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I have tried to to determine values of real $x$ for which $\sum\limits_{n=0}^{\infty}x^{\tan(n!)-n!}$ converges but I can't , Presumably the set of values $\{ n! \bmod 2\pi\mathbb Z\}$ is dense in $[0,2\pi]$ then $\tan(n!)$ will be large arbitrarily infinitely often, and the difference $\tan(n!)-n!$ would be probably small. However I don't know anything about limit of the fraction $\frac{\tan(n!)}{n!}$ for large $n$, And I have used Stiriling formula but it weren't helpful, This really forbide me to determine how $x$ values should be to get the titled series converge, then my question is :

Question: For which values of real number $x$ we have $\sum\limits_{n=0}^{\infty}x^{\tan(n!)-n!}$ converges ? We may avoid trivial case $x=0$ , In particular how we prove the set of values $\{ n! \bmod 2\pi\mathbb Z\}$ is dense in $[0,2\pi]$ ?

Note:I have used this trick for approximation : $n! = e(-1)^{n+1}/(n+1) + O(1/n^2) (\mod \mathbb Z)$ implies that $\tan(\pi n!)-n!=\pi e((-1)^{n+1}/(n+1) + O(1/n^2)) -\frac{1}{\pi})(\mod \mathbb Z)$ but this what could saying to me ? Probably conditional convergence could be happen here .

Addedendum

I'm interested to evaluate the above series because this need to prove $\{ n! \bmod 2\pi\mathbb Z\}$ is dense in $[0,2\pi]$ which is related to irrationality measure and Bounds approximation of $\pi$ which it is interesting in number theory and in the same time looking to the behavior of $\tan(n!)-n!$ for large $n$ which it help to get bounds for some arithmitic functions , Also to loook to the relationship of convergence of series to lacunar Fourier series which we have a lot of results on convergence for them

using mathematica code for $n=400$, $\tan(n!)-n!$ I have got the below plot and this is the code ,

Block[{$MaxExtraPrecision = 800},
  lst = Table[Tan[n!] - n!, {n, 0, 400}];
  ListPlot[N[Accumulate[lst], 200], PlotRange -> All]
  ]

The copy of this question is posted here

I have tried to to determine values of real $x$ for which $\sum\limits_{n=0}^{\infty}x^{\tan(n!)-n!}$ converges but I can't , Presumably the set of values $\{ n! \bmod 2\pi\mathbb Z\}$ is dense in $[0,2\pi]$ then $\tan(n!)$ will be large arbitrarily infinitely often, and the difference $\tan(n!)-n!$ would be probably small. However I don't know anything about limit of the fraction $\frac{\tan(n!)}{n!}$ for large $n$, And I have used Stiriling formula but it weren't helpful, This really forbide me to determine how $x$ values should be to get the titled series converge, then my question is :

Question: For which values of real number $x$ we have $\sum\limits_{n=0}^{\infty}x^{\tan(n!)-n!}$ converges ? We may avoid trivial case $x=0$ , In particular how we prove the set of values $\{ n! \bmod 2\pi\mathbb Z\}$ is dense in $[0,2\pi]$ ?

Note:I have used this trick for approximation : $n! = e(-1)^{n+1}/(n+1) + O(1/n^2) (\mod \mathbb Z)$ implies that $\tan(\pi n!)-n!=\pi e((-1)^{n+1}/(n+1) + O(1/n^2)) -\frac{1}{\pi})(\mod \mathbb Z)$ but this what could saying to me ? Probably conditional convergence could be happen here .

Addedendum

I'm interested to evaluate the above series because this need to prove $\{ n! \bmod 2\pi\mathbb Z\}$ is dense in $[0,2\pi]$ which is related to irrationality measure and Bounds approximation of $\pi$ which it is interesting in number theory and in the same time looking to the behavior of $\tan(n!)-n!$ for large $n$ which it help to get bounds for some arithmitic functions , Also to loook to the relationship of convergence of series to lacunar Fourier series which we have a lot of results on convergence for them

using mathematica code for $n=400$, $\tan(n!)-n!$ I have got the below plot and this is the code ,

Block[{$MaxExtraPrecision = 800},
  lst = Table[Tan[n!] - n!, {n, 0, 400}];
  ListPlot[N[Accumulate[lst], 200], PlotRange -> All]
  ]

The copy of this question is posted here

I have tried to to determine values of real $x$ for which $\sum\limits_{n=0}^{\infty}x^{\tan(n!)-n!}$ converges but I can't , Presumably the set of values $\{ n! \bmod 2\pi\mathbb Z\}$ is dense in $[0,2\pi]$ then $\tan(n!)$ will be large arbitrarily infinitely often, and the difference $\tan(n!)-n!$ would be probably small. However I don't know anything about limit of the fraction $\frac{\tan(n!)}{n!}$ for large $n$, And I have used Stiriling formula but it weren't helpful, This really forbide me to determine how $x$ values should be to get the titled series converge, then my question is :

Question: For which values of real number $x$ we have $\sum\limits_{n=0}^{\infty}x^{\tan(n!)-n!}$ converges ? We may avoid trivial case $x=0$

Note:I have used this trick for approximation : $n! = e(-1)^{n+1}/(n+1) + O(1/n^2) (\mod \mathbb Z)$ implies that $\tan(\pi n!)-n!=\pi e((-1)^{n+1}/(n+1) + O(1/n^2)) -\frac{1}{\pi})(\mod \mathbb Z)$ but this what could saying to me ? Probably conditional convergence could be happen here .

Addedendum

I'm interested to evaluate the above series because this need to prove $\{ n! \bmod 2\pi\mathbb Z\}$ is dense in $[0,2\pi]$ which is related to irrationality measure and Bounds approximation of $\pi$ which it is interesting in number theory and in the same time looking to the behavior of $\tan(n!)-n!$ for large $n$ which it help to get bounds for some arithmitic functions , Also to loook to the relationship of convergence of series to lacunar Fourier series which we have a lot of results on convergence for them

using mathematica code for n=400, $\tan(n!)-n!$ I have got the below plot and this is the code ,

Block[{$MaxExtraPrecision = 800},
  lst = Table[Tan[n!] - n!, {n, 0, 400}];
  ListPlot[N[Accumulate[lst], 200], PlotRange -> All]
  ]

The copy of this question is posted here

I have tried to to determine values of real $x$ for which $\sum\limits_{n=0}^{\infty}x^{\tan(n!)-n!}$ converges but I can't , Presumably the set of values $\{ n! \bmod 2\pi\mathbb Z\}$ is dense in $[0,2\pi]$ then $\tan(n!)$ will be large arbitrarily infinitely often, and the difference $\tan(n!)-n!$ would be probably small. However I don't know anything about limit of the fraction $\frac{\tan(n!)}{n!}$ for large $n$, And I have used Stiriling formula but it weren't helpful, This really forbide me to determine how $x$ values should be to get the titled series converge, then my question is :

Question: For which values of real number $x$ we have $\sum\limits_{n=0}^{\infty}x^{\tan(n!)-n!}$ converges ? We may avoid trivial case $x=0$ , In particular how we prove the set of values $\{ n! \bmod 2\pi\mathbb Z\}$ is dense in $[0,2\pi]$ ?

Note:I have used this trick for approximation : $n! = e(-1)^{n+1}/(n+1) + O(1/n^2) (\mod \mathbb Z)$ implies that $\tan(\pi n!)-n!=\pi e((-1)^{n+1}/(n+1) + O(1/n^2)) -\frac{1}{\pi})(\mod \mathbb Z)$ but this what could saying to me ? Probably conditional convergence could be happen here .

Addedendum

I'm interested to evaluate the above series because this need to prove $\{ n! \bmod 2\pi\mathbb Z\}$ is dense in $[0,2\pi]$ which is related to irrationality measure and Bounds approximation of $\pi$ which it is interesting in number theory and in the same time looking to the behavior of $\tan(n!)-n!$ for large $n$ which it help to get bounds for some arithmitic functions , Also to loook to the relationship of convergence of series to lacunar Fourier series which we have a lot of results on convergence for them

using mathematica code for $n=400$, $\tan(n!)-n!$ I have got the below plot and this is the code ,

Block[{$MaxExtraPrecision = 800},
  lst = Table[Tan[n!] - n!, {n, 0, 400}];
  ListPlot[N[Accumulate[lst], 200], PlotRange -> All]
  ]

The copy of this question is posted here

I have tried to to determine values of real $x$ for which $\sum\limits_{n=0}^{\infty}x^{\tan(n!)-n!}$ converges but I can't , Presumably the set of values $\{ n! \bmod 2\pi\mathbb Z\}$ is dense in $[0,2\pi]$ then $\tan(n!)$ will be large arbitrarily infinitely often, and the difference $\tan(n!)-n!$ would be probably small. However I don't know anything about limit of the fraction $\frac{\tan(n!)}{n!}$ for large $n$, And I have used Stiriling formula but it weren't helpful, This really forbide me to determine how $x$ values should be to get the titled series converge, then my question is :

Question: For which values of real number $x$ we have $\sum\limits_{n=0}^{\infty}x^{\tan(n!)-n!}$ converges ? We may avoid trivial case $x=0$

Note:I have used this trick for approximation : $n! = e(-1)^{n+1}/(n+1) + O(1/n^2) (\mod \mathbb Z)$ implies that $\tan(\pi n!)-n!=\pi e((-1)^{n+1}/(n+1) + O(1/n^2)) -\frac{1}{\pi})(\mod \mathbb Z)$ but this what could saying to me ? Probably conditional convergence could be happen here .

Addedendum

I'm interested to evaluate the above series because this need to prove $\{ n! \bmod 2\pi\mathbb Z\}$ is dense in $[0,2\pi]$ which is related to irrationality measure and Bounds approximation of $\pi$ which it is interesting in number theory and in the same time looking to the behavior of $\tan(n!)-n!$ for large $n$ which it help to get bounds for some arithmitic functions , Also to loook to the relationship of convergence of series to lacunar Fourier series which we have a lot of results on convergence for them

using mathematica code for n=400, $\tan(n!)-n!$ I have got the below plot and this is the code , In other side if we use this identity: $\cos n! = \cos (n! 2 \pi \sum_{i \geq 0} \frac{d_i}{i!}) = \cos (2 \pi N + 2 \pi \sum_{i \geq n+1} \frac{d_i }{(n+1)(n+2) \cdots i})$ as montioned in this answer , we would get the limit is $1$ , Now for $\sin(n!)$ we may look to this answer ,The limit would be $0$, Now we exploite these two result for looking to the behavior of $\tan(n!)-n!$ such that for large $n$ we have $\tan(n!)=\sin(n!)/\cos(n!)$ using the the fact $\lim\tan(n!)=\lim \sin(n!)\times \lim 1/\cos(n!),n\to \infty $ we would get $0$,then the behavior of $\tan(n!)-n!$ behave like $-n!$ in the power hence the titled series would be converge for every real $x >0$ .but this should contradict numerical result which show that the series of the general term ($\tan(n!)$) diverge .

Block[{$MaxExtraPrecision = 800},
  lst = Table[Tan[n!] - n!, {n, 0, 400}];
  ListPlot[N[Accumulate[lst], 200], PlotRange -> All]
  ]

The copy of this question is posted here

I have tried to to determine values of real $x$ for which $\sum\limits_{n=0}^{\infty}x^{\tan(n!)-n!}$ converges but I can't , Presumably the set of values $\{ n! \bmod 2\pi\mathbb Z\}$ is dense in $[0,2\pi]$ then $\tan(n!)$ will be large arbitrarily infinitely often, and the difference $\tan(n!)-n!$ would be probably small. However I don't know anything about limit of the fraction $\frac{\tan(n!)}{n!}$ for large $n$, And I have used Stiriling formula but it weren't helpful, This really forbide me to determine how $x$ values should be to get the titled series converge, then my question is :

Question: For which values of real number $x$ we have $\sum\limits_{n=0}^{\infty}x^{\tan(n!)-n!}$ converges ? We may avoid trivial case $x=0$

Note:I have used this trick for approximation : $n! = e(-1)^{n+1}/(n+1) + O(1/n^2) (\mod \mathbb Z)$ implies that $\tan(\pi n!)-n!=\pi e((-1)^{n+1}/(n+1) + O(1/n^2)) -\frac{1}{\pi})(\mod \mathbb Z)$ but this what could saying to me ? Probably conditional convergence could be happen here .

Addedendum

I'm interested to evaluate the above series because this need to prove $\{ n! \bmod 2\pi\mathbb Z\}$ is dense in $[0,2\pi]$ which is related to irrationality measure and Bounds approximation of $\pi$ which it is interesting in number theory and in the same time looking to the behavior of $\tan(n!)-n!$ for large $n$ which it help to get bounds for some arithmitic functions , Also to loook to the relationship of convergence of series to lacunar Fourier series which we have a lot of results on convergence for them

using mathematica code for n=400, $\tan(n!)-n!$ I have got the below plot and this is the code ,

Block[{$MaxExtraPrecision = 800},
  lst = Table[Tan[n!] - n!, {n, 0, 400}];
  ListPlot[N[Accumulate[lst], 200], PlotRange -> All]
  ]