Calculus 2

Calculus Level 4

∑ n = 0 ∞ 1 ( 2 n + 1 ) 1729 = ( 1 − 2 a ) ζ ( b ) \large \sum _{n=0}^\infty \dfrac 1{( 2n+1)^{1729}} = (1-2^a) \zeta (b)

The equation above holds true for integers a a and b b , find b − a b-a .

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


The answer is 3458.

Rishabh Deep Singh by Rishabh Deep Singh , 23, India

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1 solution

S = ∑ n − 1 ∞ 1 ( 2 n + 1 ) 1729 Let n = 1729 = 1 1 n + 1 3 n + 1 5 n + 1 7 n + ⋯ = 1 1 n + 1 2 n + 1 3 n + 1 4 n + ⋯ − ( 1 2 n + 1 4 n + 1 6 n + 1 8 n + ⋯ ) = 1 1 n + 1 2 n + 1 3 n + 1 4 n + ⋯ − 1 2 n ( 1 1 n + 1 2 n + 1 3 n + 1 4 n + ⋯ ) = ( 1 − 1 2 n ) ( 1 1 n + 1 2 n + 1 3 n + 1 4 n + ⋯ ) = ( 1 − 2 − n ) ζ ( n ) where ζ ( ⋅ ) denotes the Riemann zeta function. \begin{aligned} S & = \sum_{n-1}^\infty \frac 1{(2n+1)^{1729}} & \small \color{#3D99F6} \text{Let }n = 1729 \\ & = \frac 1{1^n} + \frac 1{3^n} + \frac 1{5^n} + \frac 1{7^n} + \cdots \\ & = \frac 1{1^n} + \frac 1{2^n} + \frac 1{3^n} + \frac 1{4^n} + \cdots - \left( \frac 1{2^n} + \frac 1{4^n} + \frac 1{6^n} + \frac 1{8^n} + \cdots \right) \\ & = \frac 1{1^n} + \frac 1{2^n} + \frac 1{3^n} + \frac 1{4^n} + \cdots - \frac 1{2^n}\left( \frac 1{1^n} + \frac 1{2^n} + \frac 1{3^n} + \frac 1{4^n} + \cdots \right) \\ & = \left(1 - \frac 1{2^n}\right) \left( \frac 1{1^n} + \frac 1{2^n} + \frac 1{3^n} + \frac 1{4^n} + \cdots \right) \\ & = \left(1 - 2^{-n} \right) \zeta (n) & \small \color{#3D99F6} \text{where }\zeta (\cdot) \text{denotes the Riemann zeta function.} \end{aligned}

⟹ b − a = n + n = 3458 \implies b-a = n+n = \boxed{3458}

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