I found the following problem on AoPS , and couldn't find a solution (and no headway whatsoever).
Show that $a_n = \sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{3}+\cdots +\sqrt{\frac{1}{n}}}}} < (\pi)^{\frac{1}{e}}$.
And find $\lim\limits_{n\to\infty} a_n = \lim\limits_{n\to\infty} \sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{3}+\cdots +\sqrt{\frac{1}{n}}}}}$.
The first inequality is weak, and I have no idea how to proceed. I can't find a recurrence relation either.
I am not sure if the limit (let $a = \lim a_n$) can be evaluated in closed form, though convergence is easily established and bounds can also be made as close as you desire. Lower bounds are easy as $a_n$ is monotone increasing, successive $a_n$ give better lower bounds.
To find good upper bounds, we may borrow a trick often useful for nested radicals. Note that from the well known limit (for $x> 0$): $$f(x) = \sqrt{x+ \sqrt{x + \sqrt{x+\cdots }}} \implies f(x)=\frac{1+\sqrt{1+4x}}2$$
Hence for any $k>1$, $$a = \sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{3}+\cdots}}} < \sqrt{1+\sqrt{\frac12+\sqrt{\frac13+\cdots+\sqrt{\frac1k+f\left(\frac1{k+1} \right)}}}}$$
for our purposes $k=3$ suffices and gives $a < \sqrt{1+\sqrt{\frac12+\sqrt{\frac13+f\left(\frac14\right)}}} \implies a < 1.522998 < \pi^{1/e} \approx 1.523671$