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Calculus Level 4

∑ n = 1 ∞ ∑ p = 2 x 1 n p < 2016 \displaystyle \sum _{ n=1 }^{ \infty }\sum _{ p=2 }^{ x }{ \frac { 1 }{ { n }^{ p } } } <2016

Find the largest integer value of x x that satisfies the given inequality.

This problem is original.
Picture credits: Wildschönau feiert Neues Jahr 07 by Ximeg, Wikipedia
4030 2015 2016 1482 1172

Jonas Katona by Jonas Katona , 22, USA

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1 solution

Aman Rajput
May 15, 2021

Using double summation property, change the order as upper limits are constant. ∑ p = 2 x ∑ n = 1 ∞ 1 n p \sum_{p=2}^{x}\sum_{n=1}^{\infty}\frac{1}{n^p} = ∑ p = 2 x ζ ( p ) =\sum_{p=2}^{x}\zeta(p) Solving this equation by substituting all values of x one by one as given in options we observed x = 2016 x=2016

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