The value of 3 + 1 4 + 1 3 + 1 4 + 1 3 + … ∞ is equal to

Let $x=3+\frac{1}{4+\frac{1}{3+\frac{1}{4+\frac{1}{3+\dots \infty }}}}$

So, $x=3+\frac{1}{4+\frac{1}{x}}=3+\frac{1}{\frac{4x+1}{x}}$

$\Rightarrow \left(x-3\right)=\frac{x}{(4x+1)}$

$\Rightarrow (4x+1)(x-3)=x$

$\Rightarrow 4x^2-12x+x-3=x$

$\Rightarrow 4x^2-12x-3=0$

$x=\frac{12\pm \sqrt{(12{)}^2+12\times 4}}{2\times 4}=\frac{12\pm \sqrt{12\left(16\right)}}{8}$

$=\frac{12\pm 4\times 2\sqrt{3}}{8}=\frac{3\pm 2\sqrt{3}}{2}$

$x=\frac{3}{2}\pm \sqrt{3}=1.5\pm \sqrt{3}.$

But only positive value is accepted

So, $x=1.5+\sqrt{3}$