I incorrectly proved that $$\frac32-\sum_{k=1}^\infty\frac{(k+1)!B_{2k}}{(2k)!}=\frac{\pi^2}6$$According to Wolfram Alpha, the LHS is approximately $1.34096$, while $\frac{\pi^2}6\approx1.64493$. Here is my proof:
Theorem $1$: Let $f(x)$ be an infinitely differentiable continuous function in the interval $[1,\infty]$ with the property that $\lim_{x\rightarrow0}f(x)=0$. Then $$f(x)=\sum_{k=1}^\infty\left(\frac{(-1)^{k+1}B_{2k}}{(2k)!}f^{(k+1)}(x)-f'(k+x)\right)-\frac{f'(x)}2$$Proof of Theorem $1$: The Euler-Maclaurin Summation formula states that $$\sum_{k=1}^xf(k)=\int_0^xf(t)dt+\frac12(f(x)-f(0))+\sum_{k=1}^\infty\frac{(-1)^kB_{2k}}{(2k)!}(f^{(k)}(x)-f^{(k)}(0))$$Another identity is $$\sum_{k=1}^xf(k)=\sum_{k=1}^\infty(f(k)-f(k+x))$$Which telescopes. Note that for this to be true, $f(k)\rightarrow0$. Taking the derivative of both sides and setting them equal to each other, we have $$f(x)+\frac{f'(x)}2+\sum_{k=1}^\infty\frac{(-1)^kB_{2k}}{(2k)!}f^{(k+1)}(x)=-\sum_{k=1}^\infty f'(k+x)$$Solve for $f(x)$. Q.E.D
Set $f(x)=\frac1x$. Then we have the following: $$\frac1x=\sum_{k=1}^\infty\left(\frac1{(k+x)^2}+\frac{B_{2k}(k+1)!}{(2k)!x^{k+2}}\right)+\frac1{2x^2}$$Set $x=1$: $$\frac12=\sum_{k=1}^\infty\left(\frac1{(k+1)^2}+\frac{(k+1)!B_{2k}}{(2k)!}\right)$$Splitting the sum into two parts and substituting: $$\frac12=\frac{\pi^2}6-1+\sum_{k=1}^\infty\frac{(k+1)!B_{2k}}{(2k)!}$$Add $1$ and subtract the sum on both sides: $$\frac32-\sum_{k=1}^\infty\frac{(k+1)!B_{2k}}{(2k)!}=\frac{\pi^2}6$$
I am not sure what is incorrect here. I am very sure Theorem $1$ is true, but the second part doesn't seem to be wrong either.