How can I prove that the sum has value $1$?

Question

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 $1$ ?

Answer 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

$$\sum_{k=1}^n -\frac{\ln(1-\frac{1}{(k+1)^2})}{\ln(2)}=\frac{\ln(\frac{2n+2}{n+2})}{\ln(2)}$$

Answer 2

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 $$\sum_{n\geq1}\log\left(1-\frac{1}{\left(n+1\right)^{2}}\right)=-\log\left(2\right).$$

Answer 3

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

Answer 4

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

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

\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.

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