Subsequence of $\sin n$

Question

Find one convergent subsequence of $a_n= \sin n$ and prove it.

I know that for all $n_1<n_2<n_3<\dots$, then $a_{n_1},a_{n_2},...$ is the sub-sequence of $a_n$. But I don't know how to use this definition in this problem

Answer 1

Think about it like this. $\sin x$ is the y coordinate of the intersection between the line at an angle of $x$ radians with the unit circle. Here I've drawn lines at angles of $n$ radians for $n$ going from $0$ to $10$.

If you count around the circle, you'll see that the point corresponding to an angle of $6$ radians is quite close to the point corresponding to an angle of $0$ (the $x$-axis). Maybe there's a subsequence of $\sin(n)$ converging to $\sin(0)=0$. In fact there is, and we can in fact prove this without exhibiting an actual example.

Note: feel free to skip this bit. It's just part of the thought process I had when doing the question, and it's a quickish way to see that we're not wasting time looking for a subsequence that converges to $0$ (as all we can tell from the question is that there is a subsequence converging to some value).

The idea is to prove that the set of points $(\cos(n), \sin(n))$ is dense in the unit circle. In particular, this tells us that $\sin(n)$ has a subsequence converging to any value in $[-1,1]$. We use two results: the Bolzano-Weierstrass Theorem, which tells us that any bounded sequence has a convergent subsequence, and the irrationality of $\pi$.

By the Bolzano-Weierstrass Theorem, there is a subsequence $(\cos(k_n),\sin(k_n))$ converging to some point $P$ on the unit circle. Now, for any $\varepsilon>0$, we must have some $m$ such that $0<\|(cos(m), \sin(m))-P\|<\varepsilon$.

Now we can choose $m_1,m_2$ such that $0<\|(cos(m_1), \sin(m_1))-P\|<\varepsilon$ and $0<\|(cos(m_2), \sin(m_2))-P\|<\varepsilon$ and we then get that $\|(\cos(m_1),\sin(m_1))-(\cos(m_2),\sin(m_2))\|<2\varepsilon$ by the triangle inequality. By the periodicity of $\sin$ and the continuity of the projection from $(\cos, \sin)$ on to $\sin$, we can eventually conclude that there is a subsequence $\sin(k_n)$ converging to $0$.

Now we can adapt this to show that there must be a subsequence converging to any point round the circle. The reason for this is that our subsequence $\sin(k_n)$ can never take the value $0$ (assuming that $k_n$ is positive). That's because we would then have $k_n=2r\pi$ for some integer $r$, which can't happen because $\pi$ is irrational.

What this means is that we have points at arbitrarily small but positive distance from $(1,0)$ in the sequence $(\cos(n),\sin(n))$. We can use these as sort of 'building blocks' to get a point $(\cos(m),\sin(m))$ within $\varepsilon$ of any point $Q$ on the unit circle: choose $n$ such that $\|(\cos(n),\sin(n))-(1,0)\|<\frac12\varepsilon$, and then take some multiple $an$ of $n$ such that $(\cos(an),\sin(an))$ is close to your point $Q$.

---

None of the above is necessary to do your question, and that's why I haven't bothered to be totally rigorous and follow everything through. I'm not really interested in the sequence $(\cos(n),\sin(n))$ being dense in the unit circle, but that's enough to convince me that it's not a waste of time to look for a subsequence of $\sin(n)$ converging to $0$.

This ends up being fairly simple to do. If $\sin(p)\approx0$, it means that $p\approx q\pi$ for some integer $q$. What we want is for the approximation to get better and better.

We use the Euclidean algorithm to find the continued fraction expansion of $\pi$:

$$ \pi=3+\frac1{7+\frac1{15+\frac1{1+\dots}}} $$

If you're familiar with the extended Euclidean algorithm, this is the same as trying to find the $\gcd$ of $\pi$ and $1$ - but the algorithm never stops, because $\pi$ is irrational. We start by seeing how many times $1$ goes into $\pi$:

$$ \pi = 3\times 1 + v_1 $$

where $v_1=\pi-3=0.1415926535897932\dots$. We keep going in the same way:

\begin{align} 1&=7\times v_1+v_2\\ v_1&=15\times v_2+v_3\\ v_2&=1\times v_3+v_4\\ v_3&=292\times v_4+v_5\\ &\dots \end{align}

At each step, we choose the largest integer $q_n$ such that $q_nv_n<v_{n-1}$, and let $v_{n+1}$ be the remainder. So $(q_n)=3,7,15,1,292,\dots$. You can go through the algorithm on a calculator if you want, and check that these values are correct. The steps are:

• You deal with two numbers at each step. Start with $\pi$ and $1$.
• Divide the larger number by the smaller number.
• Subtract the integer part of the resulting number; this will be $q_n$.
• The remainder is now $v_{n+1}$.
• Your new two numbers are: the smaller number from the previous step, and the new number $v_{n+1}$.

Now one important thing to notice about the $v_n$ is that they converge to $0$. Also note that we can now write, for example:

\begin{align} v_2&=1-7v_1\\ &=1-7(\pi-3)\\ &=22-7\pi \end{align}

(This gives the famous $\pi\approx\dfrac{22}7$ approximation.)

It takes longer, but we can also express $v_5$ as a linear combination of $1$ and $\pi$:

\begin{align} v_5&=v_3-292v_4\\ &=v_3-292(v_2-1v_3)=293v_3-292v_2\\ &=293(v_1-15v_2)-292v_2=293v_1-4687v_2\\ &=293v_1-4687(1-7v_1)=33102v_1-4687\\ &=33102(\pi-3)-4687=33102\pi-103993\\ \end{align}

(This gives the less famous - though more accurate - approximation $\pi\approx\dfrac{103993}{33102}$.)

Putting all this together, the fact that $v_n\to0$ means that the sequence of approximations generated in this way - $\pi-3, 7\pi-22, 106\pi-333, 113\pi-355, 33102\pi-103993, \dots$ - must converge to $0$ as well. In particular, the sequence $\sin(3), \sin(22), \sin(333), \sin(355), \sin(103993),\dots$ is a convergent subsequence of $\sin(n)$ converging to $0$ (by continuity of the $\sin$ function).

The rational approximations to $\pi$ ($\dfrac31, \dfrac{22}7, \dfrac{333}{106}, \dfrac{355}{113}, \dfrac{103993}{33102}, \dots$) are called the convergents to $\pi$. I don't think that their numerators have a special name, but you can just say: 'Let $q_n$ denote the numerator of the $n$-th convergent to $\pi$. Then $q_1<q_2<\dots$ and $\sin(q_n)\to0$.'