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Find $$\lim_{n \to \infty}\frac{\sin 1+2\sin \frac{1}{2}+\cdots+n\sin \frac{1}{n}}{n}.$$

This is a recent exam question, which I couldn't figure out in the exam. My guess is it doesn't exist.

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closed as off-topic by GNUSupporter 8964民主女神 地下教會, Namaste, Ethan Bolker, José Carlos Santos, Parcly Taxel Mar 13 '18 at 23:34

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    $\begingroup$ That is a neat exam question :) $\endgroup$ – Dominic Michaelis May 13 '13 at 5:46
  • $\begingroup$ I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 12 '18 at 16:46
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For small $|x|$ we have $\sin(x)\approx x$. Hence \[ i \cdot \sin\frac{1}{i} \approx 1 \] for big values of $i$. Hence \[ \frac{\sum_{i=1}^n i\cdot \sin\frac{1}{i} }{n}\approx \frac{n}{n}=1\]

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    $\begingroup$ Thanks. Seems so easy now. Should have figured it out. $\endgroup$ – user67773 May 13 '13 at 5:46
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    $\begingroup$ I also thought of that, but even if it is intuitive I don't think it's really rigorous.. Could you provide a more detalied answer? $\endgroup$ – Ant Nov 14 '13 at 17:39
  • $\begingroup$ @ant think of the power series of $\sin$, due to taylor we know $\sin(x)=x-\frac{\xi^3}{6}$ where $\xi\in (-x,x)$ $\endgroup$ – Dominic Michaelis Nov 14 '13 at 19:19
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    $\begingroup$ I meant, you substitute $ i \cdot \sin{\frac{1}{i}}$ with 1 for every i, but that only works for "big enough" i... isn't this something to take into consideration? $\endgroup$ – Ant Nov 14 '13 at 19:44
  • $\begingroup$ @Ant nope because of taylor we can write it as $$\sum_{i=1}^n i \cdot \sin\left( \frac{1}{i}\right) = \sum_{i=1}^n i \cdot \left( \frac{1}{i} -\frac{1}{\xi^3}\right) = \sum_{i=1}^n 1- \frac{i}{6 \cdot \xi^3}$$ And hence $$\sum_{i=1}^n i \cdot \sin\left( \frac{1}{i}\right) > \sum_{i=1}^n 1- \frac{1}{6 i^2}$$ $\endgroup$ – Dominic Michaelis Nov 15 '13 at 6:08
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Another more general approach:

$$a_n\xrightarrow[n\to\infty]{} a\implies \frac{a_1+...+a_n}n\xrightarrow[n\to\infty]{}a$$

And since

$$\lim_{n\to\infty}\,n\,\sin\frac1n=\lim_{n\to\infty}\frac{\sin\frac1n}{\frac1n}=1\;\;\ldots\ldots$$

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    $\begingroup$ These are the so called Cesaro averages. If a series converges the Cesaro averages converge to the same value. $\endgroup$ – Georgy Nov 23 '15 at 16:07
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Or we can use Stolz–Cesàro theorem to find that $$\lim_{n \rightarrow \infty}\frac{\sin 1+2\sin \frac{1}{2}+\cdots+n\sin \frac{1}{n}}{n}=\lim_{n\to \infty}\frac{(n+1)\sin {\frac{1}{n+1}}}{1}=1.$$

This is similar with @DonAntonio's solution.

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Start by writing it in summation form. That is,

$$ \lim_{n\to\infty} \sum_{i=1}^n \frac{i\sin(1/i)}{n} $$ Now, $\sin(1/i)<1/i$ for $i\in\mathbb{N}$. Furthermore, $\sin(1/i)>\frac1i-\frac1{6i^3}$ for $i\in\mathbb{N}$. Now, we squeeze the result by noting that $$ \lim_{n\to\infty} \sum_{i=1}^n \frac1n = 1 $$ and $$ \lim_{n\to\infty} \sum_{i=1}^n \frac{1-\frac1{6i^2}}n = \lim_{n\to\infty} \sum_{i=1}^n \frac1n-\frac1{6i^2n} = 1 $$ Therefore, we have that $$ \lim_{n\to\infty} \sum_{i=1}^n \frac{i\sin(1/i)}{n}=1 $$

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  • $\begingroup$ Because $\sin(x)=x-\frac{x^3}6+O(x^5)$, and so $\sin(x)/x = 1-\frac{x^2}6+O(x^4)$. $\endgroup$ – Glen O May 13 '13 at 5:48
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    $\begingroup$ Nevermind I didn't read carefully enough $\endgroup$ – Frudrururu May 13 '13 at 6:15
  • $\begingroup$ @Wishingwell the dummy variable is $i$ and not $n$. $n$ is a constant in the expression. $\endgroup$ – Milind May 13 '13 at 6:17
  • $\begingroup$ @Wishingwell : $$\sum_{i=1}^n\frac1n=\frac1n\sum_{i=1}^n1=\frac1n\cdot n=1$$ $\endgroup$ – DonAntonio May 13 '13 at 6:20