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It's a follow up of Do we have $\prod_{n=1}^{\infty}\left(1+\frac{\tan\left(n\right)+1}{n\left(\tan\left(n+1\right)+1\right)}\right)=^?0$

Conjecture/Problem:

It seems we have for $N$ sufficiently large:

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

Checking the sum until $N=800000$ there is no counter-example.

My progress:

I try to find a counter-example using the generalized triangle inequality we have:

$$\left|\sum_{n=1}^{N}\left(1+\frac{\tan\left(n\right)+1}{n\left(\tan\left(n+1\right)+1\right)}\right)\right|\leq \sum_{n=1}^{N}\left|\left(1+\frac{\tan\left(n\right)+1}{n\left(\tan\left(n+1\right)+1\right)}\right)\right|$$

So remains to find $M\geq N$ such that:

$$M+\ln^2(M)\leq \left|M+\sum_{n=1}^{M}\frac{\tan\left(n\right)+1}{n\left(\tan\left(n+1\right)+1\right)}\right|$$

Wich seems to be not true

On the other hand the function:

$$f(x)=\sum_{n=1}^{M}\left|1+\frac{\tan\left(n\right)+1}{xn\left(\tan\left(n+1\right)+1\right)}\right|$$

Seems convex on $(0,2)$ .

So keeping in mind this result (?) we can use a chord such that $x,a,b\in(0,2)$:

$$f(x)\leq \frac{f(a)-f(b)}{a-b}(x-a)+f(a)$$

But I cannot proceed further.

Edit :

It seems $\exists,n,m,M$ such that $m\leq n\leq M$ then we have :

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

Question:

How to (dis)prove it?

Any help is appreciated.

Barackouda
  • 3,879
  • Please check if you need to prove $\sum |a_n| - g(N) < 0$ or $\sum |a_n| - g(N) > 0$. – Ivan Kaznacheyeu Sep 27 '22 at 09:55
  • @IvanKaznacheyeu I need to prove the first one . – Barackouda Sep 27 '22 at 10:39
  • I suppose there will be problem with such $N$ that $|\tan(N)|$ is large. For example, $N=52174$, $\tan(N)=-181570.2957...$, $\sum_{n=1}^{N}\left|\left(1+\frac{\tan\left(n\right)+1}{n\left(\tan\left(n+1\right)+1\right)}\right)\right|-N-\ln^2(N) = 9.4...$. – Ivan Kaznacheyeu Sep 27 '22 at 15:22
  • @IvanKaznacheyeu I say for large $N$ . For example try $N=100000$ it seems to decrease but it's not fixed yet . – Barackouda Sep 27 '22 at 16:01
  • Is $N=3083975227$ large for you? – Ivan Kaznacheyeu Sep 29 '22 at 09:54
  • @YvanKaznacheyeu What you want to tell me ? With Desmos I cannot with this value of N . If you have a counter-example it would be really nice ! – Barackouda Sep 29 '22 at 10:15
  • Let $k$ is large integer such that $\tan(k) = ck$ with some $c>1$. Let $r_{k}$ is sum for $N=k$ minus $(k+\ln^2 k)$. Then $r_{k}-r_{k-1} \approx 2.794c$. Then in order $r_{k}$ to be positive we need, $c>-0.357r_{k-1}$. There is possibility that exists large $k$ such that $c$ satisfies this inequality. If such $k$ exists, this $k$ will be much larger than $N=3083975227$ ($c\approx 4.33$) so it cannot be checked with direct calculation. – Ivan Kaznacheyeu Sep 29 '22 at 15:45
  • @YvanKaznacheyeu see please this link it shows the claim for $n$ large enought https://math.stackexchange.com/questions/4542968/conjecture-about-n-tan-leftn1-right-tan-leftn-right1n-n-tann1n – Barackouda Oct 01 '22 at 16:56

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