How do I evaluate this sum :$\sum_{n=0}^{\infty} \frac{\sin(n!)}{\cos(n!)}$ if it's not open problem?

Question

I proposed this question on MSE but some comments affirmed that is unsolved problem and no answer. I would like to see what MO say about it.

How do I evaluate this sum :$$\displaystyle\sum_{n=0}^{\infty} \frac{\sin(n!)}{\cos(n!)}$$

Note : I used many criterions of convergence to test whether it converges but i didn't succeed.

Thank you for any help .

Presumably the set of values $\{ n! \bmod 2\pi\mathbb Z\}$ is dense in the interval $[0,2\pi]$, in which case $\tan(n!)$ will be arbitrarily large, infinitely often. So your series is highly unlikely to be absolutely convergent. (Whether one can prove this or not, I don't know.) Conditional convergence also seems unlikely. But it might be interesting to try to prove $\sum_{n\le X}\tan(n!)=o(X)$, or even $O(X^\epsilon)$ for some $\epsilon<1$. (I haven't thought about it, maybe this is easy, maybe not.) — Joe Silverman, Commented Jun 23, 2015 at 17:43
Actually, it's not clear that there's even a polynomial bound, i.e., $\sum_{n\le X}\tan(n!)=O(X^K)$ for some $K>0$. Rigorously, probably the best one can do is use transcendence estimates for $\pi$ to bound $n!$ away from multiples of $\pi/2$ where $\tan$ blows up, which is going to yield an effective, but horrible, upper bound in terms of $X$. — Joe Silverman, Commented Jun 23, 2015 at 17:53
On the other hand, $\sum_{n=0}^\infty \tan(\pi n! /e)$ will converge. — Robert Israel, Commented Jun 23, 2015 at 19:16
May be add "x" and change $\sin(n!)$ to $\sin(n!x)$ ? In this form it is a so called lacunar Fourier series, there are a lot of results on convergence for them. May be some result will help? — Sergei, Commented Jun 24, 2015 at 5:08

Per Alexandersson · Accepted Answer · 2015-06-23 17:44:12Z

5

I computed $f(n)=\sum_{k=0}^n \tan(k!)$ for different $n$, and got the following plot. It does not seem to have a limit.

Sums for different n

Mathematica code:

Block[{$MaxExtraPrecision = 600},
 lst = Table[Tan[n!], {n, 0, 200}];
 ListPlot[N[Accumulate[lst], 200], PlotRange -> All]
 ]

answered Jun 23, 2015 at 17:44

Per Alexandersson

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$\begingroup$ i have the same plot with wolfram alpha but what about analytical proof ? $\endgroup$
– Zeraoulia rafik
Commented Jun 23, 2015 at 17:46
2

$\begingroup$ It's probably more interesting to look at the values of something like $\sup_{n\le N} \frac{\log|f(n)|}{\log n}$ and see if it looks as if there might be a polynomial bound for $|f(n)|$. $\endgroup$
– Joe Silverman
Commented Jun 23, 2015 at 17:48
3

$\begingroup$ Joe Silverman gives a good argument. I guess to actually prove that $n!$ mod $2\pi \mathbb{Z}$ is dense is quite nontrivial, and perhaps even an open problem. $\endgroup$
– Per Alexandersson
Commented Jun 23, 2015 at 17:48
$\begingroup$ @JoeSilverman: Ah, yes, it looks like $\log|f(n)|/\log(n) \leq 1.64$ for $n \leq 600$, and $g(N)=\sup_{n\leq N} \frac{\log|f(n)|}{\log(n)}$ is very close to being constant. $\endgroup$
– Per Alexandersson
Commented Jun 23, 2015 at 17:57
2

$\begingroup$ @PerAlexandersson I upvoted to compensate for that. $\endgroup$
– Gejza Jenča
Commented Jun 29, 2015 at 20:03

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