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$$\lim_{n\rightarrow\infty}\sum_{r=1}^n \sin\left(\frac{r}{n} \right)=\lim_{n\rightarrow\infty} \left[\sin\frac{1}{n}+\sin\frac{2}{n}+\cdots+\sin(1)\right]=0+0+\cdots+\sin(1)=1$$

Could anybody explain why this is wrong? I've tried to see why this doesn't work but I don't see why not. Thank you.

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  • $\begingroup$ $\sin\left(\frac{\pi}{2}\right) = 1$. $\sin(1) \ne 1$ $\endgroup$
    – Varun Iyer
    Jan 25, 2015 at 14:56
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    $\begingroup$ $\frac{n-1}{n}$ also goes to zero?? $\endgroup$
    – velut luna
    Jan 25, 2015 at 14:57

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What you do is incorrect.

Consider the same logic applied to

$$ \lim_{n\to\infty} (\frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n})$$

According to your logic, it is $0$, but the limit is actually $\log 2$.

If the number of terms is dependent on $n$, you cannot do an individual limit like that.

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  • $\begingroup$ Thanks that was the answer I was looking for. Any chance you could explain a but further why you cannot split the limit like that when the number of terms depends on $n$? $\endgroup$
    – Redding
    Jan 25, 2015 at 15:00
  • $\begingroup$ @Redding: The example I gave you is a good example. But the simplest reason is that the theorems you use to justify taking individual limits only involve finite number of terms, independent of $n$. There are no such theorems when the number of terms is dependent on $n$. $\endgroup$
    – Aryabhata
    Jan 25, 2015 at 15:03
  • $\begingroup$ Awesome thanks! $\endgroup$
    – Redding
    Jan 25, 2015 at 15:05
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    $\begingroup$ An even simpler example is to add $n$ terms of $1/n$: $(\frac{1}{n}+\frac{1}{n}+\cdots+\frac{1}{n})$. The answer is of course $1$ for all $n$. But according to your argument, it goes to $0$ as $n$ tends to infinity because every term goes to zero. $\endgroup$
    – velut luna
    Jan 25, 2015 at 15:06
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Using the Riemann sum we have

$$\lim_{n\to\infty}\frac1n\sum_{r=1}^n\sin\left(\frac rn\right)=\int_0^1\sin(x)dx=1-\cos(1)\ne0$$ hence we see that the desired limit is infinite.

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Hint:

$${1\over n}\sum_{r=1}^n\sin(r/n)\quad\text{is a Riemann sum for}\quad\int_0^1\sin x\,dx$$

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Hint: since $$\sin{\dfrac{r}{n}}=\dfrac{r}{n}+o(1/n),n\to\infty$$ so $$\lim_{n\to\infty}\sum_{k=1}^{n}\sin{\dfrac{k}{n}}=\lim_{n\to\infty}\dfrac{1+2+\cdots+n}{n}\to\infty$$

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Have a look at the Riemann sum,

$$\lim_{n\to\infty} \frac{1}{n}\sum_{r=1}^n \sin\left(\frac{r}{n}\right) = \int_0^1 \sin x \ dx = 1 - \cos 1 > 0$$

Now, you don't have the $1/n$. So your sum must diverge to $+\infty$

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  • $\begingroup$ Can't say I've seen this result before. So if the sum is independant of $n$ we say the sum diverges to $+\infty$? Why is this? $\endgroup$
    – Redding
    Jan 25, 2015 at 15:06
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    $\begingroup$ We have that $\lim (1/n)(sum) = $a finite number. Hence it must be that $\lim sum$ is infinite, because if were finite, then $\lim (1/n)(sum)$ would be zero. (Downvoter--care to comment?) $\endgroup$
    – Simon S
    Jan 25, 2015 at 15:09
  • $\begingroup$ Hmmm okay that makes sense to me although thinking about anything multplied by $\infty$ scares me a little. Thanks. $\endgroup$
    – Redding
    Jan 25, 2015 at 15:11
  • $\begingroup$ Not sure what you mean. Whatever else, this argument doesn't depend on multiplying anything by $\infty$. $\endgroup$
    – Simon S
    Jan 25, 2015 at 15:14

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