Conjecture: $\sum_{n=1}^N\left|1+\frac{1+\tan(n)}{n\left(1+\tan(n+1)\right)}\right|-N-\ln^2N<0$, for sufficiently-large $N$

Question

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:

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

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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.