[MAT] Séries

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$S=\frac { 1 }{ 1\cdot 3\cdot 5 } +\frac { 1 }{ 7\cdot 9\cdot 11 } +\cdot \cdot \cdot  +\frac { 1 }{ (6k-5)(6k-3)(6k-1) } + \cdot \cdot \cdot$

Mostre que

$S=\frac { 1 }{ 16 } { ln{ 3 } }$

Solução de Víctor Domene:

 Por frações parciais:

$ \frac{1}{(6k-5)(6k-3)(6k-1)} = \frac{1}{8(6k-5)} + \frac{1}{8(6k-1)} - \frac{1}{4(6k-3)} $

Escrevendo em função de integrais:

$ \frac{1}{(6k-5)(6k-3)(6k-1)} = $

$\frac{1}{8} \int_{0}^1{x^{6k-6}} + \frac{1}{8} \int_{0}^1{x^{6k-2}} - \frac{1}{4}\int_{0}^1{x^{6k-4}} $

$ 8\sum_{k=1}^{\infty}{\frac{1}{(6k-5)(6k-3)(6k-1)}} =$ 

$\sum _{ k=1 }^{ \infty  }{ { \int _{ 0 }^{ 1 }{ x^{ 6k-6 } } +\int _{ 0 }^{ 1 }  }{ x^{ 6k-2 } }-2\int _{ 0 }^{ 1 }{ x^{ 6k-4 } }  }  $

$ 8\sum_{k=1}^{\infty}{\frac{1}{(6k-5)(6k-3)(6k-1)}} = $

$\int_{0}^1{(1+x^4-2x^2)} \sum_{k=1}^{\infty}{x^{6k-6}} $

$ 8\sum_{k=1}^{\infty}{\frac{1}{(6k-5)(6k-3)(6k-1)}} = $

$\int_{0}^{1}{(x^2-1)}^2 \frac{1}{1-x^6} $

$ 8\sum_{k=1}^{\infty}{\frac{1}{(6k-5)(6k-3)(6k-1)}} = \int_{0}^1{\frac{1-x^2}{1+x^2+x^4}} $

resolvendo essa integral...

$ A = \int_{0}^{1}{\frac{1-x^2}{1+x^2+x^4} \;dx} $

Por frações parciais:

$ A = \int_{0}^{1}{\frac{1-2x}{2(x^2-x+1)} + \frac{2x+1}{2(x^2+x+1)}} $

$ 2A = \int_{0}^{1}{\frac{1-2x}{x^2-x+1}} + \int_{0}^{1}{\frac{2x+1}{x^2+x+1}} $

$ t = x^2 - x +1 \iff dx = \frac{dt}{2x-1} $

$ s = x^2 + x + 1 \iff dx = \frac{ds}{2x+1} $

$ 2A = \int_{0}^{1}{\frac{1-2x}{t}} \frac{dt}{2x-1} + \int_{0}^{1}{\frac{2x+1}{s}} \frac{ds}{2x+1} $

$ 2A = -\int_{0}^{1}\frac{t}{dt} + \int_{0}^{1}\frac{s}{ds} $

$ 2A = ln(s/t) $

$ A = \frac{1}{2}\; ln \left( \frac{x^2+x+1}{x^2-x+1} \right) $

calculando A de 0 a 1:

$ A = \frac{1}{2} \; ln(3) - \frac{1}{2}\; ln(1) = \frac{ln(3)}{2} $

portanto:

$ 8\sum_{k=1}^{\infty}{\frac{1}{(6k-5)(6k-3)(6k-1)}} = \frac{ln(3)}{2} $

$\boxed{ \sum_{k=1}^{\infty}{\frac{1}{(6k-5)(6k-3)(6k-1)}} = \frac{ln(3)}{16}} $