$\lim\limits_{n \to \infty}\sqrt{1+\sqrt{\frac{1}{2^2}+\sqrt{\frac{1}{3^2}+\cdots+\sqrt{\frac{1}{n^2}}}}}.$

Question

Problem

Evaluate $\lim\limits_{n \to \infty}T_n$ where $$T_n=\sqrt{1+\sqrt{\frac{1}{2^2}+\sqrt{\frac{1}{3^2}+\cdots+\sqrt{\frac{1}{n^2}}}}}.$$

Analysis

It's obvious that $T_n$ is increasing with a greater $n$, since \begin{align*} T_{n+1}&=\sqrt{1+\sqrt{\frac{1}{2^2}+\sqrt{\frac{1}{3^2}+\cdots+\sqrt{\frac{1}{n^2}+\sqrt{\frac{1}{(n+1)^2}}}}}}\\ &>\sqrt{1+\sqrt{\frac{1}{2^2}+\sqrt{\frac{1}{3^2}+\cdots+\sqrt{\frac{1}{n^2}+0}}}}\\ &=T_n. \end{align*}

Moreover, we can prove that $T_n$ is bounded upward, since \begin{align*} T_n&=\sqrt{1+\sqrt{\frac{1}{2^2}+\sqrt{\frac{1}{3^2}+\cdots+\sqrt{\frac{1}{n^2}}}}}\\ &\leq \sqrt{1+\sqrt {1+\sqrt{1+\cdots+\sqrt{1}}}} \\ &\to \frac{\sqrt{5}+1}{2}. \end{align*} Therefore, $T_n$ is convergent as $n \to \infty$, by the monotonicity convergence theorem.

But where does it converge to on earth? Does the limit have a excact value? I have already computed the value using the former $20$ terms by Mathematica, it output:

Answer 1

You can get an upper bound by "freezing" the denominators in the radicands at some value. If you freeze at $n=1$, you get the $(1+\sqrt{5})/2$ bound with which you proved convergence. Now consider a later upper bound in the sequence, say you "freeze" at $n=4$. Thereby

$L<\sqrt{1+\sqrt{\dfrac{1}{4}+\sqrt{\dfrac{1}{9}+\sqrt{\dfrac{1}{16}+\sqrt{\dfrac{1}{16}+\sqrt{\dfrac{1}{16}+...}}}}}}$

We render

$\sqrt{\dfrac{1}{16}+\sqrt{\dfrac{1}{16}+\sqrt{\dfrac{1}{16}+...}}}=\color{blue}{\dfrac{2+\sqrt{5}}{4}}$

by the usual fixed point method. Thereby

$L<(UB)_4=\sqrt{1+\sqrt{\dfrac{1}{4}+\sqrt{\dfrac{1}{9}+\color{blue}{\dfrac{2+\sqrt{5}}{4}}}}}$

Rounding to the next higher multiple of $0.001$:

$L<1.468$

When you "fix" the radicands at increasing values of $n$, the sequence of upper bounds you get will converge to the true limit from above, allowing you to certify arbitrary accuracy with a finite input.