A series of series

Calculus Level 4

Given

$A = \displaystyle \sum_{x=1}^{\infty} \dfrac{6}{x^2}$

$B = \displaystyle \sum_{n=1}^{\infty} \dfrac{ (-1)^{n-1}A^{2n-1}}{(2n-1)\pi^{2(2n-1)}}$

and $B$ can be represented in the form $\dfrac{a \pi}{b}$ , where $a, b \in \mathbb{Z}^{+}$ and are coprime, find $a+b$ .

4 Only B doesn't converge 3 Both A and B don't converge 5 7

1 solution

Refaat M. Sayed
Nov 12, 2016

$A=\sum \limits^{\infty }_{x=1}\frac{6}{x^{2}} =6\zeta \left( 2\right) =\pi ^{2}$ Now put $A= \pi^{2}$ to get $B=\sum \limits^{\infty }_{n=1}\frac{\left( -1\right) ^{n-1}}{\left( 2n-1\right) } =\sum \limits^{\infty }_{k=0}\frac{\left( -1\right) ^{k}}{\left( 2k+1\right) } =\arctan \left( 1\right) =\frac{\pi }{4}$ So $\boxed { a+b=5}$

A series of series

1 solution

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