If-1-1-2-1-2-2-1-3-2-to-pi-2-6-then-1-1-2-1-3-2-1-5-2-equals-

Question Number 26521 by Samar khan last updated on 26/Dec/17

If \frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \dots to \infty = \frac{π^{2}}{6}, then

\frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + \dots equals

Commented by abdo imad last updated on 26/Dec/17

\sum_{n = 1}^{\propto} \frac{1}{n^{2}} = \sum_{p = 1}^{\propto} \frac{1}{{(2 p)}^{2}} + \sum_{p = 0}^{\propto} \frac{1}{{(2 p + 1)}^{2}}

\Rightarrow \sum_{p = 1}^{\propto} \frac{1}{{(2 p + 1)}^{2}} = (1 - \frac{1}{4}) \sum_{n = 1}^{\propto} \frac{1}{n^{2}} = \frac{3}{4} \frac{π^{2}}{6} = \frac{π^{2}}{8}

Answered by ajfour last updated on 26/Dec/17

(\frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + . .) + \frac{1}{4} (\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + . .) = \frac{π^{2}}{6}

s o \frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + . . = \frac{3}{4} (\frac{π^{2}}{6}) = \frac{π^{2}}{8} .

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