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how to prove that

$$\sum_{n=0}^\infty(-1)^n(\overline{H}_n-\ln2)^2=\frac{\pi^2}{24}\ ?$$

where $\overline{H}_n=\sum_{k=1}^n\frac{(-1)^{k-1}}{k}$ is the alternating harmonic number.


This problem is proposed by a friend on a Facebook group and I managed to prove the equality using only integration but can we prove it using series manipulations?

Here is my work,

In page $105$ of this paper we have

$$\overline{H}_n-\ln2=(-1)^{n-1}\int_0^1\frac{x^n}{1+x}dx$$

Therefore

$$(\overline{H}_n-\ln2)^2=\int_0^1\int_0^1\frac{(xy)^n}{(1+x)(1+y)}dxdy$$

$$\Longrightarrow \sum_{n=0}^\infty(-1)^n(\overline{H}_n-\ln2)^2=\int_0^1\int_0^1\frac{dxdy}{(1+x)(1+y)}\sum_{n=0}^\infty(-xy)^n$$

$$=\int_0^1\int_0^1\frac{dxdy}{(1+x)(1+y)(1+xy)}=\int_0^1\frac{1}{1+x}\left(\int_0^1\frac{dy}{(1+y)(1+xy)}\right)dx$$

$$=\int_0^1\frac{1}{1+x}\left(-\frac{\ln\left(\frac{1+x}{2}\right)}{1-x}\right)dx=-\int_0^1\frac{\ln\left(\frac{1+x}{2}\right)}{1-x^2}\ dx,\quad x=\frac{1-u}{1+u}$$

$$=\frac12\int_0^1\frac{\ln(1+u)}{u}du=\frac12(-\operatorname{Li}_2(-1))=\frac12(\frac12\zeta(2))=\frac{\pi^2}{24}$$


Regarding series manipulation, we know that

$$\overline{H}_{2n}=H_{2n}-H_n$$ but we need $\overline{H}_n$ , so is this identity helpful? any other ideas?

thank you

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  • $\begingroup$ I do not know where this could take us but $(H'_n-\log(2))^2=\Phi (-1,1,n+1)^2$ where appears Lerch function. $\endgroup$ – Claude Leibovici Jan 9 at 9:33
  • $\begingroup$ @Claude Leibovici thanks for the identity just not sure how to use it. Maybe this Lerch function itself can be written in a simpler more common functions. $\endgroup$ – Ali Shather Jan 9 at 21:49
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We shall use $A_n=\sum _{k=1}^n \frac{(-1)^{k+1}}{k}$ (and for completeness $A_0 = 0$) for the alternating harmonic sum.

The sum in question is

$$s = \sum_{k\ge 0} (-1)^k (A_k-\log(2))^2\\= \sum_{k\ge 0} \left((A_{2k}-\log(2))^2-(A_{2k+1}-\log(2))^2\right)$$

The summand can be expanded to

$$A_{2 k}^2-A_{2 k+1}^2-2 \log (2) A_{2 k}+2 \log (2) A_{2 k+1} \\ = -\frac{2 (A_{2 k}-\log (2))}{2 k+1}-\frac{1}{(2 k+1)^2}\tag{1}$$

Here we have used that for integer $k$

$$A_{k+1}-A_{k}=\frac{(-1)^k}{k+1}\tag{2}$$

So we have

$$s = s_1 + s_2$$

with

$$s_1 =2 \sum_{k\ge 0} \frac{ \log(2) - A_{2 k}}{2 k+1}\tag{3}$$

and

$$s_2 = -\sum_{k\ge0} \frac{1}{(2 k+1)^2}= -\frac{\pi^2}{8}\tag{4}$$

which we have summed up imediately.

In the first one we use that for integer $k$

$$A_{2 k}-\log(2) = \int_0^1 \frac{1-x^{2 k}}{x+1} \, dx-\int_0^1 \frac{1}{x+1} \, dx\\ = -\int_0^1 \frac{x^{2 k}}{x+1} \, dx\tag{5}$$

so that the sum under the $x$-integral becomes

$$ \sum_{k\ge0} \frac{2 x^{2 k} }{2 k+1}=\frac{2 \operatorname{arctanh}(x)}{x}=\frac{1}{x} \log(\frac{1+x}{1-x})\tag{6}$$

Hence the first sum is given by the integral

$$s_1 = \int_0^1 \frac{1}{x(1+x)} \log(\frac{1+x}{1-x})=\frac{\pi^2}{6}\tag{7}$$

Finally we get

$$s = s_1 + s_2 =\frac{\pi^2}{6} -\frac{\pi^2}{8}= \frac{\pi^2}{24}$$

Q.E.D.

Discussion

1) Integral representations of the alternating and odd harmonic numbers

Inserting $\frac{1}{k}=\int_0^1 x^{k-1}\,dx$ into the definition of $A_n$ and doing the sum under the integral we obtain

$$A_n=\int_0^1 \frac{1-(-1)^n x^n}{x+1}\,dx\tag{8a}$$

for $n=2k$ this simplifies to

$$A_{2k}=\int_0^1 \frac{1- x^{2k}}{x+1}\,dx\tag{8b}$$

as used in $(5)$.

Similarly we have for the odd harmonic number the representation

$$O_n = \int_0^1 \frac{x^{2 n}-1}{x^2-1} \, dx\tag{9}$$

2) The introduction of an integral in $(4)$ is a deviation from the requirement of the OP to use only series operations.

Ideally we would like to see $s_1= \sum_{k\ge 1} \frac{1}{k^2}$

Hence my "solution" is incomplete. I'll try to improve it.

EDIT

I'm almost there.

I find by series manipulations (changing the order of summation) that

$$s_1 = \sum_{n\ge1} \frac{O_{n}}{(n+1)(2n-1)}\tag{10}$$

Where the odd harmonic sum is defined as

$$O_n=\sum_{m=1}^n \frac{1}{2m-1}\tag{11}$$

It is easy to show that

$$O_n = H_{2n} -\frac{1}{2} H_{n}\tag{12}$$

so that

$$s_1 = \sum_{n\ge1} \frac{H_{2n} -\frac{1}{2} H_{n}}{(n+1)(2n-1)}\tag{13}$$

But we have (https://en.wikipedia.org/wiki/Harmonic_number)

$$H_{2z}=\log(2) + \frac{1}{2}(H_z+H_{z-\frac{1}{2}})\tag{14}$$

so that

$$s_1 = \sum_{n\ge1} \frac{\log(2) -\frac{1}{2} H_{n-\frac{1}{2}}}{(n+1)(2n-1)}\tag{15}$$

The first sum yields $2 \log(2)^2$. The second sum

$$s_{1b} =- \frac{1}{2}\sum_{n\ge1} \frac{ H_{n-\frac{1}{2}}}{(n+1)(2n-1)}= ?\tag{16}$$

Still has to be calculated.

Summarizing: the numerical result of $s_1$ agrees with $\pi^2/6$,

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  • $\begingroup$ Very nice (+1). I see that your solution is complete but I have a question about your equality $s = \sum_{k\ge 0} (-1)^k (A_k-\log(2))^2= \sum_{k\ge 0} \left((A_{2k}-\log(2))^2-(A_{2k+1}-\log(2))^2\right)$ . Did you use $\sum_{k=0}^\infty (-1)^k f(k)=\sum_{k=0}^\infty [f(2k)-f(2k+1)]?$ $\endgroup$ – Ali Shather Jan 9 at 20:52
  • $\begingroup$ @ Ali Shather 1) Yes, I split the summand into even and odd summands. 2) please leave the problem open. I'm still trying to fully match the requirements of the OP. $\endgroup$ – Dr. Wolfgang Hintze Jan 9 at 21:35
  • $\begingroup$ By the way is it known that $A_{2n}=\int_0^1\frac{1-x^{2n}}{1+x}dx?$ I've never seen this definition before. $\endgroup$ – Ali Shather Jan 9 at 21:43
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    $\begingroup$ @ Ali Shather I have derived the formula for $A_n$ now in the discussion. It is very simlar to that for $H_n$. $\endgroup$ – Dr. Wolfgang Hintze Jan 9 at 22:02
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The development of the solution using only series manipulations

I asked Cornel for a way and here you have the large steps to go for perfectly obtaining what you need. You need a mix of results and I'll provide with references for all you need.

First step

Start with Abel's summation $a_k=(-1)^k$, $b_k=(\overline{H}_k-\log(2))^2$ to show that $$ \sum_{n=0}^\infty(-1)^n(\overline{H}_n-\log(2))^2=2 \sum _{n=0}^{\infty } \frac{H_n-H_{2 n}+\log (2)}{2 n+1}-\frac{\pi ^2}{8}.$$

Second step

This is a non-obvious step. Build the series below $$\sum _{n=0}^{\infty } \frac{H_n-H_{2 n}+\log (2)}{(2 n+1) (2 n+2)}=\frac{1}{12} \left(\pi ^2+6 \log ^2(2)-12 \log (2)\right).$$ Apply Abel's summation with $a_k=1/((2k+1)(2k+2))$ and $b_k=H_k-H_{2 k}+\log (2)$ to almost magically get the same sum in the right-hand side, but with an opposite sign together with elementary series to calculate directly.

Third Step

Show that $$\sum_{n=1}^{\infty} \frac{1}{n}(H_{2n}-H_n-\log(2))=\log^2(2)-\frac{\pi^2}{12}.$$ The strategy of approaching the series is presented on page $250$ in the book (Almost) Impossible Integrals, Sums, and Series. The only thing you need differently here is the approach of $\displaystyle \sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_n}{n}$, which you want elementary by series manipulations. Therefore, see the next step.

Fourth step

Show that $$ \sum_{n=1}^{\infty}(-1)^{n-1} \frac{H_n}{n} = \frac{\pi^2}{12} - \frac{1}{2} \log^2(2),$$ which is proved completely elementarily by series manipulations in https://math.stackexchange.com/q/499689.

Putting all together leads to the desired result, that is the evaluation of that series by series manipulations.

Q.E.D.

A nice note: If we are allowed to use integrals, then we may also use a different approach to calculate the series from the Third step, by exploiting the series

$$\sum_{n=1}^{\infty} \left(2 H_{2n}-2 H_{n}+\frac{1}{2n}-2 \log(2)\right)\frac{\sin^2(2n x)}{n}=\log(\sin(x))\log(\cos(x)), \ 0<x<\frac{\pi}{2},$$

that appears on page $248$ of the previously mentioned book. Then in the extraction process of the desired value we also need to prove that

$$\int_0^{\pi/2} \log(\sin(x))\log(\cos(x))\textrm{d}x =\frac{\pi}{2}\log ^2(2)-\frac{\pi ^3}{48}.$$

At this point I recall that Paul Nahin presents in his book Inside Interesting Integrals a nice generalization for

$$\int_0^{\pi/2} \log(a\sin(x))\log(a\cos(x))\textrm{d}x, \ a>0,$$

which one may find on page $236$. Also, there could be a temptation to immediately treat the integral with $a=1$ as a particular case of the Beta function in the trigonometric form. According to Inside Interesting Integrals, the last integral together with the forms $\int_0^{\pi/2} \log^2(a\sin(x))\textrm{d}x, \ a>0,$ and $\int_0^{\pi/2} \log^2(a\cos(x))\textrm{d}x, \ a>0,$ were evaluated due to the English mathematician Joseph Wolstenholme.

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    $\begingroup$ Impressive (+1) $\endgroup$ – Ali Shather Jan 9 at 21:32

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