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 $1$ ?

 A: 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
$$\sum_{k=1}^n -\frac{\ln(1-\frac{1}{(k+1)^2})}{\ln(2)}=\frac{\ln(\frac{2n+2}{n+2})}{\ln(2)}$$
A: From the Weierstrass factorization theorem we have $$\frac{\sin\left(\pi x\right)}{\pi x}=\prod_{n\geq1}\left(1-\frac{x^{2}}{n^{2}}\right)
 $$ hence $$\prod_{n\geq1}\left(1-\frac{1}{\left(n+1\right)^{2}}\right)=\prod_{n\geq2}\left(1-\frac{1}{n^{2}}\right)=\lim_{x\rightarrow1}\frac{\sin\left(\pi x\right)}{\pi x\left(1-x^{2}\right)}=\frac{1}{2}
 $$ so $$\sum_{n\geq1}\log\left(1-\frac{1}{\left(n+1\right)^{2}}\right)=-\log\left(2\right).$$
A: HINT:
$$\ln\left(1-\dfrac1{(k+1)^2}\right)=\ln\dfrac{f(k+1)}{f(k)}=\ln f(k+1)-\ln f(k)$$ where $f(m)=\dfrac{m+1}m$
Now  put a few initial values of $k$ to recognize the Telescoping series
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\sum_{k = 1}^{\infty}\braces{%
-\,{\ln\pars{1 - 1/\bracks{k + 1}^{\,2}} \over \ln\pars{2}}} = 1:\ {\large ?}}$.

\begin{align}
&\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] = &\
-\,{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] = &\
{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}
&\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] = &\
-\,{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
  $\ds{\Gamma\pars{3} = 2! = \color{#f00}{2}}$.

