Question 6
Let $n$ be a positive integer. Prove that the equation
$$\frac1{x-1}+\frac1{2^2x-1}+\cdots+\frac1{n^2x-1}=\frac12$$
has a unique solution $x_n\gt 1$. Prove that $x_n$ tends to $4$ as $n$ tends to infinity.
$$\frac1{x-1}+\frac1{2^2x-1}+\cdots+\frac1{n^2x-1}=\frac12$$
has a unique solution $x_n\gt 1$. Prove that $x_n$ tends to $4$ as $n$ tends to infinity.