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

Question

Cómo demostrar que $$ \sum_{n = 0}^{\infty} \left(\frac{1}{3n + 1} - \frac{1}{3n + 2}\right) = \frac{\pi}{3\sqrt{3}} $$

Answer 1

Esto es $$\sum_{n=0}^\infty\int_0^1(x^{3n}-x^{3n+1})\,dx =\int_0^1\sum_{n=0}^\infty(x^{3n}-x^{3n+1})\,dx$$ (convergencia monótona). Esto es igual a $$\int_0^1\frac{1-x}{1-x^3}\,dx=\int_0^1\frac{dx}{1+x+x^2}$$ que pueden ser evaluados por métodos de "nivel A".

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\sum_{n = 0}^{\infty}\pars{{1 \over 3n + 1} - {1 \over 3n + 2}}} = {1 \over 3}\sum_{n = 0}^{\infty}\pars{{1 \over n + 1/3} - {1 \over n + 2/3}} \\[5mm] = &\ {1 \over 3}\bracks{\Psi\pars{2 \over 3} - \Psi\pars{1 \over 3}} = {1 \over 3}\,\bracks{\pi\cot\pars{\pi\,{1 \over 3}}} \label{1}\tag{1} \\[5mm] = &\ \bbx{\pi \over 3\root{3}} \approx 0.6046 \end{align}

\eqref {1} es Fórmula de reflexión de Euler para el Función Digamma $\ds{\Psi}$ .