Define $$\Pi=\Pi_1\cup\Pi_2\cup\Pi_3$$ Where, $$\Pi_1=\{6mn-m+n|m, n\in \mathbb{N}\}$$ $$\Pi_2=\{6mn-m-n|m, n\in \mathbb{N}\}$$ $$\Pi_3=\{6mn+m+n|m, n\in \mathbb{N}\}$$. Note that a twin prime pair other than $(3, 5)$ must be of the form $(6n-1, 6n+1)$. Hence if there are finitely many twin primes we must have $k\in \Pi$ for all $k>j\in\mathbb{N}$.Thus we have $\{1,2,3, ..., j\}\cup\Pi=\mathbb{N}$
We know $$\sum_{t=1}^{\infty} \frac{1}{t^2}=\frac{\pi^2}{6}$$
Thus we have $$\sum_{d=1}^{j} \frac{1}{d^2}+\sum_{d=j+1}^{\infty} \frac{1}{d^2}=\sum_{d=1}^{j} \frac{1}{d^2}+\sum_{x\in{(\Pi_1)_{>j}}} \frac{1}{x^2}+\sum_{y\in{(\Pi_2\cup\Pi_3)}_{>j},y\notin{\Pi_1}_{> j} } \frac{1}{y^2}=\frac{\pi^2}{6}$$ Now $$1+\sum_{d=2}^{j}(\frac{1}{d}-\frac{1}{1+d})\leq 1+\sum_{d=2}^{j} \frac{1}{d^2}\leq 1+\sum_{d=2}^{j}(\frac{1}{d-1}-\frac{1}{d})$$ Further more, $$\frac{3}{2}-\frac{1}{j+1}\leq \sum_{d=1}^{j} \frac{1}{d^2}\leq 2-\frac{1}{j}$$
$$\frac{\pi^2}{6}-2+\frac{1}{j}\leq \sum_{x\in{(\Pi_1)_{>j}}} \frac{1}{x^2}+\sum_{y\in{(\Pi_2\cup\Pi_3)}_{>j},y\notin{\Pi_1}_{>j} } \frac{1}{y^2}\leq \frac{\pi^2}{6}-\frac{3}{2}+\frac{1}{j+1}$$
Let the expression, $$\mu_j=\sum_{x\in{(\Pi_1)_{>j}}} \frac{1}{x^2}+\sum_{y\in{(\Pi_2\cup\Pi_3)}_{> j},y\notin{\Pi_1}_{>j} } \frac{1}{y^2}$$
Also, $$\mu^{*}_j=\sum_{x\in{(\Pi_1)_{>j}},x\notin(\Pi_2\cup\Pi_3)_{> j}} \frac{1}{x^2}+\sum_{y\in{(\Pi_2\cup\Pi_3)}_{> j} } \frac{1}{y^2}$$
Note that $\mu^{*}_j=\mu_{j}$ Thus, $$2\mu_{j}=\mu^{*}_j+\mu_{j}= \sum_{x\in{(\Pi_1)_{>j}},x\notin(\Pi_2\cup\Pi_3)_{>j}} \frac{1}{x^2}+\sum_{y\in{(\Pi_2\cup\Pi_3)}_{> j} } \frac{1}{y^2}+ \sum_{x\in{(\Pi_1)_{>j}}} \frac{1}{x^2}+\sum_{y\in{(\Pi_2\cup\Pi_3)}_{> j},y\notin{\Pi_1}_{>j} } \frac{1}{y^2}\geq\sum_{a\in(\Pi_1)_{>j}} \frac{1}{a^2}+\sum_{b\in(\Pi_2\cup\Pi_3)_{>j}} \frac{1}{b^2}=\chi_{j}$$
Now, $$\frac{1}{25}\sum_{a\in \Pi_1}\frac{1}{a^2}+\frac{1}{25}\sum_{b\in(\Pi_2\cup\Pi_3)} \frac{1}{b^2}>U=\sum_{6k-1=\text{composite}} \frac{1}{(6k-1)^2}+\sum_{6k+1=\text{composite}} \frac{1}{(6k+1)^2}=(1-\frac{1}{2^2})(1-\frac{1}{3^2})\zeta(2)-\sum_{p\in\mathbb{P}_{\geq 5}}\frac{1}{p^2}$$
$$U=\sum_{6k-1=\text{composite}} \frac{1}{(6k-1)^2}+\sum_{6k+1=\text{composite}} \frac{1}{(6k+1)^2}>\frac{1}{49}\sum_{a\in \Pi_1}\frac{1}{a^2}+\frac{1}{49}\sum_{b\in(\Pi_2\cup\Pi_3)} \frac{1}{b^2}$$
Hence, $$\chi_j=\sum_{a\in(\Pi_1)} \frac{1}{a^2}+\sum_{b\in(\Pi_2\cup\Pi_3)} \frac{1}{b^2}-\sum_{a\in(\Pi_1)_{\leq j}} \frac{1}{a^2}-\sum_{b\in(\Pi_2\cup\Pi_3)_{\leq j}} \frac{1}{b^2}$$
Now, Note that, $$\frac{3}{100}<\sum_{p\in\mathbb{P}_{\geq 5}}\frac{1}{p^2}=F<\frac{35}{100}$$
(Got the value of $F$ from Wikipedia) But, $$\sum_{a\in(\Pi_1)} \frac{1}{a^2}+\sum_{b\in(\Pi_2\cup\Pi_3)} \frac{1}{b^2}>25U>\frac{(25\times 3\times8\times\pi^2)-(216\times 8.75)}{216}>18$$
Now, $$ \sum_{a\in(\Pi_1)_{\leq j}} \frac{1}{a^2}+\sum_{b\in(\Pi_2\cup\Pi_3)_{\leq j}} \frac{1}{b^2}<\sum_{l=1}^{j} \frac{2}{l^2}\leq 4-\frac{2}{j}$$ Thus, $$\chi_{j}>14+\frac{2}{j}$$ Hence, $$\mu_{j}>7+\frac{1}{j}>\frac{\pi^2}{6}-\frac{3}{2}+\frac{1}{j+1}$$ A contradiction!!!
Can you guys check if it's correct, Thanks in advance. Please do as much as criticism in the proof, I will not mind anything cause, I need a good solution.
Thanks in advance.