HARMONIC NUMBERS AND HARMONIC SUMS

Harmonic numbers

Harmonic numbers are defined as

$H_{0}=0$

and

$H_{n}=\sum_{k=1}^{n}\frac{1}{k}$

We can derive a recurrence relation to $H_{n}$ , namely

$H_{n}=\sum_{k=1}^{n-1}\frac{1}{k}+\frac{1}{n}$

$H_{n}=H_{n-1}+\frac{1}{n}$

Multiply both sides by x^n

$H_{n}x^n=H_{n-1}x^n+\frac{x^n}{n}$

Sum both sides from to infinity

$\sum_{n=1}^{\infty}H_{n}x^n=\sum_{n=1}^{\infty}H_{n-1}x^n+\sum_{n=1}^{\infty}\frac{x^n}{n}$

$\sum_{n=1}^{\infty}H_{n}x^n-x\sum_{n=1}^{\infty}H_{n-1}x^{n-1}=\sum_{n=1}^{\infty}\frac{x^n}{n}$

$\sum_{n=1}^{\infty}H_{n}x^n-x\sum_{n=0}^{\infty}H_{n}x^{n}=-\ln(1-x)$

since $H_{0}=0$ we get

$\sum_{n=1}^{\infty}H_{n}x^n-x\sum_{n=1}^{\infty}H_{n}x^{n}=-\ln(1-x)$

$(1-x)\sum_{n=1}^{\infty}H_{n}x^n=-\ln(1-x)$

$\boxed{\sum_{n=1}^{\infty}H_{n}x^n=-\frac{\ln(1-x)}{(1-x)}}$ (1)

Sums involving Harmonic Numbers

Now nefine

$f_{p}^{k}(x)=\sum_{n=1}^{\infty}\frac{H_{n}^{k}}{n^p}x^n$

(1) can be represented by

$\boxed{f_{0}^{1}(x)=\sum_{n=1}^{\infty}H_{n}x^{n}=-\frac{\log(1-x)}{1-x}}$ (2)

Note that if we differentiate $f_{p}^{k}(x)$ w.r. to we get

$\frac{df_{p}^{k}(x)}{dx}=\sum_{n=1}^{\infty}\frac{H_{n}^{k}}{n^p}nx^{n-1}$

$\frac{df_{p}^{k}(x)}{dx}=\sum_{n=1}^{\infty}\frac{H_{n}^{k}}{n^{p-1}}x^{n-1}$

Multiplying both sides by

$x\frac{df_{p}^{k}(x)}{dx}=\sum_{n=1}^{\infty}\frac{H_{n}^{k}}{n^{p-1}}x^{n}$

$\boxed{\frac{df_{p}^{k}(x)}{dx}=\frac{f_{p-1}^{k}(x)}{x}}$

On the other hand

$f_{p}^{k}(x)=\int_{0}^{x}\frac{f_{p-1}^{k}(x)}{t}dt$

$f_{p}^{k}(x)=\int_{0}^{x}\frac{1}{t}\sum_{n=1}^{\infty}\frac{H_{n}^{k}}{n^{p-1}}t^ndt$

$f_{p}^{k}(x)=\sum_{n=1}^{\infty}\frac{H_{n}^{k}}{n^{p-1}}\int_{0}^{x}t^{n-1}dt$

$\boxed{f_{p}^{k}(x)=\sum_{n=1}^{\infty}\frac{H_{n}^{k}}{n^{p}}x^n}$

Therefore we get the recurrence relation

$\boxed{\frac{df_{p}^{k}(x)}{dx}=\frac{f_{p-1}^{k}(x)}{x} \,\,\Longleftrightarrow \,\, f_{p}^{k}(x)=\int_{0}^{x}\frac{f_{p-1}^{k}(t)}{t}dt}$ (3)

Lets now try to compute

$S=\sum_{n=1}^{\infty}\frac{H_{n}}{n^22^n}$

which corresponds to $f_{2}^{1}(x)\Big|_{x=\frac{1}{2}}$ in the notation introduced above.

From (3) we can work recursively to get

$f_{2}^{1}(x)=\int_{0}^{x}\frac{f_{1}^{1}(t)}{t}dt$

$f_{1}^{1}(t)=\int_{0}^{t}\frac{f_{0}^{1}(u)}{u}du$ (4)

$f_{2}^{1}(x)=\int_{0}^{x}\frac{1}{t}\int_{0}^{t}\frac{f_{0}^{1}(u)}{u}\,du\,dt$

And from (2) we get that

$\sum_{n=1}^{\infty}\frac{H_{n}}{n^2}x^n=\int_{0}^{x}\frac{1}{t} \Big\{-\int_{0}^{t}\frac{\ln(1-u)}{u(1-u)}\,du \Big\} \,dt$ (5)

Let start by computing the inner integral in (5), i.e.

$I=-\int_{0}^{t}\frac{\ln(1-u)}{u(1-u)}\,du$

Doing a partial fraction we get

$I=-\int_{0}^{t}\frac{\ln(1-u)}{u}\,du-\int_{0}^{t}\frac{\ln(1-u)}{(1-u)}du$

The first integral we immediatly recognize it as $Li_{2}(t)$ . For the second one, lets compute the indefinite version first and then plug the limits to the result.

$\int\frac{\ln(1-u)}{(1-u)}du$

First, lets make the change of variable $1-u=y \, \Rightarrow \, du=-dy$ . We get

$-\int\frac{\ln(y)}{y}dy$

Performing a second change of variable, $\log(y)=w \, \Rightarrow \, dw=\frac{dy}{y}$

$-\int w dw=\frac{w^2}{2}+C$

and therefore

$\int\frac{\ln(1-u)}{(1-u)}du=\frac{\log^2(1-u)}{2}+C$

Plugging the limits

$-\int_{0}^{t}\frac{\ln(1-u)}{(1-u)}du=\frac{\log^2(1-t)}{2}$

In conclusion we get that

$I=Li_{2}(t)+\frac{\log^2(1-t)}{2}$ (6)

and from equations (4) we also conclude that

$f_{1}^{1}(t)=\int_{0}^{t}\frac{f_{0}^{1}(u)}{u}du=I$

$f_{1}^{1}(t)=Li_{2}(t)+\frac{\log^2(1-t)}{2}$

and

$\boxed{\sum_{n=1}^{\infty}\frac{H_{n}}{n}x^n=Li_{2}(t)+\frac{\log^2(1-t)}{2}}$ (7)

To conclude our evaluation, we have to plug (6) back in (5)

$\sum_{n=1}^{\infty}\frac{H_{n}}{n^2}x^n=\int_{0}^{x}\frac{1}{t} \Big\{ Li_{2}(t)+\frac{\log^2(1-t)}{2} \Big\} \,dt$

$\sum_{n=1}^{\infty}\frac{H_{n}}{n^2}x^n=\int_{0}^{x}\frac{Li_{2}(t)}{t}dt+\frac{1}{2}\int_{0}^{x}\frac{\log^2(1-t)}{t} \,dt$

$\sum_{n=1}^{\infty}\frac{H_{n}}{n^2}x^n=Li_{3}(x)+\frac{1}{2}\underbrace{\int_{0}^{x}\frac{\log^2(1-t)}{t} \,dt}_{I}$ (8)

Lets now in evaluating by integrating by parts

$\int_{0}^{x}\frac{\log^2(1-t)}{t} \,dt=uv \big|_{0}^{x}-\int_{0}^{x}vdu$

$u=\log^2(1-t) \Rightarrow \, du=-2\frac{\log(1-t)}{(1-t)}dt$

$v=\log(t)$

$I=\log(x)\log^2(1-x)+2\int_{0}^{x}\log(1-t)\frac{\log(t)}{1-t}dt$

Now observe that $\frac{dLi_{2}(1-t)}{dt}=\frac{\log(t)}{(1-t)}$ (see apendix) and integrating by parts again we get

$I=\log(x)\log^2(1-x)+2 \Big[ \log(1-t)Li_{2}(1-t) \Big|_{0}^{x}+\int_{0}^{x}\frac{Li_{2}(1-t)}{1-t}dt\Big]$

$I=\log(x)\log^2(1-x)+2\log(1-x)Li_{2}(1-x) +2Li_{3}(1-t)\Big|_{0}^{x}$

$\boxed{I=\log(x)\log^2(1-x)+2\log(1-x)Li_{2}(1-x) -2Li_{3}(1-x)+2\zeta(3)}$ (9)

Plugging (9) in (8) we get the desired result

$\boxed{\sum_{n=1}^{\infty}\frac{H_{n}}{n^2}x^n=Li_{3}(x)+\frac{1}{2}\log(x)\log^2(1-x)+\log(1-x)Li_{2}(1-x) -Li_{3}(1-x)+\zeta(3)}$ (10)

If we set $x=\frac{1}{2}$ in (10) we get

$\sum_{n=1}^{\infty}\frac{H_{n}}{n^22^n}=Li_{3}(\frac{1}{2})+\frac{1}{2}\log(\frac{1}{2})\log^2(1-\frac{1}{2})+\log(1-\frac{1}{2})Li_{2}(1-\frac{1}{2}) -Li_{3}(1-\frac{1}{2})+\zeta(3)$