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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    $\begingroup$ That is a neat exam question :) $\endgroup$ – Dominic Michaelis May 13 '13 at 5:46
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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

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


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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

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.


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 $$ \lim_{n\to\infty} \sum_{i=1}^n \frac{i\sin(1/i)}{n}=1 $$

  • $\begingroup$ 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