How to find$$\lim_{n\to\infty}n^2\left(\sin(2\pi en!)-\frac{2\pi}{n}\right)$$
I am very confused. I don't know how to show the limit exists, or what it is.
Asked
Active
Viewed 925 times
4
absent1935
- 175
-
Can L'Hopital's Rule be applied here? – Bryan Mar 24 '13 at 04:20
-
@isophyxen No. What you want is Stolz Cesaro. – Pedro Mar 25 '13 at 15:23
-
Related: http://math.stackexchange.com/questions/492179 – Watson Jan 28 '17 at 19:05
1 Answers
10
Recall that $e = 1 + \dfrac1{1!} + \dfrac1{2!} + \dfrac1{3!} + \cdots + \dfrac1{n!} + \dfrac1{(n+1)!} + \dfrac1{(n+2)!} + \cdots$. Hence, $$n! e = \text{Integer} + \dfrac1{n+1} + \dfrac1{(n+1)(n+2)} + \dfrac1{(n+1)(n+2)(n+3)} + \cdots$$ Hence, \begin{align} \sin(2\pi n!e) & = \sin\left(\text{Integer} \times 2 \pi + \dfrac{2\pi}{n+1} + \dfrac{2 \pi}{(n+1)(n+2)} + \mathcal{O}(1/n^3) \right)\\ & = \sin\left(\dfrac{2\pi}{n+1} + \dfrac{2 \pi}{(n+1)(n+2)} + \mathcal{O}(1/n^3) \right)\\ & = \dfrac{2 \pi}{n+1} + \dfrac{2 \pi}{(n+1)(n+2)} + \mathcal{O}(1/n^3) \end{align} Hence, $$\sin(2 \pi n!e) - \dfrac{2 \pi}{n} = -\dfrac{2 \pi}{n(n+1)} + \dfrac{2 \pi}{(n+1)(n+2)} + \mathcal{O}(1/n^3)$$ Hence, $$n^2 \left(\sin(2 \pi n!e) - \dfrac{2 \pi}{n}\right) = 2 \pi \left(-\dfrac{n}{n+1} + \dfrac{n^2}{(n+1)(n+2)} + \mathcal{O}(1/n)\right)$$ Hence, $$\lim_{n \to \infty} n^2 \left(\sin(2 \pi n!e) - \dfrac{2 \pi}{n}\right) = 2 \pi \times 0 = 0$$
jk123
- 143