Let $x=\frac{1}{b+\cdots}.$ Consider the value of $x.$ Since $b \geq 1$ and the numerator is $1,$ it follows that $x<1$. Now, consider the value of $a$. Since $x<1,$ $1 \leq \frac{15}{11} < 2$ and $a$ is integral, $a$ must be $1.$ So we now have $\frac{15}{11} = 1 + \frac{4}{11}.$ Since we require the numerator of each nested fraction to be $1,$ we can use the fact that $\frac{m}{p} = \frac{1}{\frac{p}{m}}$ to get $\frac{15}{11} = 1 + \frac{1}{\frac{11}{4}}.$ Using this technique successively, we can decompose the expression and get $$ \begin{aligned} \dfrac{15}{11} &= 1+\dfrac{4}{11} \\[2mm] &= 1+\dfrac{1}{\dfrac{11}{4}} \\[2mm] &= 1+\dfrac{1}{2+\dfrac{3}{4}} \\[2mm] &= 1+\dfrac{1}{2+\dfrac{1}{\dfrac{4}{3}}} \\[2mm] &= 1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{3}}}. \end{aligned} $$ And hence $a=1,b=2,c=1,d=3.$