Riemann zeta function.

Calculus Level 4

Find the value of ∑ n = 1 ∞ ζ ( 2 n ) − 1 2 2 n \sum _{ n=1 }^{ \infty }{ \frac { \zeta (2n)-1 }{ { 2 }^{ 2n } } } . Answer is in a b \frac { a }{ b } form where a a and b b are relatively prime. Write answer as a + b a+b .


The answer is 7.

Shivamani Patil by shivamani patil , 21, India

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1 solution

Haroun Meghaichi
Sep 5, 2014

∑ n = 1 ∞ ζ ( 2 n ) − 1 2 2 n = ∑ n = 1 ∞ ∑ k = 2 ∞ 1 ( 2 k ) 2 n = ∑ k = 2 ∞ ∑ n = 1 ∞ 1 ( 2 k ) 2 n \sum_{n=1}^{\infty} \frac{\zeta(2n)-1}{2^{2n}} = \sum_{n=1}^{\infty} \sum_{k=2}^{\infty} \frac{1}{(2k)^{2n}} = \sum_{k=2}^{\infty} \sum_{n=1}^{\infty} \frac{1}{(2k)^{2n}} The interchange is justified by the absolute convergence, then the sum is : ∑ k = 2 ∞ 1 2 k 2 − 1 = 1 2 ∑ k = 2 ∞ 1 2 k − 1 − 1 2 k + 1 = 1 6 \sum_{k=2}^{\infty} \frac{1}{2k^2-1} = \frac{1}{2} \sum_{k=2}^{\infty} \frac{1}{2k-1}- \frac{1}{2k+1} = \frac{1}{6}

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