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My book says that the following series $$\sum_{n=0}^{\infty}\sin(n!\pi x)$$

converges for all rational values of $x$ (which is correct since then all the terms are $0$ after some definite index) and also for $x = e$, $=(2k+1)e$, $=2k/e$, $=\sin1$, $=\cos1$. I don't understand why, since it seems that the general term doesn't go to zero, or am I mistaken? Also is it probably a typo?

For $x = e$, according to Wolfram Alpha, the sum does not converge.

Mittens
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dude
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  • What book is it? – saulspatz Oct 26 '20 at 17:32
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    Knopp's theory and applications of infinite series – dude Oct 26 '20 at 17:32
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    Then I believe it. – saulspatz Oct 26 '20 at 17:36
  • Yeah probably we should believe him :). Still it happened that I found one or two typos – dude Oct 26 '20 at 17:38
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    I'm a bit late, but the $x = e$ case has been handled on this website before, with quite a lovely detailled answer: https://math.stackexchange.com/questions/2419113/proving-the-series-sum-n-0-infty-sine-pi-n-converges –  Oct 26 '20 at 18:03
  • Did Wolfram Alpha say the sum does not converge? Or did it merely fail to show a sum? – GEdgar Oct 26 '20 at 18:10
  • It says it does not converge :(. You can plot the partial sums only, no closed form or approximated sum. – dude Oct 26 '20 at 18:12
  • Note that the question asked in the title is still open, even showing convergence for the handful of values given. In particular, a characterization of the set of $x$ for which the series converges, or even just some properties like measure, would be wonderful. – Steven Stadnicki Oct 26 '20 at 18:34
  • @StevenStadnicki yes! If you want to provide your own answer, I will upvote it! I think I got how to prove it for the next values. It would be very interesting though, to see the characterization you are talking about. – dude Oct 26 '20 at 18:45
  • @Peanut If I knew an answer to the question I'd post it! I really don't know. – Steven Stadnicki Oct 26 '20 at 18:49
  • By the way, I think wolframalpha's partial sums are not to be trusted on this one, since the convergence of this series crucially depends on the precise number inserted: In another question, it was shown that the set of numbers where the above series converges has empty interior. So if your numerical approximation of $e$ is only slightly off (and of course it will be), you might start diverging. A clear telltale sign is when the plot of the partial sums starts behaving strangely chaotic at some higher integer, which happens with this sum. –  Oct 26 '20 at 18:57
  • @LukasMiristwhisky Agreed! There was indeed a sudden swing in the partial sums plot that made me suspect something was wrong with it! – dude Oct 26 '20 at 18:59
  • @LukasMiristwhisky Do you have the reference for that other question? I think that would settle what I was looking for... – Steven Stadnicki Oct 26 '20 at 20:29
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    @StevenStadnicki https://math.stackexchange.com/questions/2420161/convergence-of-sum-n-0-infty-sinx-pi-n?noredirect=1&lq=1 –  Oct 27 '20 at 08:07

3 Answers3

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Idea:

$$n!e = n!(1+1/1!+1/2!+1/3! +\cdots+1/n! + r_n).$$

Verify that $n!(1+1/1!+1/2!+1/3! +\cdots+1/n!)$ is an odd integer if $n$ is even, and is an even integer if $n$ is odd. Also, $r_n$ decreases to $0$ as $n\to \infty.$ So I think $\sum \sin(n!\pi e)$ is an alternating series whose terms in absolute value decrease to $0.$ Hence this series converges.

zhw.
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I think I can prove it for the first one, and then the others should be similar. In the case of $$\sum_{n=0}^\infty\sin(n!\pi e)$$ we need to consider the value of $n!\pi e\operatorname{mod} \pi$, and it's enough to consider the value of $n!e\operatorname{mod}1$. Now $$n!e=n!\sum_{k=0}^\infty\frac1{k!}$$ so that $$n!e\operatorname{mod} 1=n!\sum_{k=n+1}^\infty\frac1{k!}<\sum_{k=1}^\infty\frac1{(n+1)^k}=\frac1n\to0\text{ as } n\to\infty$$

saulspatz
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2

I discussed the problem with $x=(2k+1)e$ during the past spring session in my Calculus session. I think it is safe to post details here now.

$$ \begin{align} n!e&=n!\left(\sum^n_{k=0}\frac{1}{k!}+\sum^\infty_{k=n+1}\frac{1}{k!}\right)\\ &=\Big(\big(n!+ n! + n(n-1)\cdot\ldots\cdot3+\ldots+n(n-1)\big)+n+1\Big)+\Big(n!\sum^\infty_{k=n+1}\frac{1}{k!}\Big)\\ &=\big(P_n+(n+1)\Big)+a_n \end{align} $$ where $P_n=n!+ n! + n(n-1)\cdot\ldots\cdot3+\ldots+n(n-1)$ is an even number, and $a_n=n!\sum^\infty_{k=n+1}\frac{1}{k!}$.

We observe that

$$ \begin{align} \frac{1}{n+1}<a_n=\sum^\infty_{k=1}\frac{1}{(n+1)\cdot\ldots\cdot(n+k)}<\frac{1}{n+1}\sum^\infty_{k=0}\frac{1}{(n+2)^k}=\frac{n+2}{(n+1)^2} \end{align} $$ and $$a_n-a_{n+1}>\frac{1}{n+1}-\frac{n+3}{(n+2)^2}=\frac{1}{(n+1)(n+2)^2}>0$$

As a consequence, for each $m>0$, there exists $M>0$ such that $0<\sin(a_n\pi m)$ is monotone decreasing for $n\geq M$. Also, for $m$ odd $$\sin(n!m\pi e)=\sin(m P_n\pi+(n+1)m+a_nm)=\sin((n+1)m\pi+a_nm\pi)=(-1)^{n+1}\sin(ma_n\pi)$$ Therefore, the convergence of the series $\sum^\infty_{n=0}\sin(n! \pi m e)$ with $m\geq1$ an odd integer, follows from the convergence of the alternating series $\sum^\infty_{n=1}(-1)^{n+1}\sin(\pi m e a_n)$

I found a similar solution in an old book: Takeuchi, Y., Sequences and Series, Vol 1., Limusa, Mexico 1980.

I think the case $x=\sin 1$ may be treated siimlarly, but extra care is required since the factorial expansion in $\sin 1=\sum^\infty_{n=0}\frac{(-1)^k}{(2k+1)!}$ and $\cos 1=\sum^\infty_{k=0}\frac{(-1)^k}{(2k)!}$ is an alternating sequence. It might be necessary to group the terms in pairs the obtain convergent sequences with positive terms.

Mittens
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