Does the sum converge for all values of $a$?

Here is the sum (for which $a$ is the variable):

$$a+\sin(a)+\sin(\sin(a))+\cdots$$

Does the sum always converge for all values of $a$?

So far, this is what I have done:

1) I plugged in many values for $a$ and the terms themselves converge to $0$, but the sum itself converges so slowly that I can't tell.

2) I graphed $x,\sin(x), \sin(\sin(x)),\cdots \text{etc}$ and the graphs tend towards $y=0$, but tend slowly.

...and my level of mathematical knowledge is not enough to tell, or much less solve, this problem beyond this. Thank you for any help.

Solution 1:

Let $a$ be any given positive real that is $< \pi$. Your sum is precisely:

$$\sum_{n=1}^{\infty}a_n; \text{Where $a_1 := a$, and for all $n \ge 1,$ $a_{n+1} = \sin(a_n)$}$$

It should be clear that $\{a_n\}$ is non-negative and decreasing: for all $n \ge 1$, $a_{n+1} = \sin(a_n) \begin{cases} \le a_n \\ \small \text{and} \\ \ge 0 \end{cases}$. Thus $\{a_n\}$ is convergent.

The limit of $\{a_n \}$ is the unique fixed point of the function $x \mapsto \sin(x)$, namely: $0$.

Next, introduce the sequence $b_n := 1/a_n ^2$, $n \ge 1$. Due to the well-behaved nature of $a_n$ near $\infty$:

$$\lim_{n \to \infty}\left( \frac{b_n - b_{n-1}}{n-(n-1)} \right) = \lim_{n \to \infty} \left( \frac{a_{n-1} ^2 - \sin^2(a_{n-1})}{a_{n-1} ^2 \sin^2(a_{n-1})} \right) = \lim_{t \to 0} \left( \frac{t ^2 - \sin^2(t)}{t ^2 \sin^2(t)} \right) = \dots = \frac13$$

But $\{n\}$ is strictly increasing and $\lim_n n = + \infty$. Therefore by Cesàro's theorem:

$$\lim_{n \to \infty} \frac{b_n}{n} = \frac{1}{3}$$

Which shows that $b_n \sim \frac{n}{3}, n \to \infty$, or equivalently: $a_n \sim \frac{\sqrt3}{\sqrt n}$.

Because $\{a_n\}$ and $\{\frac{\sqrt3}{\sqrt n} \}$ are two asymptotic sequences, the divergence of the series having the latter as its general term implies the divergence of that having the former as its general term.

By using the the nice properties of $\sin$, we get:

$$( \forall \ a \notin \pi \mathbb Z)(\sum_{n=1}^{\infty} a_n \ \text{is divergent}) $$

Where $\pi \mathbb Z$ denotes the set of integral multiples of $\pi$.

For more clarity:

• If $a \in (- \pi, 0)$, then $a_n \equiv -b_n$, where $b_1 = -a$ and for all $n \ge 1$, $b_{n+1} := \sin(b_n)$, and we know that the series having the latter as the general term is divergent.

• If $a \in (\pi, 2\pi)$, then for all $n \ge 2$, $a_n= - b_n$, where $b_1 = a - \pi$ and for all $n \ge 1$, $b_{n+1} = \sin(b_n)$.

These two instances demonstrate how the problem can be always reduced to the one with $0 < a < \pi$.

Finally, it's clear that for $a \in \pi \mathbb Z$, $a_n \equiv 0$ and the series must converge.