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The original problem is $$\sum \frac{\sin(n)\tan(n)\ln(e-\frac{1}{n})}{n^3}$$ I know that $\sum\frac{1}{n^3}$ converges absolutely, also $\sin(n)\leq 1$ and $\ln(e-1)<\ln(e-1/n)<\ln(e)=1$ Now how to determine whether $$\sum \frac{\tan(n)}{n^3}$$ converges? (If this converged, it would assure that the original converges, because it is an upper bound and by comparison test, we are done). Otherwise, if $\sum \tan(n)/n^3$ diverged, we wish to show about the series $$\sum\frac{\sin(n)\tan(n)}{n^3}$$ whether it converges/diverges.

It seems obvious that $\tan(n)$ has no limit (also, how to formally prove this?), but what about the $n^3$ doesn't it actually send it to $0$?

I would have an argument which I am quite unsure of,... $\tan(n)$ is only "$\pm\infty$" in each $(k+\frac{1}{2})\pi$, but we can't ever reach those points, because $n$ is from $\mathbb{N}$?

Michal Dvořák
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  • @EdmundoMartins, but as the OP correctly notes, $\tan n$ doesn't "explode" because $n$ is not a rational multiple of $\pi$ (though it should be $\pi(k+1/2)$ in the exploding condition) – Yuriy S Oct 30 '18 at 11:16
  • @YuriyS, you and the OP are absolutely correct, my bad... – Edmundo Martins Oct 30 '18 at 11:19
  • Well, we can't get arbitrarily close to $k\pi$, because $\mathbb{N}$ is not dense. – Michal Dvořák Oct 30 '18 at 11:28
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    If $n$ was rational, then we could get arbitrarily close to k pi as desired, so we could always get "bigger" bound so in fact tan(n) would be unbounded, but for naturals it is not the case because N is not dense. – Michal Dvořák Oct 30 '18 at 11:32
  • No, unfortunately it is not, I can't show that :/. Only if we could get arbitrarily close to $k \pi$, i could say it is unbounded, but otherwise, nothing else I can conclude. – Michal Dvořák Oct 30 '18 at 11:34
  • Among the first $40$ terms we have the largest in absolute value to be $\tan(11) \approx -226~$ so I'm not expecting it to be bounded – Yuriy S Oct 30 '18 at 11:41
  • @YuriyS neither am I, but I wish to prove it. – Michal Dvořák Oct 30 '18 at 11:42
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    Related: https://math.stackexchange.com/q/1056119/269624 – Yuriy S Oct 30 '18 at 12:00
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    Convergence depends on the irrationality measure of $\pi$, which is unknown. For sure, $|\sin(n)\tan(n)|$ can be as large as $\Theta(n)$ if $n$ is related to a numerator of a convergent of $\pi$. On the other hand if we assume that the irrationality measure of $\pi$ is, just to say, $4$, then the main term of such series does not converge to $0$ as $n\to +\infty$. – Jack D'Aurizio Oct 30 '18 at 12:12
  • So the post math.stackexchange.com/q/1056119/269624 that $tan(n)$ for $n\in\mathbb{N}$ is unbounded, but what about the $n^3$ then? – Michal Dvořák Oct 30 '18 at 12:14
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    If we assume that the irrationality measure of $\pi$ is $2$ (a widespread belief, but just a belief) then such series is convergent by comparison with $\sum_{n\geq 1}\frac{1}{n^2}$. The $\log(e-1/n)$ term is utterly irrelevant in any case. – Jack D'Aurizio Oct 30 '18 at 12:16
  • I guess that this series is not known if convergent or not. See a slightly different but similar problem: http://mathworld.wolfram.com/FlintHillsSeries.html – Twi Oct 30 '18 at 12:50
  • Actually, since we don't know, I edited the original post, we agreed that $\ln(e-1/n)$ does nothing in the sum basically, since it is positive and bounded by $1/2$ from below and $1$ from above, so we actually wish to show whether $\sum \frac{\sin{(n)}\tan{(n)}}{n^3}$ converges. – Michal Dvořák Oct 30 '18 at 13:37

1 Answers1

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From @JackD'Aurizio's comments it is clear that the convergence of this series is a part of an open problem of irrationality measure of $\pi$.

So, considering this question is not likely to be answered, I'll just offer some illustrative numerical results:

$$S_1=\sum_{n=1}^N \frac{\tan(n)}{n^3} \\ S_2=\sum_{n=1}^N \frac{\sin(n) \tan(n)}{n^3}$$

enter image description here

The same constant values are preserved at least up to $N=10^5$, though I haven't checked further.

It's useful to also check:

$$S_1=\sum_{n=1}^N \frac{\tan(n)}{n^2} \\ S_2=\sum_{n=1}^N \frac{\sin(n) \tan(n)}{n^2}$$

enter image description here

and:

$$S_1=\sum_{n=1}^N \frac{\tan(n)}{n} \\ S_2=\sum_{n=1}^N \frac{\sin(n) \tan(n)}{n}$$

enter image description here

We see some problems with convergence for $1/n$, as expected.

Again, this is just an illustration, definitely not an answer.

Yuriy S
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    Unfortunately I do not know what is irrationality measure. But I am very curious if convergence can be proved for any larger $n$? – lesnik Oct 30 '18 at 13:24
  • @lesnik, but it can't. Unless someone proves the bound for the irrationality measure of $\pi$, as Jack's comment says. What more can we do? – Yuriy S Nov 05 '18 at 06:42