como se calcula

Question

¿Cómo se calcula el valor de

PS

Sé que esta serie convergerá ... pero no sé cómo calcular el límite.

Answer 1

Sugerencia Se puede observar que $$ \ sum_ {j = 0} ^ \ infty \ frac {(- 1) ^ j} {3j + ​​+ 1} = \ sum_ {j = 0} ^ \ infty \ int_0 ^ 1 (-1) ^ jx ^ {3j} dx = \ int_0 ^ 1 \ frac {dx} {1 + x ^ 3} = \ frac {\ pi} {3 \ sqrt {3}} + \ frac {1} {3} \ cdot \ ln 2 $$ donde hemos usado la fracción parcial descomposición $$ \ frac {1} {1 + x ^ 3} = \ frac {-x +2} {3 \ left (x ^ 2-x +1 \ right )} + \ frac {1} {3 (x +1)}. $$

Answer 2

Insinuación:

Necesitamos$$\sum_{r=0}^\infty\dfrac{(-1)^r}{3r+1}$ $

Ahora$$\sum_{r=0}^\infty\dfrac{(x^3)^r}{3r+1}=\dfrac1x\int\sum_{r=0}^\infty x^{3r}dx$ $

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove armada]{\displaystyle{#1}}\,} \newcommand{\llaves}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{j = 0}^{\infty}{\pars{-1}^{\,j} \over 3j + 1} & = \sum_{j = 0}^{\infty}\pars{{1 \over 6j + 1} - {1 \over 6j + 4}} = {1 \over 6}\sum_{j = 0}^{\infty}\pars{{1 \over j + 1/6} - {1 \over j + 2/3}} \\[5mm] & = {1 \over 6}\pars{H_{-1/3} - H_{-5/6}}\qquad\pars{~H_{z}:\ Harmonic\ Number~} \end{align}

La serie de Números que se evalúan con Gauss Digamma Teorema:

$$ \left\{\begin{array}{rcl} \ds{H_{-1/3}} & \ds{=} & \ds{-\gamma + {\root{3} \over 6}\,\pi - {3 \over 2}\,\ln\pars{3}} \\[2mm] \ds{H_{-5/6}} & \ds{=} & \ds{-\gamma - {\root{3} \over 2}\,\pi - {3 \over 2}\,\ln\pars{3} - 2\ln\pars{2}} \\[3mm] &&\gamma:\ Euler\!-\!Mascheroni\ Constant \end{array}\right. $$

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$$ \bbx{\sum_{j = 0}^{\infty}{\pars{-1}^{\,j} \over 3j + 1} = {\raíz{3} \más de 9}\,\pi + {1 \over 3}\,\ln\pars{2}} $$