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


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.
