How can we show that :- $\sum_{n=1}^{\infty}\left({\phi^2+\gamma\over n+1}-{\phi+\gamma\over n}-\ln{n+1\over n}\right)=-\phi^2$

Question

How can we prove $(1)$?

$$\sum_{n=1}^{\infty}\left({\phi^2+\gamma\over n+1}-{\phi+\gamma\over n}-\ln{n+1\over n}\right)=-\phi^2\tag1$$

$\phi$;Golden ratio

$\gamma$;Euler's constant

an attempt:

Using $$\sum_{n=1}^{\infty}\left({1\over n}-\ln{n+1\over n}\right)=\gamma\tag2$$

$(2)-(1)$

$$(\gamma+\phi^2)\sum_{n=1}^{\infty}\left({1\over n}-{1\over n+1}\right)=\gamma+\phi^2\tag3$$

Telescope sum $$\sum_{n=1}^{\infty}\left({1\over n}-{1\over n+1}\right)=1\tag4$$

This is not a proof. Any help.

Answer 1

$$ \begin{align} \phi^2 &= \phi+1 \\[4mm] \color{red}{S} &= \sum_{n=1}^{\infty}\left(\frac{\phi^2+\gamma}{n+1}-\frac{\phi+\gamma}{n}-\log{\frac{n+1}{n}}\right) \\[2mm] &= \sum_{n=1}^{\infty}\left(\frac{\phi^2+\gamma}{n+1}-\frac{\phi\color{red}{+1}+\gamma\color{red}{-1}}{n}-\log{\frac{n+1}{n}}\right) \\[2mm] &= \sum_{n=1}^{\infty}\left(\frac{\phi^2+\gamma}{n+1}-\frac{\phi^2+\gamma}{n}+\frac{1}{n}-\log{\frac{n+1}{n}}\right) \\[2mm] &= \sum_{n=1}^{\infty}\left(\frac{\phi^2+\gamma}{n+1}-\frac{\phi^2+\gamma}{n}\right)+\sum_{n=1}^{\infty}\left(\frac{1}{n}-\log{\frac{n+1}{n}}\right) \\[2mm] &= -\left(\phi^2+\gamma\right)\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right)+\,\sum_{n=1}^{\infty}\left(\frac{1}{n}-\log{\frac{n+1}{n}}\right) \\[2mm] &= -\phi^2-\gamma+\gamma=\color{red}{-\phi^2} \end{align} $$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\sum_{n = 1}^{\infty}\bracks{{\phi^{2} + \gamma \over n + 1} - {\phi + \gamma \over n} - \ln\pars{n + 1 \over n}} = -\phi^{2}\,,\qquad \left\{\begin{array}{rl} \ds{\phi:} & \ds{Golden\ Ratio.} \\[1mm] \ds{\gamma:} & \ds{Euler-Mascheroni\ Constant.} \end{array}\right.}$

With $\ds{N \geq 1}$: \begin{align} &\sum_{n = 1}^{N}\bracks{{\phi^{2} + \gamma \over n + 1} - {\phi + \gamma \over n} - \ln\pars{n + 1 \over n}} \\[5mm] = &\ \pars{\phi^{2} + \gamma}\pars{H_{N} + {1 \over N + 1} - 1} - \pars{\phi + \gamma}H_{N} - \bracks{\sum_{n = 2}^{N + 1}\ln\pars{n} - \sum_{n = 1}^{N}\ln\pars{n}} \\[5mm] = &\ \overbrace{\pars{\phi^{2} - \phi}}^{\ds{=\ 1}}\ H_{N} + {\phi^{2} + \gamma \over N + 1}- \phi^{2} - \gamma - \ln\pars{N + 1} \\[5mm] = &\ \overbrace{\braces{\vphantom{\Large A}\bracks{\vphantom{\large A}H_{N} - \ln\pars{N + 1}} - \gamma}} ^{\ds{\stackrel{\mrm{as}\ N\ \to\ \infty}{\to}\,\,\, 0}}\ +\ {\phi^{2} + \gamma \over N + 1} - \phi^{2}\,\,\, \stackrel{\mrm{as}\ N\ \to\ \infty}{\to}\,\,\, \bbx{\ds{-\phi^{2}}} \end{align}