# [SOLVED]Intriguing Harmonic Sum

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
This thread will be dedicated for a trial to prove the following

$$\displaystyle \sum_{k\geq 1} \frac{H^2_k}{k^2}=\frac{17}{4}\zeta(4)=\frac{17\pi^4}{360}$$

$$\displaystyle \mbox{where }\,\,H^2_k =\left( 1+\frac{1}{2}+\frac{1}{3}+\cdots \frac{1}{k}\right)^2$$​

In this paper the authors give solutions to the sum and others , but the process is quite complicated and uses contour integration , Fourier series and Parseal's thoerem , .... .

I believe we can solve it using elementary methods .

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#### topsquark

##### Well-known member
MHB Math Helper
This method will be dedicated for a trial to prove the following

$$\displaystyle \sum_{k\geq 1} \frac{H^2_k}{k^2}=\frac{17}{4}\zeta(4)=\frac{17\pi^4}{360}$$

$$\displaystyle \mbox{where }\,\,H^2_k =\left( 1+\frac{1}{2}+\frac{1}{3}+\cdots \frac{1}{k}\right)^2$$​

In this paper the authors give solutions to the sum and others , but the process is quite complicated and uses contour integration , Fourier series and Parseal's thoerem , .... .

I believe we can solve it using elementary methods .
The harmonic series has no explicit formula for its partial sum. But we can possibly deal with the sum
$$\displaystyle \sum_{j = 1}^k \frac{1}{jk}$$

I don't recognize the series, but if there is one and we can break it down into partial sums I think there might be an outside chance of an induction proof.

Possible do you think?

-Dan

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
I am thinking of using the formula

$$\psi_0^2(k+1) = (H_k -\gamma)^2$$

or possibly solving the integral

$$\displaystyle \left( \int^1_0 \frac{1-x^n}{1-x}\, dx \right)^2$$

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#### ZaidAlyafey

##### Well-known member
MHB Math Helper
The harmonic series has no explicit formula for its partial sum. But we can possibly deal with the sum
$$\displaystyle \sum_{j = 1}^k \frac{1}{jk}$$

I don't recognize the series, but if there is one and we can break it down into partial sums I think there might be an outside chance of an induction proof.

Possible do you think?

-Dan

I don't possibly get your idea , could you elaborate ?

#### topsquark

##### Well-known member
MHB Math Helper
The "inner" sum (before we square it and take the sum of that) is the 1/jk summation. I was thinking if we could get an explicit formula for that in terms of k we might be able to get a partial sum expression to sum over the square of and proceed inductively. If it works the outer sum would still be nasty as all (Swear) to compute, but it might be a way forward.

-Dan

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$$\displaystyle \left( \int^1_0 \frac{1-x^n}{1-x} \right)^2\, dx$$
There's a typo here. What integral is this?

-Dan

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
We can use the following generating function

$$\displaystyle \sum_{n\geq 1} H^2_n x^n = \frac{\log^2(1-x)+\operatorname{Li}_2(x)}{1-x}$$

$$\displaystyle \sum_{n\geq 1} H^2_n x^{n-1} = \frac{\log^2(1-x)}{1-x}+\frac{\log^2(1-x)}{x}+\frac{\operatorname{Li}_2(x)}{1-x}+\frac{\operatorname{Li}_2(x)}{x}$$

By integrating we get

$$\displaystyle \sum_{n\geq 1}\frac{ H^2_n }{n}x^{n} = -\frac{\log^3(1-x)}{3}+\int^x_0 \frac{\log^2(1-x)}{x}\, dx+\int^x_0 \frac{\operatorname{Li}_2(x)}{1-x}\, dx+\operatorname{Li}_3(x)$$

$$\displaystyle \int^{x}_0 \frac{\operatorname{Li}_2(x)}{1-x}\, dx= -\log(1-x) \operatorname{Li}_2(x)- \int^x_0 \frac{\log^2(1-x)}{x}$$

Hence we have

$$\displaystyle \sum_{n\geq 1}\frac{ H^2_n }{n}x^{n} = -\frac{\log^3(1-x)}{3}-\log(1-x) \operatorname{Li}_2(x)+\operatorname{Li}_3(x)$$

By integrating we get

$$\displaystyle \sum_{n\geq 1}\frac{ H^2_n }{n^2}x^{n} = -\frac{1}{3}\int^x_0 \frac{\log^3(1-x)}{x}\, dx+\frac{ \operatorname{Li}^2_2(x)}{2}+\operatorname{Li}_4(x)$$

Of course this a generalization of the harmonic sum , an attempt to solve the integral will be made later .

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Let us solve

$$\displaystyle \int^x_0 \frac{\log^3(1-t)}{t}\, dt$$

If we start by parts we get

$$\displaystyle \int^x_0 \frac{\log^3(1-t)}{t}\, dt = \log^3(1-x)\log(x) +3 \int^x_0 \frac{\log^2(1-t) \log(t)}{1-t} \, dt$$

Integrating by parts again we get

$$\displaystyle \int^x_0 \frac{\log^3(1-t)}{t}\, dt = \log^3(1-x)\log(x) +3 \left(\log^2(1-x)\operatorname{Li}_2(1-x)+2 \int^x_0 \frac{\log(1-t) \operatorname{Li}_2(1-t)}{1-t} \right)$$

Simplifying we get

$$\displaystyle \int^x_0 \frac{\log^3(1-t)}{t}\, dt = \log^3(1-x)\log(x)+3 \log^2(1-x)\operatorname{Li}_2(1-x)+6 \int^x_0 \frac{\log(1-t) \operatorname{Li}_2(1-t)}{1-t}\, dt$$

By parts for the third time

$$\displaystyle \int^x_0 \frac{\log^3(1-t)}{t}\, dt = \log^3(1-x)\log(x)+3 \log^2(1-x)\operatorname{Li}_2(1-x)-6 \operatorname{Li}_3(1-x) \log(1-x) +6\operatorname{Li}_4(1-x) -6\zeta(4)$$

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#### ZaidAlyafey

##### Well-known member
MHB Math Helper
So that becomes

\begin{align}
\sum_ {n\geq 1}\frac{ H^2_n }{n^2}x^{n} &= \frac{\pi^4}{45}-\frac{1}{3}\log^3(1-x)\log(x)- \log^2(1-x)\operatorname{Li}_2(1-x)+2 \operatorname{Li}_3(1-x) \log(1-x) \\& \, -2\operatorname{Li}_4(1-x)+\frac{ \operatorname{Li}^2_2(x)}{2}+\operatorname{Li}_4(x )
\end{align}

Putting $x =1$ most of the terms will go BAAAA.... m and we are left with

$$\displaystyle \sum_{n\geq 1}\frac{ H^2_n }{n^2} =\frac{\pi^4}{45}+\frac{\zeta^{\, 2}(2)}{2}+\zeta(4)=\frac{\pi^4}{45}+ \frac{\pi^4}{72}+\frac{\pi^4}{90}=\frac{17 \pi^4}{360}$$

Wasn't that nice !

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