Interesting Limit (4)

Calculus Level 3

lim n k = 1 n sin 2 k 1 n 2 = ? \large\lim_{n\to\infty}\sum_{k=1}^n\sin\frac{2k-1}{n^2}=\, ?


The answer is 1.

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1 solution

Brian Lie
Mar 17, 2018

L = lim n k = 1 n sin 2 k 1 n 2 = lim n 1 2 sin 1 n 2 k = 1 n 2 sin 2 k 1 n 2 sin 1 n 2 = lim n 1 2 sin 1 n 2 k = 1 n ( cos 2 k 2 n 2 cos 2 k n 2 ) = lim n 1 2 sin 1 n 2 ( 1 cos 2 n ) = lim n sin 2 1 n sin 1 n 2 = lim n 1 n 2 sin 1 n 2 ( sin 1 n 1 n ) 2 = 1 \begin{aligned} L&=\lim\limits_{n\rightarrow\infty}\sum_{k=1}^n\sin\frac{2k-1}{n^2} \\&=\lim\limits_{n\rightarrow\infty}\frac1{2\sin\frac1{n^2}}\sum_{k=1}^n 2\sin\frac{2k-1}{n^2}\sin\frac1{n^2} \\&=\lim\limits_{n\rightarrow\infty}\frac1{2\sin\frac1{n^2}}\sum_{k=1}^n\left(\cos\frac{2k-2}{n^2}-\cos\frac{2k}{n^2}\right) \\&=\lim\limits_{n\rightarrow\infty}\frac1{2\sin\frac1{n^2}}\left(1-\cos\frac2n\right) \\&=\lim\limits_{n\rightarrow\infty}\frac{\sin^2\frac 1n}{\sin\frac1{n^2}} \\&=\lim\limits_{n\rightarrow\infty}\frac{\frac 1{n^2}}{\sin\frac1{n^2}}\cdot\left(\frac{\sin \frac 1n}{\frac 1n}\right)^2 \\&=\boxed 1 \end{aligned}

@Brian Lie Could u please suggest some good books for Calculus??

Aaghaz Mahajan - 3 years, 2 months ago

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I've read Calculus by James Stewart alone but it's pretty good.

Brian Lie - 3 years, 2 months ago

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