Find the limit!

Calculus Level pending

Let γ = lim ⁡ n → inf ⁡ ∣ sin ⁡ ( 2 π e n ! ) ∣ 1 ln ⁡ ( n ) \gamma=\lim_{n\rightarrow \inf}|\sin (2\pi en!)|^{\frac{1}{\ln (n)}}

Find [ 2015 γ ] [2015\gamma] , the integer part of 2015 γ 2015\gamma .

Note:If you think the limit does not exist, enter 0


The answer is 741.

Cuize Han by Cuize Han , 30, China

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1 solution

Cuize Han
Jan 3, 2015

The key observation is that sin ⁡ ( 2 π e n ! ) = sin ⁡ ( 2 π { e n ! } ) \sin(2\pi en!)=\sin(2\pi \{en!\}) and { e n ! } = 1 n + 1 ( 1 + O ( 1 n ) ) \{en!\}=\frac{1}{n+1}(1+O(\frac{1}{n})) because e = 1 + 1 1 ! + 1 2 ! + . . . + 1 n ! + 1 ( n + 1 ) ! + . . e=1+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}+\frac{1}{(n+1)!}+.. Then it's easy to find γ = 1 e \gamma=\frac{1}{e}

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