# An equality about infinite sums involving squared harmonic numbers [duplicate]

I found the following equality. But I forgot the source as well as the solution.

If you give the proof, I appreciate it.

$$\sum_{k=1}^\infty \frac{1}{k^2}\left( 1 + \frac12 + \dots + \frac1k \right)^2 = \frac{17\pi^4}{360} = \frac{17}{4} \sum_{k=1}^\infty \frac{1}{k^4}$$

Begin by noting that $$H_n = 1 + \frac{1}{2} + \cdots + \frac{1}{n},$$ where $$H_n$$ is the $$n$$th Harmonic number. So we wish to show the following Euler sum $$\sum_{n = 1}^\infty \left (\frac{H_n}{n} \right )^2 = \frac{17 \pi^4}{360}.$$

To prepare the ground, I first show that $$\int_0^1 x^{n - 1} \ln^2 (1 - x) \, dx = \frac{H^2_n + H^{(2)}_n}{n}.\qquad (*)$$ To show this result, I will make use of the following result I proved here: $$\frac{\partial}{\partial m} \operatorname{B} (n,m) = - \operatorname{B} (n,m) \sum_{k = 0}^{n - 1} \frac{1}{k + m}.\tag1$$ Here $$\operatorname{B} (n,m)$$ is the Beta function.

Now \begin{align} \int_0^1 x^{n - 1} \ln^2 (1 - x) \, dx &= \frac{\partial^2}{\partial m^2} \left [\int_0^1 x^{n - 1} (1 - x)^{m - 1} \, dx \right ]_{m = 1}\\ &=\frac{\partial^2}{\partial m^2} \operatorname{B} (n,m) \Big{|}_{m = 1}\\ &= \frac{\partial}{\partial m} \left [\frac{\partial}{\partial m} \operatorname{B} (n,m) \right ]_{m = 1}\\ &= \frac{\partial}{\partial m} \left [- \operatorname{B} (n,m) \sum_{k = 0}^{n = 1} \frac{1}{k + m} \right ]_{m = 1}\tag2\\ &= \operatorname{B} (n,m) \left. \left (\sum_{k = 0}^{n - 1} \frac{1}{k + m} \right )^2 \right |_{m = 1} - \operatorname{B} (n,m) \frac{\partial}{\partial m} \left [\sum_{k = 0}^{n - 1} \frac{1}{k + m} \right ]_{m = 1}\tag3\\ &= \operatorname{B} (n,1) \left (\sum_{k = 0}^{n - 1} \frac{1}{k + 1} \right )^2 - \operatorname{B} (n,m) \left [H^{(2)}_{m - 1} - H^{(2)}_{m + n - 1} \right ]_{m = 1}\tag4\\ &= \operatorname{B}(n,1) \left (\sum_{k = 1}^n \frac{1}{k} \right )^2 + \operatorname{B} (n,1) H^{(2)}_n\tag5\\ &= \frac{H^2_n + H^{(2)}_n}{n},\tag6 \end{align} as required. Here $$H^{(p)}_n$$ denotes the $$n$$th generalised harmonic number of order $$p$$ such that $$H^{(1)}_n = H_n$$.

Reasons

(2) Application of the result given in (1).

(3) Product rule together with using the result given in (1).

(4) $$\displaystyle{\sum_{k = 0}^{n - 1} \frac{1}{k + m} = \psi (n + m) - \psi (m)}$$, where $$\psi (x)$$ is the digamma function. Thus

$$\displaystyle{\frac{\partial}{\partial m} \sum_{k = 0}^{n - 1} \frac{1}{k + m} = \psi^{(1)} (n + m) - \psi^{(1)} (m)} = -H^{(2)}_{m + n - 1} + H^{(2)}_{m - 1}$$ since $$\psi^{(1)} (a) = \zeta (2) - H^{(2)}_{a - 1}$$.

(5) Shifting the index in the sum by $$k \mapsto k - 1$$.

(6) Since $$\operatorname{B} (n,1) = 1/n$$.

Moving to the main event, if we divide ($$*$$) by $$n$$ before summing the result from $$1$$ to $$\infty$$ one has $$\sum_{n = 1}^\infty \int_0^1 \frac{x^{n - 1}}{n} \ln^2 (1 - x) \, dx = \sum_{n = 1}^\infty \frac{H^2_n}{n^2} + \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^2}.$$ Making use of the result $$\sum_{n = 1}^\infty \frac{H^{(p)}_n}{n^p} = \frac{\zeta^2 (p) + \zeta (2p)}{2},$$ which is proved here, on setting $$p = 2$$ we see that $$\sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^2} = \frac{7 \pi^4}{360},$$ giving $$\sum_{n = 1}^\infty \left (\frac{H_n}{n} \right )^2 = \sum_{n = 1}^\infty \int_0^1 \frac{x^{n - 1}}{n} \ln^2 (1 - x) \, dx - \frac{7 \pi^4}{360}.\qquad (**)$$ For the remaining term on the right hand side it can be knocked over with relative ease. Interchanging the sum with the integral sign we have \begin{align} \int_0^1 \ln^2 (1 - x) \, dx \sum_{n = 1}^\infty \frac{x^{n - 1}}{n} \, dx &= -\underbrace{\int_0^1 \frac{\ln^3 (1 - x)}{x} \, dx}_{x \mapsto 1 - x}\\ &= -\int_0^1 \frac{\ln^3 x}{1 - x} \, dx\\ &= -\int_0^1 \ln^3 x \sum_{n = 0}^\infty x^n \, dx\\ &= -\sum_{n = 0}^\infty \int_0^1 x^n \ln^3 x \, dx\\ &= -\sum_{n = 0}^\infty \frac{d^3}{ds^3} \left [\int_0^1 x^{n + s} \, dx \right ]_{s = 1}\\ &= -\sum_{n = 0}^\infty \frac{d^3}{ds^3} \left [\frac{1}{n + s + 1} \right ]_{s = 1}\\ &= 6 \, \underbrace{\sum_{n = 0}^\infty \frac{1}{(n + 1)^4}}_{n \mapsto n - 1}\\ &= 6 \sum_{n = 1}^\infty \frac{1}{n^4}\\ &= 6 \, \zeta (4)\\ &= \frac{\pi^4}{15}. \end{align}

Substituting the value found for this last term back into ($$**$$) we finally arrive at $$\sum_{n = 1}^\infty \left (\frac{H_n}{n} \right )^2 = \frac{\pi^4}{15} - \frac{7 \pi^4}{360} = \frac{17 \pi^4}{360},$$ as required.

• nice solution omegadot but its easier to do it this way $\frac{\partial}{\partial m} \left [\sum_{k = 0}^{n - 1} \frac{1}{k + m} \right ]_{m = 1}=- \left .\sum_{k = 0}^{n - 1} \frac{1}{(k + m)^2} \right ]_{m = 1}=-\sum_{k = 0}^{n - 1} \frac{1}{(k + 1)^2}=-\sum_{k = 1}^{n} \frac{1}{k^2}=-H_n^{(2)}$ to avoid digamma and its derivative and thats the point of deriving the derivative of beta in this form. – Ali Shadhar Jul 17 '19 at 5:00