Sum Over Sum

Level 2

1 + 1 2 5 + 1 3 5 + 1 4 5 + 1 5 5 + ⋯ 1 − 1 2 5 + 1 3 5 − 1 4 5 + 1 5 5 − ⋯ \Large \frac{1+\frac{1}{2^{5}}+\frac{1}{3^{5}}+\frac{1}{4^{5}}+\frac{1}{5^{5}}+\cdots}{1-\frac{1}{2^{5}}+\frac{1}{3^{5}}-\frac{1}{4^{5}}+\frac{1}{5^{5}}-\cdots}

Evaluate the expression above.

Bonus : Generalize the expression above by replacing the exponent 5 with an integer n > 1 n > 1 .

32 31 \frac{32}{31} 8 7 \frac{8}{7} 64 63 \frac{64}{63} 16 15 \frac{16}{15}

Aidan Poor by Aidan Poor , 18, USA

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1 solution

Chew-Seong Cheong
Oct 19, 2018

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

For n = 5 n=5 , Q ( 5 ) = 2 4 2 4 − 1 = 16 15 Q(5) = \dfrac {2^4}{2^4-1} = \boxed{\dfrac {16}{15}} .

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