Confusion about infinitely many bounded prime gaps?

Question

I was working at something and I had this idea: Let $f(x) = \frac{1}{x}$ for all $x > 0$. Let $p_n$ denote the $n$'th prime number. Then by mean value theorem $$ f(p_{n+1}) - f(p_n) = f'(c_n) (p_{n+1} - p_n)$$ where $c_n \in (p_n,p_{n+1})$. Hence $$ \frac{1}{p_{n+1}} = -\frac{1}{c_n^2} (p_{n+1} - p_n) + \frac{1}{p_n}$$ Suppose that $\exists N \in \mathbb{N}$ such that $\forall n > N$ $p_{n+1} - p_n > D$ where $D$ is a positive constant. Then $-\frac{p_{n+1} - p_n}{c_n^2} < -\frac{D}{p_{n+1}^2}$ hence $$ \frac{1}{p_{n+1}} < -D\sum_{i=N+1}^{n+1}\frac{1}{p_i^2} + \frac{1}{p_N} = -D \sum_{i=1}^{n+1} \frac{1}{p_i^2} + D\sum_{i=1}^{N} \frac{1}{p_i^2} + \frac{1}{p_N}$$ Now $Z_p(2,n+1) = \sum_{i=1}^{n+1} \frac{1}{p_i^2} $ is bounded (converges to prime zeta function of 2). Therefore

$$ \frac{1}{p_{n+1}} < -D \cdot Z_p(2,n+1) + D\cdot C_2 + \frac{1}{p_N}$$ where $C_2 = \sum_{i=1}^{N} \frac{1}{p_i^2}$. Obviously $C_2 < Z_p(2,n+1)$. We take $D$ such that $D \cdot \left( C_2 - Z_p(2,n+1)\right) + \frac{1}{p_N} < 0$ that is $$ D > \frac{1}{p_N \cdot \left( Z_p(2,n+1) - C_2\right)} > 0$$ to obtain a contradiction ! Therefore there are infinitely many $k \in \mathbb{N}$ such that $p_{n_k+1} - p_{n_k} < D$ ?!

As mentioned in the comments the question is whether $\frac{1}{p_N \cdot \left( Z_p(2,n+1) - C_2\right)}$ is bounded above for $N > 0$ ?

Edit:

The answer is NO!

Answer 1

You conclude at some point:

Therefore $$\frac{1}{p_{n+1}}<−D\cdot Z_p(2,n+1)+D\cdot C_2+\frac{1}{p_N}$$

which basically is the same as; $$\color{red}{0<}\frac{1}{p_{n+1}}<\color{red}{D\cdot \left(C_2- Z_p(2,n+1)\right)+\frac{1}{p_N}}$$

and the next thing you do is:

We take $D$ such that $D\cdot \left(C_2−Z_p(2,n+1)\right)+\frac{1}{p_N}\color{red}{<0}$ ...

Do you see the problem?