Respuesta :

Medunno13

[tex]1+\frac{1}{6}+\frac{1}{11}+\cdots=\sum^{\infty}_{n=1} \frac{1}{5n-4} \\ \\ \frac{1}{4}+\frac{1}{9}+\cdots=\sum^{\infty}_{n=1} \frac{1}{5n-1} \\ \\ S=\sum^{\infty}_{n=1} \left(\frac{1}{5n-4} -\frac{1}{5n-1} \right) \\ \\ =\frac{\pi}{5} \sqrt{\frac{1}{5} (5+2\sqrt{5})}[/tex]

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