A series involves harmonic number

Question

How do we get a closed form for $$\sum_{n=1}^\infty\frac{H_n}{(2n+1)^2}$$

Answer 1

Here's another solution. I'll denote various versions of the sum

$$ \sum_{k=1}^\infty\sum_{j=1}^k\frac1j\frac1{k^2} $$

by an $S$ with two subscripts indicating which parities are included, the first subscript referring to the parity of $j$ and the second to the parity of $k$, with '$\mathrm e$' denoting only the even terms, '$\mathrm o$' denoting only the odd terms, '$+$' denoting the sum of the even and odd terms, i.e. the regular sum, and '$-$' denoting the difference between the even and the odd terms, i.e. the alternating sum. Then

$$ \begin{align} \sum_{n=1}^\infty\frac{H_n}{(2n+1)^2} &= 2\sum_{n=1}^\infty\sum_{i=1}^n\frac1{2i}\frac1{(2n+1)^2} \\ &= 2S_{\mathrm{eo}} \\ &= 2(S_{++}-S_{\mathrm o+}-S_{\mathrm{ee}}) \\ &= 2\left(S_{++}-S_{\mathrm o+}-\frac18S_{++}\right) \\ &= 2\left(\frac38S_{++}+\left(\frac12S_{++}-S_{\mathrm o+}\right)\right) \\ &= \frac34S_{++}+S_{-+} \\ &= \frac32\zeta(3)+\sum_{k=1}^\infty\sum_{j=1}^k\frac{(-1)^j}j\frac1{k^2}\;, \end{align} $$

where I used the result $\sum_nH_n/n^2=2\zeta(3)$ from the blog post Aeolian linked to and reduced the present problem to finding the analogue of that result with the sign alternating with $j$, which we can rewrite as

$$ \begin{align} \sum_{k=1}^\infty\sum_{j=1}^k\frac{(-1)^j}j\frac1{k^2} &= \sum_{k=1}^\infty\sum_{j=1}^\infty\frac{(-1)^j}j\frac1{k^2}-\sum_{k=1}^\infty\sum_{j=k+1}^\infty\frac{(-1)^j}j\frac1{k^2} \\ &= -\zeta(2)\log2+\sum_{j=1}^\infty\frac{(-1)^j}{j+1}\sum_{k=1}^j\frac1{k^2}\;. \end{align} $$

This last double sum can be evaluated by the method applied in the blog post, making use of the fact that summing the coefficients of a power series in $x$ corresponds to dividing it by $1-x$:

$$ \begin{align} \sum_{j=1}^\infty x^j\sum_{k=1}^j\frac1{k^2}=\def\Li{\operatorname{Li}}\frac{\Li_2(x)}{1-x}\;, \end{align} $$

where $\Li_2$ is the dilogarithm. Thus

$$ \begin{align} \sum_{j=1}^\infty\frac{(-1)^j}{j+1}\sum_{k=1}^j\frac1{k^2} &= \int_0^1\sum_{j=1}^\infty (-x)^j\sum_{k=1}^j\frac1{k^2}\mathrm dx \\ &= \int_0^1\frac{\Li_2(-x)}{1+x}\mathrm dx \\ &= \left[\Li_2(-x)\log(1+x)\right]_0^1+\int_0^1\frac{\log^2(1+x)}x\mathrm dx \\ &=-\frac{\zeta(2)}2\log2+\frac{\zeta(3)}4\;, \end{align} $$

where the boundary term is evaluated using $\Li_2(-1)=-\eta(2)=-\zeta(2)+2\zeta(2)/4=-\zeta(2)/2$ and the integral in the second term is evaluated in this separate question. Putting it all together, we have

$$ \begin{align} \sum_{n=1}^\infty\frac{H_n}{(2n+1)^2} &= \frac74\zeta(3)-\frac32\zeta(2)\log2 \\ &= \frac74\zeta(3)-\frac{\pi^2}4\log2\;. \end{align} $$

I believe all the rearrangements can be justified, despite the series being only conditionally convergent in $j$, by considering the partial sums with $j$ and $k$ both going up to $M$; then all the rearrangements can be carried out within that finite square of the grid, and the sums of the remaining terms go to zero with $M\to\infty$.

Answer 2

I gave an integral representation for a more general form. Here is an integral representation for your sum

$$\sum_{n=1}^\infty\frac{H_n}{(2n+1)^2}= \frac{1}{4}\,\int_{0}^{1}\!{\frac {\ln \left( 1-z \right) \ln \left( z\right) }{z\sqrt {1-z}}}{dz}= \frac{1}{4}(7\,\zeta \left( 3 \right) -{\pi }^{2}\ln \left( 2 \right))\sim 0.393327464. $$

The above integral can be evaluated through beta function. Here is the technique from previous problems. Basically, you need to consider the integral

$$ \int_{0}^{1} z^s (1-z)^{w-1/2} dz. $$

Answer 3

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\sum_{n = 1}^{\infty}{H_{n} \over \pars{2n + 1}^{2}}:\ {\large ?}}$.

Lets consider $\ds{\fermi\pars{x}\equiv \sum_{n = 1}^{\infty}{H_{n} \over \pars{2n + 1}^{2}}\,x^{2n + 1}. \qquad\fermi\pars{1}={\large ?}\,,\quad \fermi\pars{0} = 0}$.

\begin{align} \fermi'\pars{x}&=\sum_{n = 1}^{\infty}{H_{n} \over 2n + 1}\,x^{2n} \ \imp\ \bracks{x\fermi'\pars{x}}'=\sum_{n = 1}^{\infty}H_{n}\,x^{2n} =-\,{\ln\pars{1 - x^{2}} \over 1 - x^{2}}\,,\qquad\fermi'\pars{0} = 0 \end{align} where we used the Harmonic Number Generating Function.

Then \begin{align} &x\fermi'\pars{x}=-\int_{0}^{x}{\ln\pars{1 - t^{2}} \over 1 - t^{2}}\,\dd t \\[3mm]&\imp \fermi\pars{1}=-\int_{0}^{1}{\dd x \over x}\int_{0}^{x} {\ln\pars{1 - t^{2}} \over 1 - t^{2}}\,\dd t =-\int_{0}^{1}{\ln\pars{1 - t^{2}} \over 1 - t^{2}}\int_{t}^{1}{\dd x \over x} \,\dd t \end{align}

$$\begin{array}{|c|}\hline\\ \quad\sum_{n = 1}^{\infty}{H_{n} \over \pars{2n + 1}^{2}} =\int_{0}^{1}{\ln\pars{t}\ln\pars{1 - t^{2}} \over 1 - t^{2}}\,\dd t\quad \\ \\ \hline \end{array} $$

\begin{align} &\color{#c00000}{\sum_{n = 1}^{\infty}{H_{n} \over \pars{2n + 1}^{2}}} =\int_{0}^{1}{\ln\pars{t^{1/2}}\ln\pars{1 - t} \over 1 - t}\,\half\,t^{-1/2} \,\dd t ={1 \over 4}\int_{0}^{1}{t^{-1/2}\ln\pars{t}\ln\pars{1 - t} \over 1 - t}\,\dd t \\[3mm]&={1 \over 4}\lim_{\mu\ \to\ 0 \atop{\vphantom{\LARGE A}\nu\ \to\ 0}} \partiald{}{\mu}\partiald{}{\nu}\int_{0}^{1}t^{\mu - 1/2} \pars{1 - t}^{\nu - 1}\,\dd t ={1 \over 4}\lim_{\mu\ \to\ 0 \atop{\vphantom{\LARGE A}\nu\ \to\ 0}} \partiald{}{\nu}\Gamma\pars{\nu}\partiald{}{\mu} \bracks{\Gamma\pars{\mu + 1/2} \over \Gamma\pars{\mu + \nu + 1/2}} \\[3mm]&={1 \over 4}\lim_{\nu\ \to\ 0} \partiald{}{\nu}\braces{% \Gamma\pars{\nu}\,{\Gamma\pars{1/2} \over \Gamma\pars{\nu + 1/2}} \bracks{\Psi\pars{\half} - \Psi\pars{\nu + \half}}} \\[3mm]&=-\,{1 \over 4}\,\Gamma\pars{\half}\lim_{\nu\ \to\ 0} \partiald{}{\nu}\bracks{% {\Gamma\pars{\nu + 1} \over \Gamma\pars{\nu + 1/2}}\, {\Psi\pars{1/2 + \nu} - \Psi\pars{1/2} \over \nu}} \\[3mm]&=-\,{1 \over 4}\,\Gamma\pars{\half}\lim_{\nu\ \to\ 0} \partiald{}{\nu}\braces{% {\Gamma\pars{\nu + 1} \over \Gamma\pars{\nu + 1/2}}\, \bracks{\Psi'\pars{\half} + \half\,\Psi''\pars{\half}\nu}} \\[3mm]&={\pi^{2}\gamma + \pi^{2}\Psi\pars{1/2} + 14\zeta\pars{3} \over 8} \quad\mbox{where we used}\quad\Psi\pars{1} = -\gamma\,,\quad \Psi''\pars{\half} = -14\zeta\pars{3}. \end{align}

With $\ds{\Psi\pars{\half} = -2\ln\pars{2} - \gamma}$: $$ \color{#66f}{\large\sum_{n = 1}^{\infty}{H_{n} \over \pars{2n + 1}^{2}} ={1 \over 4}\,\bracks{7\zeta\pars{3} - \pi^{2}\ln\pars{2}}} \approx {\tt 0.3933} $$

Answer 4

$$\displaystyle I=\int_0^1 \dfrac{\ln x\ln(1-x^2)}{1-x^2}dx$$

Define the function $R$ on $[0;1]$,

$$R(x)=\int_0^x\dfrac{\ln t}{1-t^2}dt=\int_0^1\dfrac{x\ln(tx)}{1-t^2x^2}dt$$

Let $\epsilon$, real, such that $0<\epsilon<1$.

\begin{align} J(\epsilon)&=\Big[\left(R(x)-R(1)\right)\ln(1-x^2)\Big]_0^{1-\epsilon}+\int_0^{1-\epsilon} \dfrac{2x\left(R(x)-R(1)\right)}{1-x^2}dx\\ &=\left(R(1-\epsilon)-R(1)\right)\ln(1-(1-\epsilon)^2)+R(1)\ln\left(1-(1-\epsilon)^2\right)+\int_0^{1-\epsilon} \dfrac{2xR(x)}{1-x^2}dx\\ &=\left(R(1-\epsilon)-R(1)\right)\ln(1-(1-\epsilon)^2)+R(1)\ln\left(1-(1-\epsilon)^2\right)+\\ &\int_0^{1-\epsilon}\left(\int_0^1\dfrac{2x^2\ln(tx)}{(1-x^2)(1-t^2x^2)}dt\right)dx\\ &=\left(R(1-\epsilon)-R(1)\right)\ln(1-(1-\epsilon)^2)+R(1)\ln\left(1-(1-\epsilon)^2\right)+\\ &\int_0^{1-\epsilon}\left(\int_0^1\dfrac{2x^2\ln x}{(1-x^2)(1-t^2x^2)}dt\right)dx+\int_0^1\left(\int_0^{1-\epsilon}\dfrac{2x^2\ln t}{(1-x^2)(1-t^2x^2)}dx\right)dt\\ &=\left(R(1-\epsilon)-R(1)\right)\ln(1-(1-\epsilon)^2)+R(1)\ln\left(1-(1-\epsilon)^2\right)+\\ &\displaystyle\int_0^{1-\epsilon}\left[\dfrac{x\ln x\ln\left(\tfrac{1+tx}{1-tx}\right)}{1-x^2}\right]_{t=0}^{t=1}dx+\int_0^1 \left[\dfrac{\ln t\ln\left(\tfrac{1-x}{1+x}\right)}{t^2-1}+\dfrac{\ln t\ln\left(\tfrac{1-tx}{1+tx}\right)}{t}-\dfrac{t\ln t\ln\left(\tfrac{1+tx}{1-tx}\right)}{1-t^2}\right]_{x=0}^{x=1-\epsilon}dt\\ &=\left(R(1-\epsilon)-R(1)\right)\ln(1-(1-\epsilon)^2)+R(1)\ln\left(1-(1-\epsilon)^2\right)+\\ &\displaystyle\int_0^{1-\epsilon}\dfrac{x\ln x\ln\left(\tfrac{1+x}{1-x}\right)}{1-x^2}dx-\ln\left(\dfrac{\epsilon}{2+\epsilon}\right)R(1)+\int_0^1\dfrac{\ln t\ln\left(\tfrac{1-t(1-\epsilon)}{1+t(1-\epsilon)}\right)}{t}dt-\\ &\int_0^1\dfrac{t\ln t\ln\left(\tfrac{1+t(1-\epsilon)}{1-t(1-\epsilon)}\right)}{1-t^2}dt \end{align}

Since,

$$\lim_{\epsilon\rightarrow 0}\left(R(1-\epsilon)-R(1)\right)\ln(1-(1-\epsilon)^2)=0$$

and,

$$\lim_{\epsilon\rightarrow 0}R(1)\ln\left(\tfrac{1-(1-\epsilon)^2}{\epsilon}\right)=R(1)\ln 2$$

then,

$$\boxed{\lim_{\epsilon\rightarrow 0}J(\epsilon)=2R(1)\ln 2+\int_0^1\dfrac{\ln t\ln\left(\tfrac{1-t}{1+t}\right)}{t}dt}$$

and then,

\begin{align} \int_0^1\dfrac{\ln t\ln\left(\tfrac{1-t}{1+t}\right)}{t}dt&=\int_0^1\dfrac{\ln t\left(\ln(1-t)-\ln(1+t)\right)}{t}dt\\ &=-2\int_0^1\left(\sum_{n=0}^{\infty}\dfrac{t^{2n}}{2n+1}\right)\ln tdt\\ &=-2\sum_{n=0}^{\infty}\left(\dfrac{1}{2n+1}\int_0^1 t^{2n}\ln tdt\right)\\ &=2\sum_{n=0}^{\infty}\dfrac{1}{(2n+1)^3}\\ &=2\left(\sum_{n=1}^{\infty}\dfrac{1}{n^3}-\sum_{n=1}^{\infty}\dfrac{1}{(2n)^3}\right)\\ &=2\left(\zeta(3)-\dfrac{1}{8}\zeta(2)\right)\\ &=\dfrac{7}{4}\zeta(3)\\ \end{align}

and,

\begin{align} \displaystyle R(1)&=\int_0^1\dfrac{\ln x}{1-x^2}dx\\ &=\int_0^1 \left(\sum_{n=0}^{\infty}x^{2n}\right)\ln xdx\\ &=\sum_{n=0}^{\infty}\left(\int_0^1 x^{2n}\ln x dx\right)\\ &=-\sum_{n=0}^{\infty}\dfrac{1}{(2n+1)^2}\\ &=\sum_{n=1}^{\infty}\dfrac{1}{(2n)^2}-\sum_{n=0}^{\infty}\dfrac{1}{n^2}\\ &=\dfrac{1}{4}\zeta(2)-\zeta(2)\\ &=-\dfrac{3}{4}\zeta(2)\\ &=-\dfrac{\pi^2}{8} \end{align}

Therefore,

$$\boxed{I=\dfrac{7}{4}\zeta(3)--\dfrac{1}{4}\pi^2\ln 2}$$

Answer 5

The following new solution is proposed by Cornel Ioan Valean. Let's prove the more general case \begin{equation*} \sum _{n=1}^{\infty}\frac{H_n}{(2n+1)^{2m}}=2m\left(1-\frac{1}{2^{2m+1}}\right)\zeta(2m+1)-2\log(2)\left(1-\frac{1}{2^{2m}}\right)\zeta(2m) \end{equation*} \begin{equation*} -\frac{1}{2^{2m}}\sum_{i=1}^{m-1}(1-2^{i+1})(1-2^{2m-i})\zeta(1+i)\zeta(2m-i). \end{equation*} Proof. Using an application of The Master Theorem of Series presented in the article A master theorem of series and an evaluation of a cubic harmonic series and in the new released book, (Almost) Impossible Integrals, Sums, and Series, $\displaystyle \sum_{k=1}^{\infty} \frac{H_k}{(k+1)(k+n+1)}=\frac{(\gamma+\psi(n+1))^2+\zeta(2)-\psi^{(1)}(n+1)}{2n}$, multiplying both sides by $n$ and differentiating both sides with respect to $n$, $(2m-1)$th times, we get \begin{equation*} \sum _{k=1}^{\infty}\frac{H_k}{(2k+1)^{2m}}=\frac{1}{(2m-1)!2^{2m+1}}\lim_{n\to-1/2}\frac{\partial^{2m-1}}{\partial n^{2m-1}}\left((\gamma+\psi(n+1))^2+\zeta(2)-\psi^{(1)}(n+1)\right) \end{equation*} \begin{equation*} =\frac{1}{(2m-1)!2^{2m+1}}\biggr(2\psi^{(2m-1)}\left(\frac{1}{2}\right)\left(\gamma+\psi\left(\frac{1}{2}\right)\right)+2\sum_{i=1}^{m-1}\binom{2m-1}{i}\psi^{(2m-i-1)}\left(\frac{1}{2}\right)\psi^{(i)}\left(\frac{1}{2}\right) \end{equation*} \begin{equation*} -\psi^{(2m)}\left(\frac{1}{2}\right)\biggr) \end{equation*} \begin{equation*} =2m\left(1-\frac{1}{2^{2m+1}}\right)\zeta(2m+1)-2\log(2)\left(1-\frac{1}{2^{2m}}\right)\zeta(2m) \end{equation*} \begin{equation*} -\frac{1}{2^{2m}}\sum_{i=1}^{m-1}(1-2^{i+1})(1-2^{2m-i})\zeta(1+i)\zeta(2m-i), \end{equation*} where in the calculations we also needed the known results, $\displaystyle \psi\left(\frac{1}{2}\right)=-\gamma-2\log(2)$ and $\displaystyle \psi^{(k)}\left(\frac{1}{2}\right)=(-1)^{k-1}k!(2^{k+1}-1)\zeta(k+1)$.

A few cases of the generalization:

For $m=1$, \begin{equation*} \sum _{n=1}^{\infty}\frac{H_n}{(2n+1)^2}=\frac{7}{4}\zeta(3)-\frac{3}{2}\log(2)\zeta(2); \end{equation*} For $m=2$, \begin{equation*} \sum _{n=1}^{\infty}\frac{H_n}{(2n+1)^4}=\frac{31}{8}\zeta(5)-\frac{21}{16}\zeta(2)\zeta(3)-\frac{15}{8}\log(2)\zeta(4); \end{equation*} For $m=3$, \begin{equation*} \sum _{n=1}^{\infty}\frac{H_n}{(2n+1)^6}=\frac{381}{64}\zeta(7)-\frac{93}{64}\zeta(2)\zeta(5)-\frac{105}{64}\zeta(3)\zeta(4)-\frac{63}{32}\log(2)\zeta(6); \end{equation*} For $m=4$, \begin{equation*} \sum _{n=1}^{\infty}\frac{H_n}{(2n+1)^8}=\frac{511}{64}\zeta(9)-\frac{381}{256}\zeta(2)\zeta(7)-\frac{441}{256}\zeta(3)\zeta(6)-\frac{465}{256}\zeta(4)\zeta(5)-\frac{255}{128}\log(2)\zeta(8); \end{equation*} For $m=5$, \begin{equation*} \sum _{n=1}^{\infty}\frac{H_n}{(2n+1)^{10}}=\frac{10235}{1024}\zeta(11)-\frac{1533}{1024}\zeta(2)\zeta(9)-\frac{1785}{1024}\zeta(3)\zeta(8)-\frac{1905}{1024}\zeta(4)\zeta(7) \end{equation*} \begin{equation*} -\frac{1953}{1024}\zeta(5)\zeta(6)-\frac{1023}{512}\log(2)\zeta(10). \end{equation*}

The other case, $\displaystyle \sum _{n=1}^{\infty}\frac{H_n}{(2n+1)^{2m-1}}$, may be treated in a similar style. Using the parity of $p$ in $\displaystyle \sum _{n=1}^{\infty}\frac{H_n}{(2n+1)^p}, p\ge2$, allows you to put the closed-forms of the generalizations in more elegant ways.

Answer 6

using the following identity proved by Random Variable here $$S= \sum_{n=1}^{\infty} \frac{H_{n}}{ (n+a)^{2}}= \left(\gamma + \psi(a) \right) \psi_{1}(a) - \frac{\psi_{2}(a)}{2} \, , \quad a >0.$$

take $\ a=1/2$ $$S= \sum_{n=1}^{\infty} \frac{H_{n}}{ (2n+1)^{2}}=\frac74\zeta(3)-\frac32\ln2\zeta(2)$$

a similar identity was proved here by the mathematician Anthony Sofo, when he published some related work in 2011.