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$$\sum_{n=1}^\infty \sin^{[n]}(1)$$ 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

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  • $\begingroup$ There are higher order ratio tests that may be useful (I have no clue) $\endgroup$ – Mark Mar 16 '17 at 5:52
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    $\begingroup$ 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
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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 $-\frac16 x^2(x-2)(x+3) > 0$.

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    $\begingroup$ 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
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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.

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  • $\begingroup$ This is also very nice! Thank you ! $\endgroup$ – userX Mar 19 '17 at 2:57
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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 $(1)$.

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