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 :) $\endgroup$ Commented May 13, 2013 at 5:46
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    $\begingroup$ 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. $\endgroup$ Commented Mar 12, 2018 at 16:46

4 Answers 4


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
    Commented Nov 23, 2015 at 16:07

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
    Commented May 13, 2013 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
    Commented Nov 14, 2013 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 $\xi\in (-x,x)$ $\endgroup$ Commented Nov 14, 2013 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
    Commented Nov 14, 2013 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 $$\sum_{i=1}^n i \cdot \sin\left( \frac{1}{i}\right) > \sum_{i=1}^n 1- \frac{1}{6 i^2}$$ $\endgroup$ Commented Nov 15, 2013 at 6:08

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
    Commented May 13, 2013 at 5:48
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    $\begingroup$ Nevermind I didn't read carefully enough $\endgroup$
    – user43138
    Commented May 13, 2013 at 6:15
  • $\begingroup$ @Wishingwell the dummy variable is $i$ and not $n$. $n$ is a constant in the expression. $\endgroup$ Commented May 13, 2013 at 6:17
  • $\begingroup$ @Wishingwell : $$\sum_{i=1}^n\frac1n=\frac1n\sum_{i=1}^n1=\frac1n\cdot n=1$$ $\endgroup$
    – DonAntonio
    Commented May 13, 2013 at 6:20