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Numerical calculations suggest that $$\sum_{n=1}^\infty\frac{\ln(n+2)}{n\,(n+1)}\,\stackrel{\color{gray}?}=\,\int_0^1\frac{x\,(\ln x-1)}{\ln(1-x)}\,dx=1.553767373413083673460727...$$ How can we prove it? Is there a closed form expression for this value?

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  • $\begingroup$ Did you try to chop up the integral into pieces and match every term in your series to a piece? E.g. for which $t\in (0,1)$ we have the integral from $t$ to $1$ or from $0$ to $t$ equal to $\frac{ln(3)}{2}$? $\endgroup$ – Emre May 29 '16 at 22:15
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    $\begingroup$ Wolfram alpha gives a slightly smaller value for LHS, 1.52239, where it gives your result for the RHS $\endgroup$ – Emre May 29 '16 at 22:20
  • $\begingroup$ @E.Girgin That is odd. When I calculate the LHS in Mathematica, I get the number shown in my question. $\endgroup$ – Vladimir Reshetnikov May 29 '16 at 22:30
  • $\begingroup$ Maybe Wolframalpha is not reliable enough. $\endgroup$ – Emre May 29 '16 at 22:31
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    $\begingroup$ @DavidH: en.wikipedia.org/wiki/%C3%89pater_la_bourgeoisie :D $\endgroup$ – Jack D'Aurizio May 30 '16 at 3:02
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By Frullani's theorem $$ \log(n+2)=\int_{0}^{+\infty}\frac{1-e^{-(n+1)x}}{xe^x}\,dx \tag{1}$$ hence by multiplying both sides by $\frac{1}{n(n+1)}$ and summing over $n\geq 1$ we get: $$ \sum_{n\geq 1}\frac{\log(n+2)}{n(n+1)}=\int_{0}^{+\infty}\frac{(1-e^{-x})\left(1-\log(1-e^{-x})\right)}{xe^x}\,dx\tag{2} $$ then, by substituting $x=-\log(u)$: $$ \sum_{n\geq 1}\frac{\log(n+2)}{n(n+1)}=\int_{0}^{1}\frac{(u-1)\left(1-\log(1-u)\right)}{\log(u)}\,du\tag{3} $$ and the claim follows by substituting $v=(1-u)$.


I do not think there is a nice closed form but in terms of a derivative of some special zeta function, due to $\log(n+2) = \left.\frac{d}{d\alpha}(n+2)^{\alpha}\right|_{\alpha=0^+}$, but for sure $$ \int_{0}^{1}\frac{-x}{\log(1-x)}\,dx = \int_{0}^{1}\frac{x-1}{\log x}\,dx = \log 2$$ and the Taylor series of $\frac{x}{\log(1-x)}$ depends on Gregory coefficients.

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  • $\begingroup$ Shouldn't that be du in (3)? $\endgroup$ – Paul LeVan May 30 '16 at 6:13
  • $\begingroup$ @PaulLeVan: yep, fixed. $\endgroup$ – Jack D'Aurizio May 30 '16 at 6:40
  • $\begingroup$ Please can you add a comment with your function $f(x)$ in Frullani's theorem. I know what are $a$ and $b$ but now not $f$. Thanks in advance. $\endgroup$ – user243301 May 30 '16 at 7:25
  • $\begingroup$ @user243301: In the application of Frullani's theorem in this answer we have $f(x) = e^{-x}$ and $a = 1, b = n + 2$ $\endgroup$ – Paramanand Singh May 30 '16 at 8:24
  • $\begingroup$ Thanks a lot, I was wrong with my triple $(f,a,b)$. Very thanks @ParamanandSingh $\endgroup$ – user243301 May 30 '16 at 9:01
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Additionnally, by applying Theorem $2$ (here), one obtains a closed form in terms of the poly-Stieltjes constants.

Proposition. We have

$$ \begin{align} \int_0^1\frac{x\,(\ln x-1)}{\ln(1-x)}\,dx= \ln 2+\gamma_1(2,0)-\gamma_1(1,0) \tag1 \\\\\sum_{n=1}^\infty\frac{\ln(n+2)}{n\,(n+1)}=\ln 2+\gamma_1(2,0)-\gamma_1(1,0) \tag2 \end{align}$$

where

$$ \gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right). $$

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

Another possible integral representation: \begin{align} \color{#f00}{\sum_{n = 1}^{\infty}{\ln\pars{n + 2} \over n\pars{n + 1}}} & = \sum_{n = 1}^{\infty}{1 \over n\pars{n + 1}}\ \overbrace{\bracks{\pars{n + 1}\int_{0}^{1}{\dd t \over 1 + \pars{n + 1}t}}} ^{\ds{\ln\pars{n + 2}}} \\[3mm] & = \int_{0}^{1} \sum_{n = 0}^{\infty}{1 \over \pars{n + 2 + 1/t}\pars{n + 1}}\,{\dd t \over t} = \int_{0}^{1} {\Psi\pars{2 + 1/t} - \Psi\pars{1} \over 1 + 1/t}\,{\dd t \over t} \\[3mm] & \stackrel{t\ \to 1/t}{=}\ \color{#f00}{\int_{1}^{\infty} {\Psi\pars{2 + t} - \Psi\pars{1} \over t + 1}\,{\dd t \over t} = \gamma\ln\pars{2} + \int_{1}^{\infty} {\Psi\pars{2 + t} \over t\pars{t + 1}}\,\dd t} \end{align}

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