k= a_k\, \zeta(4k)$$ where the coefficients are $$\left\{\frac{7}{4},\frac{13}{12},\frac{703}{691},\frac{14527}{144 68},\frac{524354}{523833},\frac{3546333857}{3545461365},\frac{ 6785975897}{6785560294},\frac{30837755428255}{30837284164868} \right\}$$

As @Notamathematician commented, the numerators correspon to sequence $A348830$ in $OEIS$. Sequence $A348829$ gives the difference between numerator and denominator.

Edit

Interesting is the odd case

$$U_k=\sum\limits_{n=1}^{\infty}\frac{H_{n}^{(2k+1)}}{n^{2k+1}}=\frac 12 \big(\zeta(2k+1)\big)^2+ \,\zeta(2(2k+1)\big) $$

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