Show that $\lim\limits_{n \to +\infty}(\sin(\frac{1}{n^2})+\sin(\frac{2}{n^2})+\cdots+\sin(\frac{n}{n^2})) = \frac{1}{2}$

Question

Show that the sequence defined as $$x_n = \sin\left(\frac{1}{n^2}\right)+\sin\left(\frac{2}{n^2}\right)+\cdots+\sin\left(\frac{n}{n^2}\right)$$ converges to $\frac{1}{2}$.

My attempt was to evaluate this limit by using squeeze theorem. I managed to show that $x_n < \frac{n+1}{2n}$ by using $\sin(x) < x$, but I haven't been able to find a sequence smaller than $x_n$ that also converges to $\frac{1}{2}$. I tried showing by induction that $x_n > \frac{1}{2}-\frac{1}{n}$, but I got nowhere with that.

Any help would be appreciated.

Answer 1

Start with $x-x^3/6<\sin(x)<x$; there are many nice proofs here.

Then we have $$ \sum_{k=1}^{n}(k/n^2)-(k/n^2)^3/6 <\sum_{k=1}^{n}\sin(k/n^2) < \sum_{k=1}^{n}k/n^2 $$Using the fact that $\sum_{k=1}^{n}k^p = \frac{n^{p+1}}{p+1}+\text{ lower order terms}$, we have $$ \frac{n^2+\cdots}{2 n^2}- \frac{n^4+\cdots}{24 n^6} <\sum_{k=1}^{n}\sin(k/n^2) < \frac{n^2+\cdots}{2 n^2}, $$and the result follows by squeezing. We can actually do a bit better with the sums using closed-form identites: $$ \frac{n^2+n}{2 n^2}- \frac{n^4+2n^3+n^2}{24 n^6} <\sum_{k=1}^{n}\sin(k/n^2) < \frac{n^2+n}{2 n^2} $$

Answer 2

For a given $x_n$, the largest argument inside of $\sin$ is $1/n$. You can bound $\sin$ on $[0,1/n]$ by below with the linear function that passes by $(0,0)$ and $(1/n,\sin(1/n)$. You can therefor bound $x_n$ by below with

$$y_n=(n\sin(1/n))(\sum_{k=1}^n \frac{k}{n^2})=\frac{\sin(1/n)(n+1)}{2}. $$

As $n \to \infty$, $y_n \to 1/2$.

Hope this helps!

Answer 3

If you do not know Taylor series, use the equivalent $$\sin(x)\sim x \qquad \text{when} \qquad x \text{ is small}$$

So $$x_n \sim \frac {1} {n^2}+\frac {2} {n^2}+\cdots+\frac {n-1} {n^2}+\frac {n} {n^2}=\frac {1+2+\cdots+(n-1)+n} {n^2}$$

I am sure that you can take it from here.

Answer 4

If questioner or others have not encountered Taylor series yet, it is notable that an elementary trig identity (for the sin of a sum) applied to the $k$th term in the series can be shown to be $k$ times the first term in the series, thus it can be deduced that the limit of the series is equal to the limit of $$\frac{n(n+1)}{2}\cdot \sin(1/n^2)$$ Therefore, the limit becomes $$\frac{1}{2}\cdot \lim (u\cdot \sin(1/u)) = \frac{1}{2}$$

Answer 5

Because $\sin x$ is convex on $x \in (0, \pi)$ then the sum of $\sin (kx)$ inside that interval will be not less than $n$ times the linear average on the secant $(\sin x+sin (nx))/2$ and not greater than $n$ times the linear average on the tangent $n \sin((n+1)/2 x)$, so in this case

$$ \eqalign{ & n\left( {{{\sin \left( {1/n^{\,2} } \right) + \sin \left( {n/n^{\,2} } \right)} \over 2}} \right) \le x_{\,n} = \cr & = \sin \left( {{1 \over {n^{\,2} }}} \right) + \sin \left( {{2 \over {n^{\,2} }}} \right) + \cdots + \sin \left( {{n \over {n^{\,2} }}} \right) \le n\sin \left( {{{\left( {n + 1} \right)/2} \over {n^{\,2} }}} \right) \cr} $$

Answer 6

Another method, because it wasn't mentioned, is using Lagrange's trigonometric identities (which should not be difficult to prove using complex numbers, like here) $$\sum _{k=1}^{n}\sin(k\theta )= {\frac {1}{2}}\cot {\frac {\theta }{2}}-{\frac {\cos \left(\left(n+{\frac {1}{2}}\right)\theta \right)}{2\sin \left({\frac {\theta }{2}}\right)}}=...$$ with $\theta=\frac{1}{n^2}$. Given $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$ $$...={\frac {1}{2}}\cot {\frac {\theta }{2}}-\frac{1}{2}\left(\cos\left(n\theta\right)\cot\frac{\theta}{2}-\sin(n\theta)\right)=\\ \frac{1}{2}\cdot\left(\sin(n\theta)+\cot\frac{\theta}{2}\cdot\left(1-\cos\left(n\theta\right)\right)\right)=\\ \frac{1}{2}\cdot\left(\sin\frac{1}{n}+\cot\frac{1}{2n^2}\cdot\left(1-\cos\frac{1}{n}\right)\right)=...$$ considering that $1-\cos{a}=2\sin^2\frac{a}{2}$ $$...=\frac{1}{2}\cdot\left(\sin\frac{1}{n}+2\cdot\cot\frac{1}{2n^2}\cdot \sin^2\frac{1}{2n}\right)=...$$ also considering $\color{red}{\lim\limits_{x\to0}\frac{\sin x}{x}=1}$ $$...=\frac{1}{2}\cdot\left(\sin\frac{1}{n}+2\cdot\cos\frac{1}{2n^2}\cdot \frac{\frac{1}{2n^2}}{\sin\frac{1}{2n^2}}\cdot \frac{\sin^2\frac{1}{2n}}{\frac{1}{4n^2}}\cdot\frac{\frac{1}{4n^2}}{\frac{1}{2n^2}}\right)=\\ \frac{1}{2}\cdot\left(\sin\frac{1}{n}+\cos\frac{1}{2n^2}\cdot \color{red}{\left(\frac{\frac{1}{2n^2}}{\sin\frac{1}{2n^2}}\right)}\cdot \color{red}{\left(\frac{\sin\frac{1}{2n}}{\frac{1}{2n}}\right)^2}\right)\to \frac{1}{2}, n\to\infty$$

Answer 7

More generally, if $f(0)=0$ and $f$ is twice-differentiable at $0$,$$f(\tfrac{k}{n^2})\in\tfrac{k}{n^2}f^\prime(0)+O\left(\tfrac{k^2}{n^4}\right)\implies\lim_{n\to\infty}\sum_{k=1}^nf(\tfrac{k}{n^2})=\lim_{n\to\infty}\sum_{k=1}^n\tfrac{k}{n^2}f^\prime(0)=\tfrac12f^\prime(0),$$because $\sum_{k=1}^n\frac{k^2}{n^4}\sim\tfrac{1}{3n}\in o(1)$.