Evaluating $\sum_{n=0}^\infty\frac{(1/2)_n}{n!}(H_n-H_{n-1/2})$

harmonic-numbershypergeometric functionpochhammer-symbolsequences-and-seriesspecial functions

I am seeking a closed-form (a form in terms of known special functions) to the sum
$$
\sum_{n=0}^\infty\frac{(1/2)_n}{n!}(H_n-H_{n-1/2}).
$$

Context:

I am searching for closed-forms to special cases of the double integral
$$
I(\alpha_1,\alpha_2,\zeta,\nu)=\nu^2\int_0^1\int_0^1
{_2F_1}\left({1+\nu,1+\nu \atop \alpha_1};\frac{(1-x)(1-y)}{(1-(1-\zeta)x)(1-(1-\zeta)y)}\right)
{_2F_1}\left({1-\nu,1-\nu \atop \alpha_2};(1-x)(1-y)\right)\left((1-(1-\zeta)x)(1-(1-\zeta)y\right)^{-\nu-1}\,\mathrm dx\mathrm dy,
$$
which converges for $\alpha_1+\alpha_2>2$, $-\alpha_2<2\nu<\alpha_1$. Also, we require $\zeta\geq 0$, $\alpha_1,\alpha_2>0$, and $\nu\in\Bbb R$. This double integral represents the second moment of a continuous random variable I've been studying.

Substituting $\alpha_1=1$, $\alpha_2=3/2$, $\zeta=1$, and $\nu=-1/2$, I was able to reduce this double integral to
$$
I(1,3/2,1,-1/2)=\frac{1}{2}\sum_{n=0}^\infty\frac{(1/2)_n}{n!}(H_n-H_{n-1/2}),
$$
where $H_z$ is the Harmonic number. I can provide details to this calculation if desired. But how to proceed? The difference of the harmonic numbers can be reduce to the form $H_n-H_{n-1/2}=a_n+\log 4$, which may be helpful.

Best Answer

From the integral representation \begin{equation} H_p=\int_0^1\frac{1-t^p}{1-t}\,dt \end{equation} we have \begin{align} H_n-H_{n-1/2}&=\int_0^1\frac{t^{n-1/2}-t^n}{1-t}\,dt\\ &=\int_0^1\frac{t^{n-1/2}(1-\sqrt{t})}{1-t}\,dt\\ &=2\int_0^1\frac{u^{2n}}{1+u}\,du \end{align} The proposed series can be written as \begin{align} S&=\sum_{n=0}^\infty\frac{(1/2)_n}{n!}(H_n-H_{n-1/2})\\ &=2\sum_{n=0}^\infty\frac{(1/2)_n}{n!}\int_0^1\frac{u^{2n}}{1+u}\,du \end{align} by swapping integration and summation \begin{align} S&=2\int_0^1\frac{du}{1+u}\sum_{n=0}^\infty\frac{(1/2)_n}{n!}u^{2n}\\ &=2\int_0^1\frac{du}{(1+u)\sqrt{1-u^2}}\\ &=\int_0^1v^{-1/2}\,dv\\ &=2 \end{align} (the last integral was obtained by changing $u=(1-v)/(1+v)$).

Related Solutions

Challenging sum: Compute $\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n)^2}$

We will look into the integral \begin{align*} I = &\int_{0}^{1}\arcsin^4 x\frac{ \ln x}{\sqrt{1-x^2}}\ \mathrm dx \end{align*} taking the @nospoon's novel approach presented here. Using the MacLaurin series of $\arcsin^4 x$ $$ \arcsin^4 x =\frac 3 2 \sum_{n=1}^\infty \frac{4^{n}H_{n-1}^{(2)}}{n^2{2n \choose n}}x^{2n} $$ and the fact that $$ \small\operatorname{B}(n+\tfrac 1 2,\tfrac 1 2) = \int_0^1 x^{n-1/2}(1-x)^{-1/2}\ \mathrm dx = 2\int_0^{\frac\pi 2} \sin^{2n}\theta\ \mathrm d\theta = \frac{\pi}{4^n}{2n \choose n},\tag{$\small x\mapsto \sin^2\theta$} $$ \begin{align*}\small \psi(n+\tfrac 12 ) -\psi(n+1) =&\small \sum_{k=1}^\infty \frac 1{\scriptsize k+n} - \frac 1{\scriptsize k+n-\tfrac 1 2} \\ =&\small\sum_{k=1}^\infty \left(\frac 1{\scriptsize k} - \frac 1{\scriptsize k-\tfrac 1 2}\right)-\sum_{k=1}^n\frac 1 {\scriptsize k} + \sum_{k=1}^n\frac 1{\scriptsize k-\tfrac 1 2}\\ =&\small-2\ln 2 -H_n +2(H_{2n}-\tfrac 1 2H_n)\\ =&\small 2(H_{2n}-H_n-\ln 2), \end{align*} \begin{align*} \Longrightarrow \ {\int_{ 0}^{1 }x^{2n}\frac{ \ln x}{\sqrt{1-x^2}}\ \mathrm dx} = & \frac 1 4\int_{0 }^{1 }x^{n-1/2} { \ln x \over \sqrt{1-x}}\ \mathrm dx\tag{$\small x^2\mapsto x$}\\ =& \frac 1 4 \left[\frac{\partial }{\partial x}\operatorname{B}(x,y) \right]_{x=n+1/2,y=1/2}\\ =&\frac 1 4\Big[ \operatorname{B}(x,y)\big[\psi(x) -\psi(x+y) \big]\Big]_{x=n+1/2,y=1/2}\\ =& \frac 1 4 \operatorname{B}(n+\tfrac 1 2,\tfrac 1 2)\big[\psi(n+\tfrac 12 ) -\psi(n+1) \big]\\ =& \frac{\pi}2\frac{{2n \choose n}}{4^{n}} \left(H_{2n} - H_n -\ln 2\right), \end{align*} where $\operatorname{B}(x,y)$ and $\psi(x)$ are the Beta and digamma function, respectively, we have \begin{align*} I = &\frac 3 2\sum_{n=1}^\infty \frac{4^{n}H_{n-1}^{(2)}}{n^2{2n \choose n}}\int_{0}^{1}x^{2n}\frac{ \ln x}{\sqrt{1-x^2}}\ \mathrm dx \\ =&\frac {3\pi}4 \sum_{n=1}^\infty \frac{H^{(2)}_{n-1}}{n^2}\left(H_{2n} - H_n -\ln 2\right) \\ =&\frac {3\pi}4\sum_{n=1}^\infty \frac{H^{(2)}_{n-1}H_{2n}}{n^2}-\frac {3\pi}4\underbrace{\sum_{n=1}^\infty \frac{H^{(2)}_{n-1}H_{n}}{n^2}}_{=-2\zeta(5) +2\zeta(2)\zeta(3)}-\frac {3\pi\ln 2}4\underbrace{\sum_{n=1}^\infty \frac{H^{(2)}_{n-1}}{n^2}}_{=\frac{3}4 \zeta(4)}\\ =&\frac{3\pi}{4} \sum_{n=1}^\infty \frac{H^{(2)}_{n}H_{2n}}{n^2} -\frac{3\pi}4\underbrace{\sum_{n=1}^\infty \frac{H_{2n}}{n^4}}_{=\frac{37}{4}\zeta(5)-4\zeta(2)\zeta(3)} +\frac{3\pi}2 \zeta(5) -\frac{\pi^3}4\zeta(3) -\frac{\pi^5\ln 2}{160}\\ =&\boxed{3\pi S -\frac{87\pi}{16} \zeta(5) +\frac{\pi^3}{4}\zeta(3) -\frac{\pi^5\ln 2}{160}} \end{align*} where $S = \sum_{n=1}^\infty \frac{H_{2n}H^{(2)}_{n}}{4n^2}$ is the sum in question, and the known values of several Euler sums $$ \sum_{n=1}^\infty \frac{H^{(2)}_{n-1}H_{n}}{n^2}=-2\zeta(5) +2\zeta(2)\zeta(3),\tag{1} $$ $$\sum_{n=1}^\infty \frac{H^{(2)}_{n}}{n^2}=\frac{7}4 \zeta(4),\tag{2} $$ \begin{align*}\sum_{n=1}^\infty \frac{H_{2n}}{n^4} =& 8\sum_{n=1}^\infty \frac{H_{n}}{n^4}-8\sum_{n=1}^\infty \frac{(-1)^{n-1} H_{n}}{n^4}\\ =&8\big(3\zeta(5)-\zeta(2)\zeta(3)\big)-8\left(\frac{59}{32}\zeta(5)-\frac 1 2\zeta(2)\zeta(3)\right)\\ =&\frac{37}4\zeta(5) - 4\zeta(2)\zeta(3)\tag{3} \end{align*} are used.

Note: $(1)$ is in @nospoon's answer here, $(2)$ can be found here, and for $(3)$ you can see Euler's formula and here.

Evaluation of $I$: By making substitution $x = \sin \theta$ and using the Fourier series of $$ \ln (\sin\theta) = -\ln 2 -\sum_{k=1}^\infty \frac{ \cos(2k \theta)}{k}, $$ we get \begin{align*} I =& \int_{0}^{\frac\pi 2} \theta^4 \ln(\sin\theta)\ \mathrm d\theta\\ =&\int_{0}^{\frac\pi 2} \theta^4\left(-\ln 2 -\sum_{k=1}^\infty \frac{ \cos(2k \theta)}{k}\right)\ \mathrm d\theta\\ =& -\ln 2\int_0^{\frac \pi 2}\theta^4\ \mathrm d\theta-\sum_{k=1}^\infty \frac{1}{k}\underbrace{\int_{0}^{\frac\pi 2}\theta^4 \cos(2k \theta) \ \mathrm d\theta}_{\text{IBP}\times 4}\\ =& -\frac{\pi^5\ln 2}{160}-\sum_{k=1}^\infty \frac{1}{k}\cdot\left(-\frac{\pi^3}{8}\frac{(-1)^{k-1}}{k^2} +\frac{3\pi}{4}\frac{(-1)^{k-1}}{k^4}\right)\\ =&-\frac{\pi^5\ln 2}{160}+\frac{\pi^3}8\underbrace{\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^3}}_{=\frac 3 4 \zeta(3)} - \frac{3\pi}4\underbrace{\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^5}}_{=\frac{15}{16}\zeta(5)}\\ =&\boxed{-\frac{\pi^5\ln 2}{160}+\frac{3\pi^3}{32}\zeta(3) -\frac{45\pi}{64}\zeta(5).} \end{align*}

Combining these, we get the equation $$ 3\pi S-\frac{87\pi}{16} \zeta(5) +\frac{\pi^3}{4}\zeta(3) -\frac{\pi^5\ln 2}{160}=-\frac{\pi^5\ln 2}{160} +\frac{3\pi^3}{32}\zeta(3)-\frac{45\pi}{64}\zeta(5), $$hence it follows $$ \boxed{S = \frac{101}{64}\zeta(5) -\frac{5\pi^2}{96}\zeta(3).} $$

Addendum: By considering MacLaurin series of \begin{align*} \ln(1-x)\ln(1+x) =&-\sum_{k=1}^\infty \left(\frac{ H_{2k}}k-\frac{H_k}{k} + \frac1{2k^2}\right)x^{2k} \end{align*} and \begin{align*} \frac{H_k}{k^2} + \frac{H_k^{(2)}}{k} -\frac{\zeta(2)}{k} =& \frac{\partial }{\partial k}\left[-\frac{H_k}{k}\right]\\ =& \int_0^1 x^{k-1}\ln x\ln(1-x)\ \mathrm dx\\ =&4\int_0^1 x^{2k-1}\ln x \ln(1-x^2)\ \mathrm dx \end{align*} we have that \begin{align*} &\int_{0}^{1}\ln(1-x)\ln(1+x) \frac{\ln x\ln(1-x^2)}x \ \mathrm dx \\&=-\sum_{k=1}^\infty \left(\frac{ H_{2k}}k-\frac{H_k}{k} + \frac1{2k^2}\right)\int_{0}^{1}x^{2k-1} \ln x \ln(1-x^2)\ \mathrm dx \\ &=-\frac 1 4\sum_{k=1}^\infty \left(\frac{ H_{2k}}k-\frac{H_k}{k} + \frac1{2k^2}\right)\left(\frac{H_k}{k^2} + \frac{H_k^{(2)}}{k} -\frac{\zeta(2)}{k}\right). \end{align*} The integral can be attacked by considering algebraic identity $$ ab(a+b) = \frac 1 3 (a+b)^3 - \frac {a^3}3 -\frac{b^3}3 $$ with $a=\ln(1-x)$ and $b=\ln(1+x)$, and extant results.

For the sum, after expanding the summand, the only tricky part is $$ \sum_{k=1}^\infty\frac{H_{2k}H_k}{k^3}, $$ which can be found here. Then, the sum $\sum_{k=1}^\infty \frac{H_{2k}H_k^{(2)}}{4k^2}$ can be evaluated by solving the equation obtained.

Challenging Sum: compute $\sum_{n=1}^\infty\frac{H_n}{2n+1}\left(\zeta(3)-H_n^{(3)}\right)$

A second solution in large steps by Cornel Ioan Valean

Let's start with the following useful identity which is easily derived by using recurrence relations and simple rearrangements, manipulations with sums, that is

Let $n$ be a non-negative integer number. Then, we have $$\int_0^1 x^{2n}\frac{\log(1+x)}{1+x}\textrm{d}x$$ $$=\frac{1}{2}H_{2n}^2-2\log(2) H_{2n}+\frac{1}{2}H_{2n}^{(2)}-\frac{1}{4}H_n^2-\frac{1}{4}H_n^{(2)}+\log (2)H_n+\frac{1}{2} \log ^2(2)-\sum_{k=1}^{n-1}\frac{H_k}{2 k+1},$$ where $H_n^{(m)}=1+\frac{1}{2^m}+\cdots+\frac{1}{n^m}$ represents the $n$th generalized harmonic number of order $m$.

By multiplying both sides of the identity above by $1/n^3$ and considering the summation from $n=1$ to $\infty$, we get

$$\sum_{n=1}^{\infty} \frac{1}{n^3}\sum_{k=1}^{n-1}\frac{H_{k}}{2 k+1}=\sum_{k=1}^{\infty} \sum_{n=k+1}^{\infty}\frac{1}{n^3}\frac{H_{k}}{2 k+1}=\underbrace{\sum_{k=1}^{\infty}\frac{H_{k}}{2 k+1}\left(\zeta(3)-H_k^{(3)}\right)}_{\text{The desired series}}$$ $$=\frac{1}{2}\sum_{n=1}^{\infty}\frac{H_{2n}^2}{n^3}-2\log(2) \sum_{n=1}^{\infty}\frac{H_{2n}}{n^3}+\frac{1}{2}\sum_{n=1}^{\infty}\frac{H_{2n}^{(2)}}{n^3}-\frac{1}{4}\sum_{n=1}^{\infty}\frac{H_n^2}{n^3}-\frac{1}{4}\sum_{n=1}^{\infty} \frac{H_n^{(2)}}{n^3}$$ $$+\log (2)\sum_{n=1}^{\infty} \frac{H_n}{n^3}+\frac{1}{2}\log ^2(2)\sum_{n=1}^{\infty}\frac{1}{n^3}-\int_0^1 \frac{\log(1+x)}{1+x}\operatorname{Li}_3(x^2)\textrm{d}x,$$

where we see all the series in the right-hand side are easily reducible to known series which may also be found in the book (Almost) Impossible Integrals, Sums, and Series.

On the other hand, with simple integration by parts, we obtain $$\int_0^1 \frac{\log(1+x)}{1+x}\operatorname{Li}_3(x^2)\textrm{d}x$$ $$=\frac{1}{2}\log^2(2)\zeta(3)-2\int_0^1 \frac{\log^2(1+x)\operatorname{Li}_2(x)}{x}\textrm{d}x-2\int_0^1 \frac{\log^2(1+x)\operatorname{Li}_2(-x)}{x}\textrm{d}x,$$ where the last integrals may be found calculated in the paper The calculation of a harmonic series with a weight $5$ structure, involving the product of harmonic numbers, $H_n H_{2n}^{(2)}$.

A note: The sister of the result above (easy to obtain by recurrence relations and very useful),

$$\int_0^1 x^{2n-1} \frac{\log(1+x)}{1+x}\textrm{d}x$$ $$=2\log(2) H_{2n}-\log(2)H_n+\frac{1}{4}H_n^2+\frac{1}{4}H_n^{(2)}-\frac{1}{2}H_{2n}^2-\frac{1}{2} H_{2n}^{(2)}+\frac{H_{2n}}{2n}-\frac{H_n}{2n} $$ $$ -\frac{1}{2}\log^2(2)+\sum_{k=1}^{n-1}\frac{H_k}{2 k+1}. $$

Best Answer

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Challenging sum: Compute $\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n)^2}$

Challenging Sum: compute $\sum_{n=1}^\infty\frac{H_n}{2n+1}\left(\zeta(3)-H_n^{(3)}\right)$

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