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Number Theory Level 3

x = 1 + 1 2 2 + 1 3 2 + 1 4 2 + ⋯ + 1 201 5 2 \large x=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots+\frac{1}{2015^2} What is ⌈ x ⌉ \lceil x \rceil ?

Note: ⌈ ⋅ ⌉ \lceil \cdot \rceil denotes least integer greater than or equal to x x .

5 4 2 The series doesn't converge 3 1 0

Vilakshan Gupta by Vilakshan Gupta , 19, India

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1 solution

Chan Tin Ping
Dec 31, 2017

x = 1 + ∑ k = 2 2015 1 k 2 < 1 + ∑ k = 2 2015 1 k 2 − 1 = 1 + ∑ k = 2 2015 1 ( k − 1 ) ( k + 1 ) = 1 + ∑ k = 2 2015 1 2 ( 1 k − 1 − 1 k + 1 ) = 1 + 1 2 ( 1 + 1 2 − 1 100 − 1 101 ) = 1.74 \begin{aligned} x&=1+\sum_{k=2}^{2015} \frac{1}{k^2} \\ &<1+\sum_{k=2}^{2015} \frac{1}{k^2-1} \\ &=1+ \sum_{k=2}^{2015} \frac{1}{(k-1)(k+1)} \\ &=1+ \sum_{k=2}^{2015} \frac{1}{2} (\frac{1}{k-1}-\frac{1}{k+1}) \\ &=1+\frac{1}{2}(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}) \\ &=1.74 \end{aligned} Hence, x < 1.74 x<1.74 and it is obvious that 1 < x 1<x . So ⌈ x ⌉ = 2 \lceil x \rceil=\large 2 .

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