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Find the limit of $$ \lim_{n \to \infty} \left( \sin \frac{n}{n^2+1^2} + \sin \frac{n}{n^2+2^2} + \dotsb + \sin \frac{n}{n^2+n^2} \right) $$

I think it is a Riemann Integral but I didn't find the expression of Riemann Integral.

Mike Pierce
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james brown
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2 Answers2

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For $n\gg 0$, $\frac{n}{n^2+k^2}\approx 0$ and hence $\frac{\sin\frac{n}{n^2+k^2}}{\frac n{n^2+k^2}}\approx 1$. This allows us to forget about the sine. Then the summands become $$ \frac{n}{n^2+k^2}=\frac1n\cdot\frac{n^2}{n^2+k^2}=\frac1n\frac{1}{1+(k/n)^2}=\frac{f(k/n)}n$$ with $f(x)=\frac1{1+x^2}$, and there you have your Riemann sum

Hagen von Eitzen
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    More specifically, since $\sin{x} = x + O(x^3)$ and $\frac{n}{n^2+k^2}<\frac{1}{n}$ for all $k$, it follows that each term is within $O(\frac{1}{n^3})$ of $\frac{n}{n^2+k^2}$, and hence the sum is within $O(\frac{1}{n^2})$ of the corresponding sum. – Joey Zou Nov 03 '15 at 15:23
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    I felt this is not a rigorous proof, since the number of summands is not fixed. You have to take account in the error when you approximate $\sin x$ by $x$. – Zhanxiong Nov 03 '15 at 15:25
  • @Solitary Yeah, but not really, because the summands are all positive: For $0<\epsilon<1$ we know that for $n>n_0(\epsilon)$ we have $0<(1-\epsilon)\frac{f(k/n)}n<\sin\frac{n}{n^2+k^2}<(1+\epsilon)\frac{f(k/n)}n$ and hence $(1-\epsilon)\frac1n\sum_k f(k/n)<\sum_k\sin\frac{n}{n^2+k^2}<(1+\epsilon)\frac1n\sum_k f(k/n)$. – Hagen von Eitzen Nov 03 '15 at 17:06
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Use the inequality for $x \in (0, \pi/2)$: $$x\cos x < \sin x < x.$$ The upper bound of the summation is $$\sum_{i = 1}^n \frac{n}{n^2 + i^2}.\tag{1}$$ While since $\cos x$ is decreasing on $(0, \pi/2)$, a lower bound of the summation is $$\cos\left(\frac{n}{n^2 + 1}\right)\sum_{i = 1}^n \frac{n}{n^2 + i^2}. \tag{2}$$ Since $\lim_{n \to \infty} \cos(n/(n^2 + 1)) = 1$, $(1)$ and $(2)$ have the same limit as $n \to \infty$. The squeeze principle then identifies the limit of the original sum is the same as the limit of $(1)$, which you can handle it with Riemann sum.

Zhanxiong
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