Obtain the Cholesky decompositions of

$$H_{3}=\left(\begin{array}{ccc} 1 & \frac{1}{2} & \frac{1}{3} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \end{array}\right), \quad H_{4}=\left(\begin{array}{cccc} 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6} \\ \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & \lambda \end{array}\right) .$$

What is the minimum value of $\lambda$ for $H_{4}$ to be positive definite? Verify that if $\lambda=\frac{1}{7}$ then $H_{4}$ is positive definite.