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

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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7  
That is a neat exam question :) –  Dominic Michaelis May 13 '13 at 5:46

4 Answers 4

up vote 21 down vote accepted

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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2  
Thanks. Seems so easy now. Should have figured it out. –  Uma kant May 13 '13 at 5:46
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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? –  Ant Nov 14 '13 at 17:39
    
@ant think of the power series of $\sin$, due to taylor we know $\sin(x)=x-\frac{\xi^3}{6}$ where $\xi\in (-x,x)$ –  Dominic Michaelis Nov 14 '13 at 19:19
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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? –  Ant Nov 14 '13 at 19:44
    
@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 $$\sum_{i=1}^n i \cdot \sin\left( \frac{1}{i}\right) > \sum_{i=1}^n 1- \frac{1}{6 i^2}$$ –  Dominic Michaelis Nov 15 '13 at 6:08

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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10  
(+1) Very nice. –  BU982T May 13 '13 at 7:55
2  
Really Don $+1$ –  El Angel Exterminador May 16 '13 at 15:32

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

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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