# Does $\sum a_n$ converge if $a_n = \sin( \sin (...( \sin(x))...)$

Does $\sum a_n$ converge if $a_n = \sin( \sin (\cdots( \sin(x))\cdots)$, $\sin$ being applied n times and $x \in (0, \pi/2)$?

What about $\sum a_n^r$ for $r \in \mathbb{R^+}$?

• I suspect that the sequence of partial sums $S_N$ increases roughly as fast as the square root of $N$. I have not rigorously shown this yet. Mar 5, 2015 at 18:20

• Classicaly, $\displaystyle a_n=\sqrt\frac{3}{n}+o\left(\sqrt\frac{3}{n}\right)$

The series $\sum a_n$ is therefore divergent.

• Using the previous asympotical estimate, $\sum a_n^r$ converges iff $1/2-r> 1$

Proof of the first claim:

Notation: Given two sequences $u_n$ and $v_n$, we note $u_n\sim v_n$ iff $\lim_{n\to \infty}\frac{u_n}{v_n}=1$

Now, back to the asymptotics:

\begin{align} \frac{1}{a_{n+1}^2}-\frac{1}{a_n^2}=\frac{1}{\sin(a_n)^2}-\frac{1}{a_n^2}&=\frac{1}{a_n^2-\frac{a_n^4}{3}+o(a_n^4)}-\frac{1}{a_n^2}\\&=\frac{1}{a_n^2}\left(\frac{1}{1-\frac{a_n^2}{3}+o(a_n^2)}-1\right)\\&=\frac13+o(1) \\\end{align}

By Cesaro mean theorem, $\displaystyle \frac{1}{n}\sum_{k=1}^n \left(\frac{1}{a_{k+1}^2}-\frac{1}{a_k^2}\right)=\frac13+o(1)$

Hence $\displaystyle a_n^2=\frac{1}{\frac{n}{3}+o(n)}=\frac{3}{n}\left(1+o(1)\right)$

Hence $\displaystyle a_n=\sqrt\frac{3}{n}+o\left(\sqrt\frac{3}{n}\right)$

The estimate can be refined as $$a_n=\sqrt\frac{3}{n}-\frac{3\sqrt{3}}{10}\frac{\log(n)}{n\sqrt{n}}+o\left(\frac{\log(n)}{n\sqrt{n}}\right)$$

Proof of the estimate on partial sums

Lemma: Let $u_n$ and $v_n$ be sequence such that $u_n\sim v_n$, and $\forall n\in \mathbb N,v_n\geq 0$. If $\sum_{n\geq 1}v_n$ diverges, so does $\sum_{n\geq 1}v_n$ and $\sum_{k=1}^nu_k \sim \sum_{k=1}^nu_k$

We've proven so far that $\displaystyle a_n\sim \sqrt\frac{3}{n}$.

Hence, by the lemma, $\displaystyle S_n \sim \sum_{k=1}^n \sqrt\frac{3}{k}$

To estimate the last sum, an integral method comes in handy: indeed

$$\forall n\geq 2, \int_n^{n+1}\frac{1}{\sqrt{t}}dt \leq \frac{1}{\sqrt{n}} \leq \int_{n-1}^n \frac{1}{\sqrt{t}}dt$$

Summing these inequalities yields $$\sum_{k=1}^n \sqrt\frac{3}{k} \sim 2\sqrt{3n}$$

Hence $S_n \sim 2\sqrt{3n}$

• LeGransDOOM. I suspected that the partial sums converges as fast (or faster) than $\sqrt(N)$. But have not yet proven this rigorously. How does one prove your "classical" result for the series terms? Mar 5, 2015 at 18:54
• I found the classical result on this site. For others, here is a straightforward proof! math.stackexchange.com/questions/3215/… Mar 5, 2015 at 19:23
• Great, now we might close this as a duplicate. Mar 5, 2015 at 19:24
• @Dr.MV For your information, the partial sums satisfy $S_n = \frac{2}{3}\sqrt{n} + o(\sqrt{n})$ Mar 5, 2015 at 19:29
• Well done! A very straightforward proof indeed!! Note that I corrected the typos at the end of the proof since the partial sums go like $2\sqrt(3n)$ and not $\frac23 \sqrt{n}$. Mar 11, 2015 at 5:49

Since $sin(x)$ has values only in $[0,1]$, the function $sin(sin(x))$ has values on $[sin(0),sin(1)]$. Therefore $a_{n+1}<a_n$. Estimating the upper bound:

$|\sum a_n| \leq \sum |a_n| \leq \sum sin(sin(...(1)))$ (used triangle inequality and sine is applied $n-1$ times).

Moreover it holds $sin(x) \leq x$ such that (especially for small arguments in the sine) $sin(sin(...(x))) \leq x$. Therefore the upper bound can be estimated:

$|\sum a_n| \leq \sum 1= \infty$.

Because the multiple sines decreasing slowly, this series diverges. If the $a_n$ are exponentiated, the $a_n^r$ become smaller when $r$ become larger; it remains the upper bound in order $x^r$ for $x \leq 1$ and when summing over it, the sum will diverge as in the case for $r=1$.

• Your conclusion is invalid. A diverging upper bound doesn't imply that the series itself diverges. Mar 5, 2015 at 17:41
• The terms $a_n$ are positive. How did the triangle inequality facilitate analysis here? Mar 5, 2015 at 17:44
• I suspect that the sequence of partial sums $S_N$ increases roughly as fast as the square root of N . I have not rigorously shown this yet. Mar 5, 2015 at 18:20