Demostrando esta fórmula $1+\sum_{n=0}^{\infty }\frac{1}{\pi \left(2n+\frac{3}{4}\right)\left(2n+\frac{5}{4}\right)}=\sqrt2$

Question

Intenté probar esta fórmula pero no pude hacerlo.

$$1+\sum_{n=0}^{\infty }\frac{1}{\pi \left(2n+\frac{3}{4}\right)\left(2n+\frac{5}{4}\right)}=\sqrt{2}$$

Answer 1

$$ \begin{align} 1+\frac1{4\pi}\sum_{n=0}^\infty\frac1{(n+\frac38)(n+\frac58)} &=1+\frac1\pi\sum_{n=0}^\infty\left(\frac1{n+\frac38}-\frac1{n+\frac58}\right)\tag{1}\\ &=1+\frac1\pi\sum_{n=0}^\infty\left(\vphantom{\frac1{n+\frac38}}\right.\underbrace{\frac1{n+\frac38}}_{n+\frac38\text{ for }n\ge0}+\underbrace{\frac1{-n-1+\frac38}}_{n+\frac38\text{ for }n\le-1}\left.\vphantom{\frac1{n+\frac38}}\right)\tag{2}\\ &=1+\frac1\pi\color{#C00000}{\sum_{n=-\infty}^\infty\frac1{n+\frac38}}\tag{3}\\ &=1+\frac1\pi\color{#C00000}{\pi\cot\left(\frac38\pi\right)}\tag{4}\\[6pt] &=1+\sqrt2-1\tag{5}\\[12pt] &=\sqrt2\tag{6} \end{align} $$ Explicación:
$(1)$ : fracciones parciales
$(2)$ : reescribir los términos y establecer el siguiente paso
$(3)$ : se escribe como una suma sobre $\mathbb{Z}$
$(4)$ : uso $\color{#C00000}{(7)}$ de esta respuesta
$(5)$ : evaluar $\cot\left(\frac38\pi\right)$

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin {align}& \color {#66f}{ \large 1 + \sum_ {n\ =\ 0}^{ \infty }{1 \over \pi\pars {2n + 3/4} \pars {2n + 5/4}}} =1 + {1 \over 4 \pi } \sum_ {n\ =\ 0}^{ \infty }{1 \over \pars {n + 3/8} \pars {n + 5/8}} \\ [5mm]&=1 + {1 \over 4 \pi } \bracks { \Psi\pars {3/8} - \Psi\pars {5/8} \over 3/8 - 5/8} \end {align} donde $\ds{\Psi\pars{z}}$ es el Función Digamma $\color{#000}{\bf 6.3.1}$ .

Entonces \begin {align}& \color {#66f}{ \large 1 + \sum_ {n\ =\ 0}^{ \infty }{1 \over \pi\pars {2n + 3/4} \pars {2n + 5/4}}} =1 + {1 \over \pi } \bracks { \Psi\pars {5 \over 8} - \Psi\pars {3 \over 8}} \end {align}

Con Fórmula de reflexión de Euler $\color{#000}{\bf 6.3.7}$ $\ds{\pars{~\Psi\pars{1 - z} = \Psi\pars{z} + \pi\cot\pars{\pi z}~}}$ lo conseguiremos: \begin {align}& \color {#66f}{ \large 1 + \sum_ {n\ =\ 0}^{ \infty }{1 \over \pi\pars {2n + 3/4} \pars {2n + 5/4}}} =1 + {1 \over \pi } \bracks { \pi\cot\pars { \pi\ ,{3 \over 8}}} =1 + \cot\pars {3 \pi \over 8} \\ [5mm]&=1 + {1 \over \tan\pars {135^{ \circ }/2}} =1 + {1 \over 1/ \pars { \root {2} - 1}} = \color {#66f}{ \large\root {2}}\,. \quad { \tt\pars {~ \mbox {Ver una prueba más abajo}~}. \end {align}

También, \begin {align} -1&= \tan\pars {135^{ \circ }} = {2 \tan\pars {135^{ \circ } /2} \over 1 - \tan ^{2} \pars {135^{ \circ } /2}} \quad\imp\quad \tan ^{2} \pars {135^{ \circ } \over 2} - 2 \tan\pars {135^{ \circ } \over 2} - 1 = 0 \\ [5mm]& \imp\quad\tan\pars {135^{ \circ } \over 2} ={2 + \sqrt {8} \over 2} = 1 + \root {2} = {1 \over \root {2} - 1} \end {align}

Answer 3

$$\begin{align} 1+\sum_{n=0}^{\infty }\frac{1}{\pi \left(2n+\frac{3}{4}\right)\left(2n+\frac{5}{4}\right)} &=1+\frac1{4\pi}\sum_{n=0}^\infty\frac1{\left(n+\frac38\right)\left(n+\frac58\right)}\tag{1}\\ &=1+\frac1\pi\sum_{n=0}^\infty\left(\frac1{n+\frac38}-\frac1{n+\frac58}\right)\tag{2}\\ &=1+\frac1\pi\sum_{n=0}^\infty\int_0^1\left(x^{n-5/8}-x^{n-3/8}\right)\,\mathrm dx\tag{3}\\ &=1+\frac1\pi\int_0^1\left(\frac{x^{-5/8}-x^{-3/8}}{1-x}\right)\,\mathrm dx\tag{4}\\ &=1+\frac{1}{\pi} \Big[\pi\Big(\sqrt2-1\Big)\Big]\tag{5}\\ &=1+\sqrt2-1\tag{6}\\ &=\sqrt2\tag{7}\\ \end{align}$$

$$\large1+\sum_{n=0}^{\infty }\frac{1}{\pi \left(2n+\frac{3}{4}\right)\left(2n+\frac{5}{4}\right)} =\sqrt2 $$

Puedo añadir explicaciones si es necesario