Do we have $\prod_{n=1}^{\infty}\left(1+\frac{\tan\left(n\right)+1}{n\left(\tan\left(n+1\right)+1\right)}\right)=^?0$

Question

It seems too good to be true so let me propose it:

Do we have:

$$\prod_{n=1}^{\infty}\left(1+\frac{\tan\left(n\right)+1}{n\left(\tan\left(n+1\right)+1\right)}\right)=^?0$$

I try to plot the equation:

$$1+\frac{\left(\tan\left(x\right)+1\right)}{x\left(\tan\left(x+1\right)+1\right)}=0$$

And the root seems to be always a non integer.

On the other hand:

$$\prod_{n=1}^{10000}\left(1+\frac{\tan\left(n\right)+1}{n\left(\tan\left(n+1\right)+1\right)}\right)\simeq −3.5722175948×10^{−11} $$

Using WolframAlpha:

So is my closed form true? If yes how can we show it ?

Answer 1

Some relevant remark :

Let $x_i$ be real number $x_i\neq 1$ then if we have :

$$\sum_{n=1}^{m}\left|1+x_{m}\right|-m\left|1+\prod_{n=1}^{m}x_{i}\right|\leq 0$$

Then we have :

$$\prod_{n=1}^{m}\left|1+x_{m}\right|-\left|1+\prod_{n=1}^{m}x_{i}\right|^{m}\leq 0\tag{J}$$

the nice things the product telescope and we got :

$$\left|\prod_{n=1}^{\infty}\left(1+\frac{\tan\left(n\right)+1}{n\left(\tan\left(n+1\right)+1\right)}\right)\right|\leq 1\tag{I}$$

Then it seems we have for $N$ sufficiently large :

$$\sum_{n=1}^{N}\left|\left(1+\frac{\tan\left(n\right)+1}{n\left(\tan\left(n+1\right)+1\right)}\right)\right|-N-\ln^2(N)<0$$

Using this possible result gives $(I)$

We can also choose $N\varepsilon$ wich conclude if true the problem .

Edit : 29/09/2022 CORRECTED 01/10/2022

See also Show $\tan(n) < n^q$, conjectured $q < 1.1$ ; @Marco the answer gives for $n$ large enought :

$$\tan(n)<n^{7.11}$$

Suppose we have using the conjecture @Andreas and $m$ large enought and $n=2k+1$ $k$ a positive integer :

$$\prod_{k=0}^{n-k}\left|\frac{\left(\tan\left(k+m\right)+1\right)}{(k+m)\left(\tan\left(k+m+1\right)+1\right)}\right|\le\frac{(n-k)!(\left(m\right)^{1.1}+1)(\left(m+2\right)^{1.1}+1)\cdots\left(2(n-k-1)+m\right)^{1.1}+1)}{\left(n-k+m\right)!\left|1+\tan\left(m+ 2(n-k)\right)\right|},0\le k\le n-1$$

Then we have after telescoping then we have using majorization +exponential and dividing by $\left(2(n-k-1)+m\right)^{1.1}+1)$ to deconstruct the majorization starting from the middle :

$$\sum_{k=0}^{n}\left|\frac{\left(\tan\left(k+m\right)+1\right)}{(k+m)\left(\tan\left(k+1+m\right)+1\right)}\right|\le\sum_{k=0}^{\frac{n-1}{2}}\frac{(2k-2+m)^{1.1}+1}{(k+m)}\frac{1}{|1+\tan\left(2k+m\right)|}+1\tag{R}$$

Then we have using triangle inequality :

$$\sum_{k=0}^{n}\left|1+\frac{\left(\tan\left(k+m\right)+1\right)}{(k+m)\left(\tan\left(k+1+m\right)+1\right)}\right|\le\sum_{k=0}^{\frac{n-1}{2}}\frac{(2k-2+m)^{1.1}+1}{(k+m)}\frac{1}{|1+\tan\left(2k+m\right)|}+n/2+1$$

Unfortunelately one of part is greater than $\ln^2(N)$.

Edit 01/10/2022:

I finally get something less than $\ln^2(N)$ for $N$ large enought see Conjecture about $|n\tan\left(n+1\right)+\tan\left(n\right)+1+n|-|n\tan(n+1)+n|(1+\frac{1}{\sqrt{n}})<0$ for integer $n,n>M$

Ps: This is the general idea but perhaps there is some mistakes .

Answer 2

Because $n$ is a natural integer versus irrational $\pi$, then $n+k\pi$ for all $n$ has a uniform distribution over $[-\pi/2,\pi/2]$ for some $k\in \mathbb Z$.

So, $\tan(n)$ values between $[0,1]$ for $n+k\pi\in[0,\pi/2]$ have the same distribution of $\cot(m)$ values between $[0,1]$ for $m+l\pi \in [\pi/2,\pi]$. Also the same for the negative ones.

We are not interested to know when the multiplicative becomes greater than $1$ or less than $-1$, because it makes the total product bigger than or at least equal to $1$ or $-1$.

1. For the case $1+\frac{\tan\left(n\right)+1}{n\left(\tan\left(n+1\right)+1\right)}\ge1$, we have:

1.1 $\tan(n)+1\ge0$ and $\tan(n+1)+1\ge0$ which means $n+k\pi \in [-\pi/4,\pi/2-1]$.

1.2 $\tan(n)+1\le0$ and $\tan(n+1)+1\le0$ which is impossible, because $1_{rad}>57.2958_{deg}>45_{deg}$ and makes the other one out of $[-\pi,-\pi/2]$ region. You can use the trigonometry unit circle to prove it.

2. For the case $1+\frac{\tan\left(n\right)+1}{n\left(\tan\left(n+1\right)+1\right)}\le-1$, we have $\frac{\tan\left(n\right)+1}{\tan\left(n+1\right)+1}\le-2n$. But for large $n$, this is very rare (the probability is about zero). Because either $n+k\pi\rightarrow-\pi/2^+$ or $(n+1)+k\pi\rightarrow0^+$.

So, the union of the regions having an absolute value of greater than or equal to one is $n+k\pi \in [-\pi/4,\pi/2-1]\cup\{-\pi/2\}\cup\{0\}$, but this region doesn't cover more than half of the whole space $[-\pi/2,\pi/2]$ (about $77.7_{deg}$ vs. $90_{deg}$).

Then, the probability/occurrence of being less than $1$ is more than being greater than or equal to $1$.

Answer 3

This answer is still partial as contains a mistake (see the edit below) but it may help the thinking process. It also points out some problems of a different answer.

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I will build upon the first answer, which shows good intuition but fails to prove that the limit is zero. This will be a heuristic discussion, which I think could be made rigorous by means of theorems like the mean value theorem about the quasi-periodic motion.

I will call the following function $$ f(x):= 1+\frac{\tan(x)+1}{x(\tan(x+1)+1)} $$ whenever it is defined. It’s just the function that appears in your infinite product evaluated at $n$.

My claim is the following. I claim that the above product converges to zero if the following condition holds: $$ \int_0^\pi x\ln(f(x))dx <0 $$, while it diverges if the above integral is strictly positive. You can thus prove directly that this integral is negative to prove your claim, or compute it numerically (or compute it exactly if you manage). A small remark before “proving” it. Although the function $f$ is unbounded at some values of $x$, the logarithm makes the singularities narrower so that the integral is well defined and finite.

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Let’s start from the heuristics of the first answer: what I am going to use is the fact that for integer $n$, we can write $n=\theta+k\pi$ in a unique way with $k$ an integer and $\theta\in[0,\pi)$, and considering all the natural integers, the sequence $\theta(n)$ will lie uniformly on the interval $[0,\pi]$ (whatever that means rigorously). I will then take $\Theta$ to be a random variable with uniform distribution over the interval $[0,\pi)$ and think of the sequence $\theta(n)$ as a sequence of infinite realizations of this random variable.

Consider now our problem. Taking the natural logarithm of (the absolute value of) the above expression and using the properties of the logarithm, we equivalently ask whether $$ \lim_{N\to +\infty} \sum_{n=1}^N\ln |f(n)|=-\infty. $$ Of course, we are using the fact that $f(n)$ is never zero or infinite for integer $n$.

Since $f$ is periodic of period $\pi$, the above condition rewrites as $$ \lim_{N\to +\infty} \sum_{n=1}^N\ln |f(\theta(n))|=-\infty. $$ If we then treat $\theta$ as a random variable, we can compute the expected value $$ \mathbb E(\ln(f(\Theta)))=: E. $$ This quantity is something very concrete, and it turns out that $E$ is exactly the integral that I wrote above.

If $E$ is less than zero, looking at $\theta(n)$ as independent realizations of the variable $\Theta$ then we expect, as a consequence of the law of large numbers, $$ \lim_{N\to +\infty} \frac{1}{N}\sum_{n=1}^N\ln |f(\theta(n))|=E. $$ and my claim follows. Analogously we can treat the case $E>0$.

If I am not mistaken, this actually tells us that we expect to have a convergence rate to zero of the order $e^{NE}$.

I am sure I have done some mistakes. My point is that we shouldn’t just look at the number of occurrences of terms less than one with respect to all the occurrences, but rather look at what is the average contribution of these occurrences.

Edit. I have just realized that $f$ is actually not periodic because of that $n$, so this argument does not work and has to be substantially modified. Nevertheless, I hope I have showed that one needs better tools to study the problem.