Infinite Seriesss

Algebra Level 3

1 5 + 2 5 + 3 5 + 4 5 + 5 5 + 6 5 + 7 5 + . . . 1 5 + 3 5 + 5 5 + 7 5 + . . . = \frac{1^{-5}+2^{-5}+3^{-5}+4^{-5}+5^{-5}+6^{-5}+7^{-5}+...}{1^{-5}+3^{-5}+5^{-5}+7^{-5}+...}=

\infty 1 1 32 33 \frac{32}{33} 31 32 \frac{31}{32} 32 31 \frac{32}{31}

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2 solutions

Chew-Seong Cheong
Sep 14, 2019

Q = 1 5 + 2 5 + 3 5 + 4 5 + 5 5 + 6 5 + 7 5 + 1 5 + 3 5 + 5 5 + 7 5 + Riemann zeta function ζ ( n ) = k = 1 1 k n = ζ ( 5 ) 1 5 + 2 5 + 3 5 + ( 2 5 + 4 5 + 6 5 + ) = ζ ( 5 ) 1 5 + 2 5 + 3 5 + 2 5 ( 1 5 + 2 5 + 3 5 + ) = ζ ( 5 ) ζ ( 5 ) 2 5 ζ ( 5 ) = 1 1 1 32 = 32 31 \begin{aligned} Q & = \frac {\color{#3D99F6}1^{-5} + 2^{-5} + 3^{-5} + 4^{-5} + 5^{-5} + 6^{-5} + 7^{-5} + \cdots}{{1^{-5} + 3^{-5} + 5^{-5} + 7^{-5} + \cdots}} & \small \color{#3D99F6} \text{Riemann zeta function }\zeta(n) = \sum_{k=1}^\infty \frac 1{k^n} \\ & = \frac {\zeta(5)}{1^{-5} + 2^{-5} + 3^{-5} + \cdots - \left(2^{-5} + 4^{-5} + 6^{-5} + \cdots \right)} \\ & = \frac {\zeta(5)}{1^{-5} + 2^{-5} + 3^{-5} + \cdots - 2^{-5} \left(1^{-5} + 2^{-5} + 3^{-5} + \cdots \right)} \\ & = \frac {\zeta(5)}{\zeta(5) - 2^{-5} \zeta(5)} = \frac 1{1-\frac 1{32}} = \boxed {\frac {32}{31}} \end{aligned}


Reference: Riemann zeta function

Let the sum be S S . Then S = 1 + S 32 S=1+\dfrac{S}{32} (breaking even and odd terms of the numerator and taking 2 5 2^{-5} common from the even terms) , or S = 32 31 S=\dfrac{32}{31}

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