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:
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