((1+(1/2^2 )+(1/3^2 )+...)/4^2 )+((1+(1/2^3 )+(1/3^3 )+(1/4^3 )+..)/4^3 )+((1+(1/2^4 )+(1/3^4 )+...)/4^4 )+..=(3/4)log(2)−(π/8)


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Question Number 137059 by Dwaipayan Shikari last updated on 29/Mar/21

$\frac{1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . .}{4^{2}} + \frac{1 + \frac{1}{2^{3}} + \frac{1}{3^{3}} + \frac{1}{4^{3}} + . .}{4^{3}} + \frac{1 + \frac{1}{2^{4}} + \frac{1}{3^{4}} + . . .}{4^{4}} + . . = \frac{3}{4} l o g (2) - \frac{π}{8}$
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