Summation Contest Conceptual Problem-1

Calculus Level 5

∑ n = 2 ∞ ζ ( n ) 2 n = ln ⁡ A \large \sum _{ n=2 }^{ \infty }{ \dfrac { \zeta ( n ) }{ { 2 }^{ n } } } = \ln A

Find A A .

Hint : Try using the generating function .

Notation : ζ ( ⋅ ) \zeta(\cdot) denotes the Riemann zeta function .

View the Brilliant Summation Contest Season-1 for concepts. Try more contest conceptual problems here .


The answer is 2.

Aditya Kumar by Aditya Kumar , 22, India

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2 solutions

Chew-Seong Cheong
Jul 11, 2016

S = ∑ n = 2 ∞ ζ ( n ) 2 n = ∑ n = 2 ∞ ∑ k = 1 ∞ 1 k n 2 n = ∑ n = 2 ∞ ∑ k = 1 ∞ 1 ( 2 k ) n = ∑ k = 1 ∞ ∑ n = 2 ∞ 1 ( 2 k ) n = ∑ k = 1 ∞ 1 ( 2 k ) 2 1 − 1 2 k = ∑ k = 1 ∞ 1 ( 2 k ) 2 − 2 k = ∑ k = 1 ∞ 1 ( 2 k ) ( 2 k − 1 ) = ∑ k = 1 ∞ ( 1 2 k − 1 − 1 2 k ) = 1 − 1 2 + 1 3 − 1 4 + ⋯ = ln ⁡ ( 2 ) \begin{aligned} S & = \sum_{n=2}^\infty \frac {\zeta (n)}{2^n} = \sum_{n=2}^\infty \frac {\sum_{k=1}^\infty \frac 1{k^n}}{2^n} = \sum_{n=2}^\infty \sum_{k=1}^\infty \frac 1{(2k)^n} \\ & = \sum_{k=1}^\infty \sum_{n=2}^\infty \frac 1{(2k)^n} = \sum_{k=1}^\infty \frac {\frac 1{(2k)^2}}{1-\frac 1{2k}} = \sum_{k=1}^\infty \frac 1{(2k)^2-2k} \\ & = \sum_{k=1}^\infty \frac 1{(2k)(2k-1)} = \sum_{k=1}^\infty \left( \frac 1{2k-1} - \frac 1{2k} \right) \\ & = 1 - \frac 12 + \frac 13 - \frac 14 + \cdots = \ln (2) \end{aligned}

⟹ A = 2 \implies A = \boxed{2}

15 Helpful 0 Interesting 0 Brilliant 0 Confused

Relevant wiki: Riemann Zeta Function

As hinted,

∑ k = 2 ∞ ζ ( k ) x k − 1 = − γ − ψ ( 1 − x ) \displaystyle \sum_{k=2}^{\infty} \zeta(k)x^{k-1} = -\gamma-\psi(1-x)

Putting x = 1 2 x=\frac{1}{2} and dividing both sides by 2 and substituting ψ ( 1 2 ) = − 2 ln ⁡ 2 − γ \displaystyle \psi(\frac{1}{2}) = -2\ln 2 - \gamma we have the answer ln ⁡ 2 \boxed{\ln 2}

9 Helpful 0 Interesting 0 Brilliant 0 Confused

HHhahah! nice generating function

Pi Han Goh - 4 years, 11 months ago

Can you provide the proof?

Aditya Kumar - 4 years, 11 months ago

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∑ k ≥ 2 ζ ( k ) ( − x ) k = ∑ k ≥ 2 ∑ s ≥ 1 ( − 1 ) k x k s k \displaystyle \sum_{k \ge 2} \zeta(k)(-x)^k = \sum_{k \ge 2}\sum_{s\ge 1} \frac{(-1)^k x^k}{s^k}

Interchanging summations we have ,

∑ s ≥ 1 ∑ k ≥ 2 ( − x ) k s k = ∑ s ≥ 1 x 2 s 2 1 + x s = ∑ s ≥ 1 x 2 s ( s + x ) \displaystyle \sum_{s \ge 1} \sum_{k \ge 2} \frac{(-x)^k}{s^k} = \sum_{s \ge 1} \frac{\frac{x^2}{s^2}}{1+\frac{x}{s}} = \sum_{s \ge 1} \frac{x^2}{s(s+x)}

So, ∑ k ≥ 2 ζ ( k ) ( − x ) k = x ∑ s ≥ 1 ( 1 s − 1 s + x ) = − x H x \displaystyle \sum_{k \ge 2} \zeta(k)(-x)^{k} = x\sum_{s\ge 1} (\frac{1}{s} - \frac{1}{s+x}) = -x H_x

We know , H x = ψ ( 1 + x ) + γ \displaystyle H_x = \psi(1+x) + \gamma

Substituting we have , ∑ k ≥ 2 ζ ( k ) ( − x ) k = x ( ψ ( 1 + x ) + γ ) \displaystyle \sum_{k \ge 2} \zeta(k)(-x)^{k} = x(\psi(1+x)+\gamma)

Now changing x → − x x \to -x & dividing by x x we have ,

∑ k = 2 ∞ ζ ( k ) x k − 1 = − ψ ( 1 − x ) − γ \displaystyle \sum_{k=2}^{\infty} \zeta(k)x^{k-1} = -\psi(1-x) -\gamma & thus proved.

Aditya Narayan Sharma - 4 years, 11 months ago

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