Series with digammas

Question

(Inspired by a comment in answer https://math.stackexchange.com/a/699264/442.)

corrected

Let $\Psi(x) = \Gamma'(x)/\Gamma(x)$ be the digamma function. Show $$ \sum_{n=1}^\infty (-1)^n\left(\Psi\left(\frac{n+1}{2}\right) -\Psi\left(\frac{n}{2}\right)\right) = -1 $$ As noted, it agrees to many decimals. But care may be required since the convergence is only conditional.

added
Both solutions are good. But could use explanation for exchange of summation.
Note that $$ 2\sum_{n=1}^\infty\sum_{m=0}^\infty \left(\frac{(-1)^n}{2m+n} -\frac{(-1)^n}{2m+n+1}\right) = 2\sum_{m=0}^\infty\sum_{n=1}^\infty \left(\frac{(-1)^n}{2m+n} -\frac{(-1)^n}{2m+n+1}\right) = -1 $$ is correct. But "Fibini" justification fails, since $$ \sum_{m=0}^\infty\sum_{n=1}^\infty \left|\frac{(-1)^n}{2m+n} -\frac{(-1)^n}{2m+n+1}\right| = \infty $$ Similarly in Random Variable's solution, the exchange $\sum_{n=1}^\infty \int_0^1 = \int_0^1 \sum_{n=1}^\infty$, although correct, cannot be justified by Fubini.

Answer 1

Using the integral representation $$\psi(s+1) = -\gamma +\int_{0}^{1} \frac{1-x^{s}}{1-x} \, dx ,$$ we get

$$ \begin{align} \sum_{n=1}^{\infty} (-1)^{n} \left(\psi \left(\frac{n}{2} \right)- \psi \left(\frac{n+1}{2}\right) \right) &= \sum_{n=1}^{\infty} (-1)^{n} \int_{0}^{1} \frac{x^{(n+1)/2-1} - x^{n/2-1}}{1-x} \, dx \\ &= \int_{0}^{1} \frac{1}{1-x} \sum_{n=1}^{\infty} (-1)^{n} \left(x^{n/2-1/2} - x^{n/2-1} \right) \, dx \\ &= \int_{0}^{1} \frac{1}{1-x} \left(- \frac{1}{1+\sqrt{x}}+ \frac{1}{(1+\sqrt{x})\sqrt{x}} \right) \, dx \\ &= 2 \int_{0}^{1} \frac{u}{1-u^{2}} \left(- \frac{1}{1+u} + \frac{1}{(1+u)u} \right) \, du \\&= \int_{0}^{1} \frac{1-u}{(1-u^{2})(1+u)} \, du\\ &= 2 \int_{0}^{1} \frac{1}{(1+u)^{2}} \, du \\ &=1. \end{align}$$

EDIT:

Daniel Fischer was kind enough to point out to me that we can use the dominated convergence theorem to justify interchanging the order of integration and summation.

Specifically, we may write $$ \begin{align} \sum_{n=1}^{\infty} (-1)^{n} \int_{0}^{1} \frac{x^{(n+1)/2-1} - x^{n/2-1}}{1-x} \, dx &= \lim_{N \to \infty}\sum_{n=1}^{N} (-1)^{n} \int_{0}^{1} \frac{x^{(n+1)/2-1} - x^{n/2-1}}{1-x} \, dx \\ &=\lim_{N \to \infty}\sum_{n=1}^{N} (-1)^{n+1} \int_{0}^{1} \frac{x^{n/2-1}} {1+\sqrt{x}} \, dx \\ &= \lim_{N \to \infty} \int_{0}^{1} \sum_{n=1}^{N} (-1)^{n+1} \frac{x^{n/2}} {x(1+\sqrt{x})} \, dx. \end{align}$$

But for $x \in [0,1]$, $$\left|\sum_{n=1}^{N} (-1)^{n+1} \frac{x^{n/2}} {x(1+\sqrt{x})} \right| \le \frac{\sqrt{x}}{x(1+\sqrt{x})},$$ which is integrable on $[0,1]$.

This, combined with the fact that $$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^{n/2}} {x(1+\sqrt{x})} $$ converges pointwise on $[0, 1)$, allows us to move the limit inside the integral.

Answer 2

We have: $$ \sum_{m\geq 0}\frac{1}{(m+a)(m+b)}=\frac{\psi(a)-\psi(b)}{a-b} \tag{1}$$ hence it is enough to compute: $$ \sum_{n\geq 1}\frac{(-1)^{n+1}}{n}\sum_{m\geq 0}\frac{1}{\left(m+\frac{n}{2}\right)\left(m+\frac{n+1}{2}\right)}\tag{2} $$ or: $$ \sum_{n\geq 1}\frac{(-1)^{n+1}}{n}\sum_{m\geq 0}\int_{0}^{+\infty}e^{-\left(m+\frac{n}{2}\right)x}-e^{-\left(m+\frac{n+1}{2}\right)x}\,dx\tag{3} $$ or: $$ \sum_{m\geq 0}\int_{0}^{+\infty}\log(1+e^{-x/2})e^{-(m+1/2)x}(e^{x/2}-1)\,dx \tag{4}$$ that is equivalent to: $$ \int_{0}^{+\infty}\frac{e^{x/2}\log(1+e^{-x/2})}{1+e^{x/2}}\,dx = \zeta(2)-\log^2(2) \tag{5}$$ through dilogarithms, to have (for the original question): $$ \sum_{n\geq 1}\frac{(-1)^n}{n}\left(\psi\left(\frac{n}{2}\right)-\psi\left(\frac{n+1}{2}\right)\right)= \color{red}{\zeta(2)-\log^2(2)}.\tag{6}$$ In the same spirit,

$$\begin{eqnarray*}\sum_{n\geq 1}(-1)^{n+1}\left(\psi\left(\frac{n}{2}\right)-\psi\left(\frac{n+1}{2}\right)\right)&=&\frac{1}{2}\sum_{n\geq 1}\sum_{m\geq 0}\frac{(-1)^n}{\left(m+\frac{n}{2}\right)\left(m+\frac{n+1}{2}\right)}\\[0.2cm]&=&2\sum_{n\geq 1}\sum_{m\geq 0}\left(\frac{(-1)^n}{2m+n}-\frac{(-1)^n}{2m+n+1}\right)\\[0.2cm]&=&-2\sum_{m\geq 0}\int_{0}^{+\infty}e^{-(2m+1)x}\tanh(x/2)\,dx\\[0.2cm]&=&-2\int_{0}^{+\infty}\frac{e^x}{(1+e^x)^2}\,dx\\[0.2cm]&=&\color{red}{-1}.\tag{7}\end{eqnarray*}$$

Answer 3

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

The Digamma Function has an interesting identity: $\ds{\Psi\pars{z + 1 \over 2} - \Psi\pars{z \over 2} = 2\sum_{k = 0}^{\infty}{\pars{-1}^{k} \over k + z}}$ such that

\begin{align} &\color{#f00}{\sum_{n = 1}^{\infty}\pars{-1}^{n} \bracks{\Psi\pars{n + 1 \over 2} -\Psi\pars{n \over 2}}} = \sum_{n = 1}^{\infty}\pars{-1}^{n}\bracks{% 2\sum_{k = 0}^{\infty}{\pars{-1}^{k} \over k + n}} \\[3mm] = &\ 2\sum_{n = 1}^{\infty}\pars{-1}^{n} \sum_{k = 0}^{\infty}\pars{-1}^{k}\int_{0}^{1}x^{k + n - 1}\,\dd x = -2\int_{0}^{1}\sum_{n = 1}^{\infty}\pars{-x}^{n - 1} \sum_{k = 0}^{\infty}\pars{-x}^{k}\,\dd x \\[3mm] = &\ -2\int_{0}^{1}{1 \over 1 + x}\,{1 \over 1 + x}\,\dd x = \color{#f00}{-1} \end{align}

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