Convergence of $\sin{\pi\sqrt{n}}$

Question

Revising for an exam:

Let $a_n = \sin{(\pi\sqrt{n})}.$ Show that:

(i) $a_{n+1} - a_{n} \rightarrow 0$

(ii) The sequence $(a_n)$ is bounded.

(iii) $(a_n)$ does not converge.

My attempt:

(i) ???

(ii) min($\sin(x)$) = -1, max($\sin{x}$) = 1, so $-1 \leq a_n \leq 1, \forall n \in \mathbb{N}$. Thus 1 is an upper bound and -1 is a lower bound.

(iii) $a_n$ has a monotonic subsequence which converges to 1 by the Bolanzo-Weierstrass theorem. Note that the subsequence $a_{n^2}$ converges to 0. Since $0 \neq 1$, $a_n$ does not converge.

Answer 1

Hint: (for (i)) $$ \sin a - \sin b = 2\sin\frac{a-b}{2}\cos\frac{a+b}{2} $$ and the product of two sequences, one converging to $0$ and the other bounded, converges to $0$.

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In more detail: $$ a_{n+1}-a_n = 2\sin\left(\pi\frac{\sqrt{n+1}-\sqrt{n}}{2}\right)\cos\left(\pi\frac{\sqrt{n+1}+\sqrt{n}}{2}\right) $$ The second factor is bounded as $\cos$ is, and the first goes to $0$ as

• $\sqrt{n+1}-\sqrt{n} = \sqrt{n}\left(\sqrt{1+\frac{1}{n}}-1\right) = \frac{1}{2\sqrt{n}}+o\left(\frac{1}{\sqrt{n}}\right)\xrightarrow[n\to\infty]{}0$.

• $\sin u\xrightarrow[u\to 0]{}0$

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Edit: For part (iii), that I hadn't realized was "still open" as well.

Suppose by contradiction $a_n\to\ell\in\mathbb{R}$.

1. As you noticed by looking at the subsequence $(a_{n^2})_n$, we necessarily have $\ell=0$.

2. Now, this implies that $a^2_n \xrightarrow[n\to\infty]{} 0$ as well, and using $\cos^2+\sin^2=1$ we get $\cos(\pi\sqrt{n})^2 \xrightarrow[n\to\infty]{} 1$.

3. Suppose for now we have shown that $$\cos(\pi\sqrt{n}) \xrightarrow[n\to\infty]{} \beta\in\{-1,1\} \tag{$\dagger$}$$

4. From the above, we have $$ a_{n+1}-a_n = b_n\cos\left(\pi\frac{\sqrt{n+1}+\sqrt{n}}{2}\right) $$ where $b_n\operatorname*{\sim}_{n\to\infty}\frac{\pi}{\sqrt{n}}$. Let's deal with the other term: as $$ \pi\frac{\sqrt{n+1}+\sqrt{n}}{2} = \pi\sqrt{n}+\frac{\pi}{4\sqrt{n}} + o\left(\frac{1}{\sqrt{n}}\right) $$ we get $$\begin{align} \cos\left(\pi\frac{\sqrt{n+1}+\sqrt{n}}{2}\right) &= \cos\left(\pi\sqrt{n}+\frac{\pi}{4\sqrt{n}} + o\left(\frac{1}{\sqrt{n}}\right)\right)\\ &= \cos \pi\sqrt{n} \cos\left(\frac{\pi}{4\sqrt{n}} + o\left(\frac{1}{\sqrt{n}}\right)\right) - \sin \pi\sqrt{n} \sin\left(\frac{\pi}{4\sqrt{n}} + o\left(\frac{1}{\sqrt{n}}\right)\right)\\ &= \cos \pi\sqrt{n} \cos(o(1)) - \sin \pi\sqrt{n}\sin(o(1)) \\ &\xrightarrow[n\to\infty]{} \beta\cdot 1 - 0\cdot 0 = \beta. \end{align}$$ Putting it all together, this leads to $$ a_{n+1}-a_n \operatorname*{\sim}_{n\to\infty}\frac{\beta\pi}{\sqrt{n}} $$ which by comparison implies that the series $$\sum_{n=0}^{\infty} (a_{n+1}-a_n)$$ diverges to $\infty$ (or $-\infty$, depending on $\beta$). But this is a contradiction, since this is a telescoping series, equal (by assumption on $(a_n)_{n\in\mathbb{N}}$ converging) to $\ell - a_0 = 0$. $\square$

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The remaining issue, of course, is that we don't actually have proven ($\dagger$). But it is enough for our purposes (handwaving a bit here, but it's not hard to make it formal) to have either: $$\cos(\pi\sqrt{n}) \xrightarrow[n\to\infty]{} \beta\in\{-1,1\}$$ or two sequances $(k_n)_n$, $(m_n)_n$ partitioning the natural numbers ($\mathbb{N} = \bigcup_n \{k_n\}\cup\{m_n\}$) such that $$\cos(\pi\sqrt{k_n}) \xrightarrow[n\to\infty]{} -1, \qquad \cos(\pi\sqrt{\ell_n}) \xrightarrow[n\to\infty]{} -1$$ which are the only two cases that can happen knowing that $\cos^2(\pi\sqrt{n}) \xrightarrow[n\to\infty]{}1$.

Indeed, the first case we took care of; and for the second case, we can now restrict the above argument to (that's the handwavy part) to either $(a_{k_{n+1}} - a_{k_n})_n$ or $(a_{m_{n+1}} - a_{m_n})_n$, knowing that at least one of the two the series $\sum_{n} \frac{1}{\sqrt{k_n}}$ and $\sum_{n} \frac{1}{\sqrt{m_n}}$ has to diverge.

Answer 2

(i) Since $\left|\sin(x)\right|\le\left|x\right|$ $$ \begin{align} \left|a_{n+1}-a_n\right| &=\left|\sin\left(\pi\sqrt{n+1}\right)-\sin\left(\pi\sqrt{n}\right)\right|\\[6pt] &=2\left|\cos\left(\pi\frac{\sqrt{n+1}+\sqrt{n}}2\right)\sin\left(\pi\frac{\sqrt{n+1}-\sqrt{n}}2\right)\right|\\[3pt] &\le2\cdot1\cdot\frac\pi2\left(\sqrt{n+1}-\sqrt{n}\right)\\[3pt] &=\frac\pi{\sqrt{n+1}+\sqrt{n}}\\[3pt] &\le\frac\pi{2\sqrt{n}}\tag{1} \end{align} $$

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(ii) Since $\left|\sin(x)\right|\le1$, we have $$ \begin{align} \left|a_n\right| &=\left|\sin\left(\pi\sqrt{n}\right)\right|\\ &\le1\tag{2} \end{align} $$

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(iii) The limit of the subsequence $$ \begin{align} \lim\limits_{n\to\infty}a_{n^2} &=\lim\limits_{n\to\infty}\sin\left(\pi\sqrt{n^2}\right)\\[3pt] &=\lim\limits_{n\to\infty}0\\[3pt] &=0\tag{3} \end{align} $$ Since $$ \begin{align} \left|\,n+\tfrac12-\sqrt{n^2+n}\,\right| &=\frac{\frac14}{n+\frac12+\sqrt{n^2+n}}\\ &\le\frac1{8n}\tag{4} \end{align} $$ and $\cos(x)\ge1-\frac12x^2$, we have that $$ \begin{align} \left|a_{n^2+n}\right| &=\left|\sin\left(\pi\sqrt{n^2+n}\right)\right|\\ &\ge\left|\cos\left(\pi\left(n+\frac12-\sqrt{n^2+n}\right)\right)\sin\left(\pi\left(n+\frac12\right)\right)\right|\\ &-\left|\sin\left(\pi\left(n+\frac12-\sqrt{n^2+n}\right)\right)\cos\left(\pi\left(n+\frac12\right)\right)\right|\\ &=\left|\cos\left(\pi\left(n+\frac12-\sqrt{n^2+n}\right)\right)\right|\\ &\ge1-\frac12\frac{\pi^2}{64n^2}\\[6pt] &\gt0.9\tag{5} \end{align} $$ for $n\ge1$.

If the sequence converged, then the limit must be the limit of the subsequence computed in $(3)$. However, $(5)$ precludes the limit from being $0$.

Answer 3

To answer the point $iii)$, note that

$$\lim_{n\to \infty} \sin ^2\left(\pi \sqrt{n^2-n}\right)=1,$$ since we have that

$$\sqrt{1-\frac{1}{n}}\approx 1-\frac{1}{2n},$$ when $n$ is large.

Q.E.D. (which you combine with $\sin{\pi\sqrt{n^2}}=0$ to show divergence)

Answer 4

Hint:

$$\sin(\pi\sqrt{n+1})-\sin(\pi\sqrt n)=2\sin\left(\frac{\sqrt{n+1}-\sqrt n}2\right)\cos\left(\frac{\sqrt{n+1}+\sqrt n}2\right).$$

As $$\sqrt{n+1}-\sqrt n=\frac1{\sqrt{n+1}+\sqrt n},$$ and $$\sin(x)\le x$$ in the first quadrant, the expression decreases to $0$ like $\dfrac1{\sqrt n}$.

Answer 5

(i) The mean value theorem shows

$$\tag 1 a_{n+1}-a_n = (\cos c_n)(\pi \sqrt {n+1} - \pi \sqrt n).$$

Verify that $\pi \sqrt {n+1} - \pi \sqrt n \to 0.$ Because $\cos c_n$ is bounded, $(1)\to 0$ as desired.

(ii) Obvious.

(iii) Lemma: Let $x_1< x_2 < \cdots\to \infty,$ with $x_{n+1}-x_n \to 0.$ Then $\sin x_n$ is dense in $[-1,1].$

Proof: It suffices to show $e^{ix_n}$ is dense in the unit circle. But think about about it: As $n\to \infty, e^{ix_n}$ makes infinitely many orbits around the circle (because $x_n \to \infty$), in steps of arc length $x_{n+1}-x_n.$ Those arc lengths $\to 0.$ Thus if $A$ is any open arc on the circle, $e^{ix_n}$ has to land in $A$ infinitely many times; you can't avoid $A$ in an orbit once the steps have length less than the length of $A.$ Thus $e^{ix_n}$ is dense in the unit circle as desired.

With $x_n = \pi \sqrt n,$ we have the hypotheses of the lemma. Thus $a_n = \sin x_n$ is dense in $[-1,1].$ So certainly $a_n$ can't converge.

Answer 6

(iii) Assume instead that $a_n\to l$ for some $l\in\mathbb{R}.\;\;$ As noted by the OP, $l=0$ since $a_{n^2}\to 0$.

For any $k\ge1, \;\;\left(2k+\frac{5}{6}\right)^2-\left(2k+\frac{1}{6}\right)^2=(4k+1)(\frac{2}{3})>1$, so there is an $n_k\in\mathbb{N}$ such that

$\hspace{.4 in}\left(2k+\frac{1}{6}\right)^2<n_k<\left(2k+\frac{5}{6}\right)^2$.

Then $2k+\frac{1}{6}<\sqrt{n_k}<2k+\frac{5}{6}\implies2\pi k+\frac{\pi}{6}<\pi \sqrt{n_k}<2\pi k+\frac{5\pi}{6}\implies \sin\pi\sqrt{n_k}>\frac{1}{2}$;

and this gives a contradiction since $a_{n_k}=\sin\pi\sqrt{n_k}\not\to0$

Answer 7

The first two parts have already been answered in several answer in a pretty standard way:

• From $|\sin\pi\sqrt{n+1}-\sin\pi\sqrt{n}| \le \pi(\sqrt{n+1}-\sqrt{n}) = \frac\pi{\sqrt{n+1}+\sqrt{n}}$ we get that $a_{n+1}-a_n \to 0$. (We have used this inequality: Show that $|\sin{a}-\sin{b}| \le |a-b| $ for all $a$ and $b$.)
• Clearly, $|\sin\pi\sqrt{n}|\le1$.

Let me try to argue that $a_n=\sin\pi\sqrt{n}$ is not convergent slightly differently than in other answers.

The first observation is that for $n=k^2$ we have $a_n=\sin\pi\sqrt{n}=\sin\pi{k}=0$. The value zero is attained for $n=(k+1)^2$ again.

What happens for $k^2 \le n \le (k+1)^2$? We can notice that for any such $n$ we have $$\sqrt{n+1}-\sqrt{n} = \frac1{\sqrt{n+1}+\sqrt{n}} \le \frac1{2\sqrt{n}}\le\frac1{2k}.$$ This means that the numbers $\sqrt{n}$ for $n$'s in this interval are increasing monotonically from $k$ to $(k+1)$ and the difference between two consecutive terms is at most $1/(2k)$. So there exists an $n_k$ such that $$k+\frac12-\frac1k \le \sqrt{n_k} \le k+\frac12+\frac1k,$$ since every interval of the length $\frac1{2k}$ with the endpoints between $k$ and $(k+1)$ contains some $\sqrt{n}$.

For any $n_k$ with the above properties we have $$|a_{n_k}|=|\sin\pi{\sqrt{n_k}}| \ge \left|\sin\pi\left(k+\frac12+\frac1k\right)\right| = \left|\sin\pi\left(\frac12+\frac1k\right)\right|.$$ Since $$\lim_{k\to\infty} \left|\sin\pi\left(\frac12+\frac1k\right)\right| = \left|\sin\frac\pi2\right|=1$$ we see that $\lim\limits_{k\to\infty} a_{n_k} \ne 0$.

We found two subsequences such that one of them converges to zero and the other one does not. So the sequence $(a_n)$ is not convergent.

Answer 8

(i) $a_{n+1}-a_{n} \to 0$

Instead of $\sin(\pi\sqrt{n})$ we will be looking at $e^{i\pi\sqrt{n}} = \cos(\pi\sqrt{n}) + i\sin(\pi\sqrt{n})$

So we have $$\lim\limits_{n \to \infty} e^{i\pi\sqrt{n+1}} - e^{i\pi\sqrt{n}} =$$ $$\lim\limits_{n \to \infty} e^{i\pi\sqrt{n}}(e^{i\pi(\sqrt{n+1}-\sqrt{n})} - 1) $$

Now we have that $\lim\limits_{n \to \infty} \sqrt{n+1}-\sqrt{n} = 0$ because for each $\epsilon$ there is $m$ so that for all $n>m$ we have $ \sqrt{n+1}-\sqrt{n} < \epsilon $ In order to find this $m$ it is sufficient to solve $ \sqrt{m+1}-\sqrt{m} = \epsilon $ for which we have

$$m=\frac{(\epsilon^2-1)^2}{4\epsilon^2}$$ so as soon as $n > \frac{(\epsilon^2-1)^2}{4\epsilon^2}$ we have $ \sqrt{n+1}-\sqrt{n} < \epsilon $

Since $e^{i\pi\sqrt{n}}$ is bounded we can insert the limit further

$$\lim\limits_{n \to \infty} e^{i\pi\sqrt{n}}(e^{i\pi(\sqrt{n+1}-\sqrt{n})} - 1)= e^{i\pi\sqrt{n}}\lim\limits_{n \to \infty}(e^{i\pi(\sqrt{n+1}-\sqrt{n})} - 1)= e^{i\pi\sqrt{n}}(e^{i\pi\cdot 0} - 1)=0$$

From there $\lim\limits_{n \to \infty} \sin(\pi\sqrt{n}) =0$

(ii) The sequence is bounded

For each element we have $-1 \leq a_{n} \leq 1$ because $-1 \leq sin(x) \leq 1$. Observe that inequality is strictly $\leq$ because there are infinite square roots that are integers and $sin(k\pi)=0$ for k integer.

(iii) Sequence does not converge

We will use the integral test, thus avoiding the problem of square root. Infinite series converges if and only if its associated integral converges.

In our case the integral is $\int\limits_{1}^{\infty} e^{i\pi\sqrt{x}} dx$ and we are interested in its imaginary part.

The integral is solved first by substitution $t=\sqrt{x}$, $dt=\frac{1}{2\sqrt{x}dx}$ which makes $2tdt=dx$ having

$$\int\limits_{1}^{\infty} e^{i\pi\sqrt{x}} dx=\int\limits_{1}^{\infty} e^{i\pi t} 2tdt$$

then partial integration $u=2t$,$dv=e^{i\pi t} dt$ finally having

$$\int\limits_{1}^{\infty} e^{i\pi\sqrt{x}} dx=\frac{2e^{i\pi\sqrt{x}}(1-i\pi\sqrt{x})}{\pi^2}|_{1}^{\infty}$$

Since we are interested in imaginary part we can extract it as

$$Im[\int\limits_{1}^{\infty} e^{i\pi\sqrt{x}} dx]=\frac{2\sin(\pi\sqrt{x})}{\pi^2}-\frac{2\cos(\pi\sqrt{x})\sqrt{x}}{\pi} |_{1}^{\infty}$$

and this does not converge so the series does not converge as well.