# A series involving the harmonic numbers : $\sum_{n=1}^{\infty}\frac{H_n}{n^3}$

Let $H_{n}$ be the nth harmonic number defined by $H_{n} := \sum_{k=1}^{n} \frac{1}{k}$.

How would you prove that

$$\sum_{n=1}^{\infty}\frac{H_n}{n^3}=\frac{\pi^4}{72}?$$

Simply replacing $H_{n}$ with $\sum_{k=1}^{n} \frac{1}{k}$ does not seem like a good starting point. Perhaps another representation of the nth harmonic number would be more useful.

• See here for another answer to this question; it's a little roundabout, but it only requires basic complex analysis. – Nick Strehlke Dec 31 '13 at 21:23

The Euler sum $\sum_{n=1}^{\infty} \frac{H_{n}}{n^{q}}$, where $q$ is an odd positive integer greater than $1$, can also be evaluated using this approach. See here.

Using the integration representation $$H_{n} = \int_{0}^{1} \frac{1-t^{n}}{1-t} \, dt \ ,$$

we have

\begin{align} \sum_{n=1}^{\infty} \frac{H_{n}}{n^{3}} &= \sum_{n=1}^{\infty} \frac{1}{n^{3}} \int_{0}^{1} \frac{1-t^{n}}{1-t} \ dt \\ &= \int_{0}^{1} \frac{1}{1-t} \sum_{n=1}^{\infty} \frac{1-t^{n}}{n^{3}} \\ &=\int_{0}^{1} \frac{\zeta(3) - \text{Li}_{3}(t)}{1-t} \ dt \tag{1} \\ &=- \Big(\zeta(3)-\text{Li}_{3}(t)\Big) \ln(1-t) \Bigg|_{0}^{1} - \int^{1}_{0} \frac{ \text{Li}_{2}(t) \log(1-t)}{t} \ dt \\ &= -\int_{0}^{1} \frac{ \text{Li}_{2}(t) \log(1-t)}{t} \ dt \\ &= \int_{0}^{1} \text{Li}_{2}(t) \, d \big(\text{Li}_{2}(t)\big) \\ &= \frac{\big(\text{Li}_{2}(t)\big)^{2}}{2} \Bigg|^{1}_{0} \\ &= \frac{\zeta^{2}(2)}{2} \\ &= \frac{\pi^{4}}{72}. \end{align}



• How do you justify swapping the sum and integral on line 2? – PeterM Feb 17 '13 at 2:05
• $\displaystyle \frac{1-t^{n}}{n^{3} (1-t)}$ is nonnegative for all $t$ and all $n$ – Random Variable Feb 17 '13 at 5:26

I will try to reduce the sum to an integral: $$\sum_{n=1}^\infty \frac{H_n}{n^3} = \sum_{n=1}^\infty H_n \frac{1}{\Gamma(3)} \int_0^\infty x^2 \mathrm{e}^{-n x} \mathrm{d} x = \frac{1}{2} \int_0^\infty x^2 \sum_{n=1}^\infty H_n \mathrm{e}^{-n x} \mathrm{d} x \tag{1}$$ We now make use of the following generating function: $$\sum_{n=1}^\infty H_n z^n = \sum_{n=1}^\infty H_n \Delta_n \left(\frac{z^n}{z-1} \right)$$ where $\Delta_n f_n = f_{n+1}-f_n$. We can now use summation by parts: $$\sum_{n=1}^m a_n \Delta_n b_n = b_{m+1} a_m - b_1 a_1 - \sum_{n=1}^{m-1} b_{n+1} \Delta_n a_n$$ with $b_n = \frac{z^n}{z-1}$ and $a_n = H_n$, and using $\Delta_n H_n = \frac{1}{n+1}$, we get $$\sum_{n=1}^\infty H_n z^n = \sum_{n=1}^\infty H_n \Delta_n \left(\frac{z^n}{z-1} \right) = -1 - \sum_{n=1}^\infty \frac{z^{n+1}}{z-1} \frac{1}{n+1} = \frac{\log(1-z)}{z-1} \tag{2}$$ Now, using $(2)$ in $(1)$: $$\sum_{n=1}^\infty \frac{H_n}{n^3} = -\frac{1}{2} \int_0^\infty x^2 \frac{\log\left(1-\mathrm{e}^{-x}\right)}{1-\mathrm{e}^{-x}} \mathrm{d}x \stackrel{t=\exp(-x)}{=} -\frac{1}{2} \int_0^1 \frac{\log(1-t)}{1-t} \frac{\log^2(t)}{t} \mathrm{d}t \tag{3}$$ The latter integral can be evaluated using derivatives of the Euler beta function: $$\int_0^1 \frac{\log(1-t)}{1-t} \frac{\log^2(t)}{t} \mathrm{d}t = \lim_{\alpha \downarrow 0} \lim_{\beta \downarrow 0} \frac{\mathrm{d}}{\mathrm{d} \alpha} \frac{\mathrm{d}^2}{\mathrm{d} \beta^2} \int_0^1 \left(1-t\right)^{\alpha-1} t^{\beta-1} \mathrm{d} t = \lim_{\alpha \downarrow 0} \lim_{\beta \downarrow 0} \frac{\mathrm{d}}{\mathrm{d} \alpha} \frac{\mathrm{d}^2}{\mathrm{d} \beta^2} \frac{\Gamma(\alpha) \Gamma(\beta)}{\Gamma(\alpha + \beta)}$$ Using $$\frac{\Gamma(\alpha) \Gamma(\beta)}{\Gamma(\alpha + \beta)} = \left(\frac{1}{\alpha} + \frac{1}{\beta} \right) \frac{\Gamma(\alpha+1) \Gamma(\beta+1)}{\Gamma(\alpha + \beta+1)} = \left(\frac{1}{\alpha} + \frac{1}{\beta} \right) \left( 1 - \frac{\pi^2}{6} \alpha \beta + \left(\alpha \beta^2 + \beta \alpha^2\right) \zeta(3) - \frac{\pi^4}{360} \left(4 \alpha \beta^3 + \alpha^2 \beta^2 + 4 \alpha^3 \beta\right) + \cdots \right)$$ Differentiating we get the result: $$\lim_{\alpha \downarrow 0} \lim_{\beta \downarrow 0} \frac{\mathrm{d}}{\mathrm{d} \alpha} \frac{\mathrm{d}^2}{\mathrm{d} \beta^2} \frac{\Gamma(\alpha) \Gamma(\beta)}{\Gamma(\alpha + \beta)} = -\frac{\pi^2}{36}$$ yielding with eq. $(3)$: $$\sum_{n=1}^\infty \frac{H_n}{n^3} = \frac{\pi^4}{72}$$

• How do you justify swapping the sum and integral in (1)? – PeterM Feb 17 '13 at 2:09
• @PeterM I guess Tonelli's theorem will do – Sasha Feb 17 '13 at 16:00

Apparently Euler showed in 1775 that: $$2 \sum_{n=1}^{\infty}\frac{H_n}{n^q} = (q+2)\zeta(q+1)- \sum_{m=1}^{q-2}\zeta(m+1)\zeta(q-m)$$ In your case, $q=3$, so that: $$2 \sum_{n=1}^{\infty}\frac{H_n}{n^3} = 5\zeta(4)- \zeta(2)^2 = 5\frac{\pi^4}{90}-\frac{\pi^4}{36}=\frac{\pi^4}{36}$$ The original proof of euler's in english can be found here.

• The final link is dead now. – Meow Apr 20 '14 at 14:14
• Another closed form of the above sum is investigated at this MSE link. – Marko Riedel Apr 27 '14 at 21:54


\begin{align}\color{#66f}{\large\sum_{n = 1}^{\infty}{H_{n} \over n^{3}}} &=\sum_{n = 1}^{\infty}{1 \over n^{3}}\ \overbrace{\quad\sum_{k = 1}^{\infty}\pars{{1 \over k} - {1 \over k + n}}\quad} ^{\ds{=\ H_{n}}}\ =\ \sum_{n = 1}^{\infty}\sum_{k = 1}^{\infty}{1 \over n^{2}k\pars{k + n}} \\[3mm]&=\half\sum_{n = 1}^{\infty}\sum_{k = 1}^{\infty} \bracks{{1 \over n^{2}k\pars{k + n}} + {1 \over k^{2}n\pars{n + k}}} =\half\sum_{n = 1}^{\infty}\sum_{k = 1}^{\infty} {k + n \over n^{2}k^{2}\pars{k + n}} \\[3mm]&=\half\sum_{n = 1}^{\infty}\sum_{k = 1}^{\infty}{1 \over n^{2}k^{2}} =\half\pars{\sum_{n = 1}^{\infty}{1 \over n^{2}}}^{2} =\half\pars{\pi^{2} \over 6}^{2}=\color{#66f}{\Large{\pi^{4} \over 72}} \approx 1.3529 \end{align}

• Nice answer. +1 I also liked your evaluation of that bino-harmonic sum. math.stackexchange.com/questions/815103/… Your approach to that sum simplified the evaluation considerably. – Random Variable Jul 23 '14 at 19:30
• @RandomVariable Thanks. $H_{\rm n}$ is somehow tricky. – Felix Marin Jul 24 '14 at 19:14
• Very slick. I am impressed by this direct approach through series manipulation. – Joel Jul 13 '15 at 17:21

Appealing to the integral identity I introduced in a previous problem for the more general case of your sum

$$B(p,q) = \sum_{k=1}^{\infty} \dfrac{H_k^{(p)}}{k^q}=\frac{(-1)^q}{\Gamma(q)}\int_{0}^{1}\!{\frac {\left(\ln\left(u\right)\right)^{q-1}{Li_{p}(u)} }{ u\left( u-1 \right)}}{du},$$

where $Li_{p}(u)$ is the polylogarithm function, we can have the following integral representation for the sum

$$B(1,3) = \sum_{k=1}^{\infty} \dfrac{H_k}{k^3}=-\frac{1}{\Gamma(3)}\int_{0}^{1}\!{\frac {\ln^2\left(u\right){\ln(1-u)} }{ u\left( 1-u \right)}}{du}$$

$$= -\frac{1}{\Gamma(3)}\lim_{w\to 0}\lim_{s \to 0}\frac{d}{dw} \frac{d^2}{ds^2}\int_{0}^{1} u^{s-1}\, (1-u)^{w-1}$$

$$= -\frac{1}{2}\lim_{w\to 0}\lim_{s \to 0}\frac{d}{dw} \frac{d^2}{ds^2}\beta(s, w)=\frac{\pi^4}{72},$$

where $\beta(s,w)=\frac{\Gamma(s)\Gamma(w)}{\Gamma(s+w)}$ is the beta function.

Note: $$Li_{1}(x) = -\ln(1-x).$$

We proved here

$$\sum_{k=1}^\infty\frac{H_k}{k^n}=\frac12\sum_{i=1}^{n-2}(-1)^{i-1}\zeta(n-i)\zeta(i+1),\quad n=3,5,7, ...$$

Set $$n=3$$ we get $$\sum_{k=1}^\infty \frac{H_k}{k^3}=\frac54\zeta(4)$$

where $$\zeta(4)=\frac{\pi^4}{90}$$