6

$$\sum \limits_{n=1}^\infty \left(\frac1n{\sum \limits_{k=1}^n \frac1k}\right)^3$$

I want to see what this converges to. Obviously it converges on inspection.

Any suggestions on methods to use?

Did
  • 279,727
Joshua A
  • 121
  • 8

2 Answers2

5

$$S:=\sum_{n=1}^\infty\frac{(H_n)^3}{n^3}$$ is convergent since the terms are equivalent to $\,\left(\dfrac{\ln(n)}n\right)^3\,$ as $\;n\to \infty\,$ and given by $$S=\frac {31}{5040}\pi^6-\frac 52\zeta(3)^2\\\approx 2.3009545517005250398$$ I don't know a complete proof but we may start with the following class of Euler sums : $$s_h(m,n):=\sum_{k=1}^\infty\frac{(H_k)^m}{(k+1)^n}$$ ($H_k$ is the $k$-th harmonic number)

and rewrite your sum as : \begin{align} S&=1+\sum_{k=1}^\infty\frac{(H_{k+1})^3}{(k+1)^3}\\ &=1+\sum_{k=1}^\infty\frac{\left(H_k+\frac 1{k+1}\right)^3}{(k+1)^3}\\ &=1+s_h(3,3)+3\,s_h(2,4)+3\,s_h(1,5)+\zeta(6)-1\\ \end{align} Using the expressions tabulated in the $1994$ paper from Bailey, Borwein and Girgensohn "Experimental Evaluation of Euler Sums" this becomes : $$S=\left(-\frac{33}{16}\zeta(6)+2\,\zeta(3)^2\right)+3\,\left(\frac 23\zeta(6)-\frac 13\zeta(2)\zeta(4)+\frac 13\zeta(2)^3-\zeta(3)^2\right)\\+3\,\left(\frac 52\,\zeta(6)-\frac 12(2\,\zeta(4)\zeta(2)+\zeta(3)^2)\right)+\zeta(6)$$

$s_h(3,3)$ was only an experimental result when the paper initially appeared but proved since them (as indicated in the updated paper). Anyway a more direct derivation should remain of interest !

Proofs concerning $\;\displaystyle \sum_{n=1}^\infty\frac{(H_n)^2}{n^2}\;$ are provided here.

Community
  • 1
Raymond Manzoni
  • 43,021
  • 5
  • 86
  • 140
  • Brilliant!!!!!! – Joshua A Nov 04 '14 at 09:22
  • 1
    Thanks @900sit-upsaday ! (even if the hard work belongs to others...). I computed the next series of the kind $;\displaystyle S_m:=\sum_{n=1}^\infty\frac{(H_n)^m}{n^m};$ (i.e. $S_4$ and $S_5$) to $200$ digits but it seems that the result is more complicated than simple sums of products of $\zeta(\text{integer}),$ (to order $2,m$). Fine continuation anyway, – Raymond Manzoni Nov 04 '14 at 23:11
1

It converges quite quickly, so using a computer can give you the sum.

For example in R

n <- 1:2^20
H <- cumsum(1/n)
S <- cumsum((H/n)^3)
S[2^20]

gives about 2.300955 though I suspect it is in fact just over $2.30095455$.

Alternatively use a good expansion of the Harmonic numbers, but I doubt this would be much more informative.

Henry
  • 157,058
  • Any idea of what it could be ? If the exponent is $2$ instead of $3$, the result would have been the beautiful $\frac{17 \pi ^4}{360}$ which is very close to twice your result ! – Claude Leibovici Oct 25 '14 at 11:38