$$\sum_{n=1}^\infty \sin^{[n]}(1)$$ Where by $\sin^{[n]}(1)$ we mean $ \sin\left(\sin\left(\dots\sin(1)\right)\right)$ composed $n$ times.

Have tried the divergence test, which fails. Have tried Ratio test, also fails, as the limit is 1. Integral test, or root test do not seem promising. Help is appreciated

| cite | improve this question | | | | |
  • $\begingroup$ There are higher order ratio tests that may be useful (I have no clue) $\endgroup$ – Mark Mar 16 '17 at 5:52
  • 4
    $\begingroup$ math.stackexchange.com/questions/14004/… $\endgroup$ – Michael Biro Mar 16 '17 at 5:54
  • $\begingroup$ @Mark Ratio test will return one, hence, it is inconclusive. :P $\endgroup$ – Simply Beautiful Art Mar 16 '17 at 13:07
  • $\begingroup$ Actually, the initial sine can be of anything real because it will always be in [-1,1]. This might be more interesting if we allow a complex sine. $\endgroup$ – richard1941 Mar 22 '17 at 22:00
  • $\begingroup$ @Mark: I was unable to apply Raabe's Test successfully. $\endgroup$ – robjohn Feb 3 at 16:57

Once we prove the inequality $$\sin x > \frac{x}{1+x} \qquad (\star)$$ for $x \in (0,1]$, we can inductively show that $\sin^{[n]}(1) > \frac1n$ when $n\ge 2$. We have $\sin \sin 1 \approx 0.75 > \frac12$, and whenever $\sin^{[n]}(1) > \frac1n$, we have $$\sin^{[n+1]}(1) = \sin \sin^{[n]}(1) > \sin \frac1n > \frac{1/n}{1+1/n} = \frac1{n+1}.$$ Therefore, by the comparison test, $$\sum_{n=1}^\infty \sin^{[n]}(1) = \sin 1 + \sum_{n=2}^\infty \sin^{[n]}(1) > \sin 1 + \sum_{n=2}^\infty \frac1n,$$ which diverges.

To prove $(\star)$... well, to be honest, I just graphed both sides. But we can prove that $\sin x > x - \frac{x^3}{6}$ on the relevant interval by thinking about the error term in the Taylor series, and $x - \frac{x^3}{6} > \frac{x}{1+x}$ can be rearranged to $(x+1)(x -\frac{x^3}{6}) - x > 0$, which factors as $-\frac16 x^2(x-2)(x+3) > 0$.

| cite | improve this answer | | | | |
  • 10
    $\begingroup$ Let $f(x) = (x+1)\sin(x)-x$. We want to show $f(x) \ge 0$. Note $f(0) = 0$ and $f'(x) = \sin(x)+(x+1)\cos(x)-1$. It suffices to show $f'(x) \ge 0$. So it suffices to show $\sin(x)+\cos(x) \ge 1$ on $[0,1]$ (since $x\cos(x) \ge 0$). But by multiplying by $\frac{\sqrt{2}}{2}$, this is equivalent to $\sin(x+\frac{\pi}{4}) \ge \frac{\sqrt{2}}{2}$. And this is true for $x \in [0,\frac{\pi}{2}]$, so we're done since $\frac{\pi}{2} > 1$. $\endgroup$ – mathworker21 Mar 16 '17 at 6:07

Another possibility is to try to find an equivalent of the general term of this sequence : $\begin{cases} a_0=1\\ a_{n+1}=sin(a_n) \end{cases}$

Note that $f(x)=sin(x)$ has derivative $f'(x)=cos(x)$ which is positive on $[0,a_0]$ and also $<1$ on $]0,a_0[$ so $f$ is a contraction. From there it is easy to prove that $a_n\to 0$.

This means that $a_{n+1}\sim a_n$ when $n\to\infty$.

In this kind of problem we always search for an $\alpha$ such that $\mathbf{(a_{n+1})^\alpha-(a_n)^\alpha}$ does not depend of $\mathbf{a_n}$ (in the $\sim$ sense of the expression) so we are able to solve the recurrence.

From Taylor expansion $\displaystyle{(a_{n+1})^\alpha = \bigg(a_n - \frac{a_n^3}{6}+o(a_n^4)\bigg)^\alpha}=(a_n)^\alpha\bigg(1 - \frac{a_n^2}{6}+o(a_n^3)\bigg)^\alpha=(a_n)^\alpha\big(1-\frac{\alpha}{6}a_n^2+o(a_n^3)\big)$

So $(a_{n+1})^\alpha-(a_n)^\alpha=-\frac{\alpha}{6}(a_n)^{\alpha+2}+o((a_n)^{\alpha+3})\quad$ we see that we need $\alpha=-2$

Let's put $b_n=1/(a_n)^2,\qquad$ $b_n\to\infty$

We have $b_{n+1}-b_n=\frac 13+o(1/b_n)$ thus $b_n\sim\frac n3\qquad$

(more precisely $b_n=n/3+o(\ln(n))$ but it is not important at this point).

Finally $a_n\sim\sqrt\frac 3n,$ which is a term of a divergent serie so $\sum a_n$ diverges as well.

| cite | improve this answer | | | | |
  • $\begingroup$ This is also very nice! Thank you ! $\endgroup$ – userX Mar 19 '17 at 2:57

I wrote this answer for another question, but this question was pointed out in a comment, so I posted it here.

$|\,a_{n+1}|=|\sin(a_n)\,|\le|\,a_n|$. Thus, $|\,a_n|$ is decreasing and bounded below by $0$. Thus, $|\,a_n|$ converges to some $a_\infty$, and we must then have $\sin(a_\infty)=a_\infty$, which means that $a_\infty=0$.

Using $\sin(x)=x-\frac16x^3+O\!\left(x^5\right)$, we get $$ \begin{align} \frac1{a_{n+1}^2}-\frac1{a_n^2} &=\frac1{\sin^2(a_n)}-\frac1{a_n^2}\\ &=\frac{a_n^2-\sin^2(a_n)}{a_n^2\sin^2(a_n)}\\ &=\frac{\frac13a_n^4+O\!\left(a_n^6\right)}{a_n^4+O\!\left(a_n^6\right)}\\ &=\frac13+O\!\left(a_n^2\right)\tag1 \end{align} $$ Stolz-Cesàro says that $$ \lim_{n\to\infty}\frac{\frac1{a_n^2}}n=\frac13\tag2 $$ That is, $$ \bbox[5px,border:2px solid #C0A000]{a_n\sim\sqrt{\frac3n}}\tag3 $$ which means that the series diverges.

Motivation for $\boldsymbol{(1)}$

Note that $$ a_{n+1}-a_n=\sin(a_n)-a_n\sim-\frac16a_n^3\tag4 $$ which is a discrete version of $$ \frac{\mathrm{d}a_n}{a_n^3}=-\frac{\mathrm{d}n}6\tag5 $$ whose solution is $$ \frac1{a_n^2}=\frac{n-n_0}3\tag6 $$ so that $$ \frac1{a_{n+1}^2}-\frac1{a_n^2}=\frac13\tag7 $$ which suggests $(1)$.

| cite | improve this answer | | | | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.