How can we show that :- $\sum_{n=1}^{\infty}\left({\phi^2+\gamma\over n+1}-{\phi+\gamma\over n}-\ln{n+1\over n}\right)=-\phi^2$ 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.
 A: $$ 
\begin{align} 
\phi^2 &= \phi+1 \\[4mm] 
\color{red}{S} &= \sum_{n=1}^{\infty}\left(\frac{\phi^2+\gamma}{n+1}-\frac{\phi+\gamma}{n}-\log{\frac{n+1}{n}}\right) \\[2mm] 
&= \sum_{n=1}^{\infty}\left(\frac{\phi^2+\gamma}{n+1}-\frac{\phi\color{red}{+1}+\gamma\color{red}{-1}}{n}-\log{\frac{n+1}{n}}\right) \\[2mm] 
&= \sum_{n=1}^{\infty}\left(\frac{\phi^2+\gamma}{n+1}-\frac{\phi^2+\gamma}{n}+\frac{1}{n}-\log{\frac{n+1}{n}}\right) \\[2mm] 
&= \sum_{n=1}^{\infty}\left(\frac{\phi^2+\gamma}{n+1}-\frac{\phi^2+\gamma}{n}\right)+\sum_{n=1}^{\infty}\left(\frac{1}{n}-\log{\frac{n+1}{n}}\right) \\[2mm] 
&= -\left(\phi^2+\gamma\right)\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right)+\,\sum_{n=1}^{\infty}\left(\frac{1}{n}-\log{\frac{n+1}{n}}\right) \\[2mm] 
&= -\phi^2-\gamma+\gamma=\color{red}{-\phi^2} 
\end{align} 
$$ 
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\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}
