Tessellate S.T.E.M.S (2019) - Mathematics - Category C - Set 1 - Objective Problem 5

Calculus Level 5

Let { x n } n ∈ N \{x_n\}_{n \in \mathbb N} be the sequence defined as x n = sin ⁡ ( 2 π n ! e ) ∀ n ∈ N x_n = \sin(2 \pi n! e)\ \forall n \in \mathbb{N} . Compute lim ⁡ n → ∞ x n \displaystyle \lim_{n \to \infty} x_n .

does not exist 0 1 e \dfrac{1}{e} 1

Tessellate S.T.E.M.S. Mathematics by Tessellate S.T.E.M.S. Ma… , 26

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2 solutions

Brian Moehring
Oct 15, 2018

This is a slightly simpler version of this problem .

In particular, all we need to show is that 0 ≤ n ! e − ⌊ n ! e ⌋ = ∑ k = n + 1 ∞ n ! k ! ≤ ∑ k = 1 ∞ 1 ( n + 1 ) k = 1 n → n → ∞ 0 0 \leq n!e - \lfloor n!e\rfloor = \sum_{k=n+1}^\infty \frac{n!}{k!} \leq \sum_{k=1}^\infty \frac{1}{(n+1)^k} = \frac{1}{n} \xrightarrow{n\to\infty} 0 as then lim ⁡ n → ∞ sin ⁡ ( 2 π n ! e ) = lim ⁡ n → ∞ sin ⁡ ( 2 π ( n ! e − ⌊ n ! e ⌋ ) ) = sin ⁡ ( 0 ) = 0 \lim_{n\to\infty} \sin\left(2\pi n!e\right) = \lim_{n\to\infty} \sin\Big(2\pi \big(n!e - \lfloor n!e\rfloor\big)\Big) = \sin(0) = \boxed{0}

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This is a problem asked in iitjee sample exams. I solved it long back. It appears to be impossible thing when we look at it for the first time.

Srikanth Tupurani - 2 years, 7 months ago
Younes Bouhafid
Oct 17, 2018

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