limits - $limlimits_{n to infty}sqrt{1+sqrt{frac{1}{2^2}+sqrt{frac{1}{3^2}+cdots+sqrt{frac{1}{n^2}}}}}.$

Evaluate $\lim\limits_{n \to \infty}T_n$ where

$$T_n=\sqrt{1+\sqrt{\frac{1}{2^2}+\sqrt{\frac{1}{3^2}+\cdots+\sqrt{\frac{1}{n^2}}}}}.$$

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: