How can I prove that the sum has value $1$? The answer of this question
Typicality of boundedness of entries of continued fraction representations
involves the infinite sum $$\sum_{k=1}^\infty -\frac{\ln(1-\frac{1}{(k+1)^2})}{\ln(2)}=1$$
Wolfram does not display $1$ if I enter the sum.
The sum converges very slowly :
I worked out
$$\sum_{k=n}^\infty -\frac{\ln(1-\frac{1}{(k+1)^2})}{\ln(2)}\approx\frac{1}{n\ln(2)}$$ for large $n$ as follows :
With the help of Wolfram I found out $-\frac{\ln(1-\frac{1}{(k+1)^2})}{\ln(2)}\approx \frac{1}{k^2\ln(2)}$ for large $k$. So, to show the approximation, I only need to show $$\sum_{k=n}^\infty \frac{1}{k^2}\approx \frac{1}{n}$$ , which follows from the definite integrals squeezing the sum.

Is my argumentation correct ?
How can I show that the sum has value $1$ ?
I found an easier proof thanks to Graubner's answer based on the infinite product idendity

$$\prod_{k=1}^n\frac{(k+1)^2}{(k+1)^2-1}=\frac{2(n+1)}{n+2}$$ which can be easily proved by induction.

Hence, we get the partial sum

$$\sum_{k=1}^n -\frac{\ln(1-\frac{1}{(k+1)^2})}{\ln(2)}=\frac{\ln(\frac{2n+2}{n+2})}{\ln(2)}$$
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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  $\ds{\sum_{k = 1}^{\infty}\braces{%
-\,{\ln\pars{1 - 1/\bracks{k + 1}^{\,2}} \over \ln\pars{2}}} = 1:\ {\large ?}}$.


\begin{align}
&amp;\sum_{k = 1}^{\infty}\braces{%
-\,{\ln\pars{1 - 1/\bracks{k + 1}^{\,2}} \over \ln\pars{2}}} =
-\,{1 \over \ln\pars{2}}\sum_{k = 1}^{\infty}
\ln\pars{k^{2} + 2k \over \bracks{k + 1}^{\,2}}
\\[5mm] = &amp;\
-\,{1 \over \ln\pars{2}}\sum_{k = 1}^{\infty}
\bracks{-\int_{0}^{1}{\dd t \over \pars{k + 1}^{2} - t}} =
{1 \over \ln\pars{2}}\int_{0}^{1}\sum_{k = 0}^{\infty}
{1 \over \pars{k + 2 - \root{t}}\pars{k + 2 + \root{t}}}\,\dd t
\\[5mm] = &amp;\
{1 \over \ln\pars{2}}\int_{0}^{1}
{\Psi\pars{2 - \root{t}} - \Psi\pars{2 + \root{t}} \over -2\root{t}}\dd t
\end{align}
$\ds{\Psi}$ is the
Digamma Function.
Note that $\ds{\Psi\pars{z} \equiv \totald{\ln\pars{\Gamma\pars{z}}}{z}}$. $\ds{\Gamma}$ is the
Gamma Function.

With $\ds{\root{t} \mapsto t}$
:
\begin{align}
&amp;\sum_{k = 1}^{\infty}\braces{%
-\,{\ln\pars{1 - 1/\bracks{k + 1}^{\,2}} \over \ln\pars{2}}} =
-\,{1 \over \ln\pars{2}}\int_{0}^{1}
\bracks{\Psi\pars{2 - t} - \Psi\pars{2 + t}}\dd t
\\[5mm] = &amp;\
-\,{1 \over \ln\pars{2}}\bracks{\vphantom{\Large A}%
-\ln\pars{\Gamma\pars{2 - t}} - \ln\pars{\Gamma\pars{2 + t}}}_{\ 0}^{\ 1} =
{1 \over \ln\pars{2}}
\ln\pars{\Gamma\pars{1}\Gamma\pars{3} \over \Gamma^{2}\pars{2}} = \bbx{\ds{1}}
\end{align}


  because $\ds{\Gamma\pars{1} = \Gamma\pars{2} = 1}$ and
  $\ds{\Gamma\pars{3} = 2! = \color{#f00}{2}}$.