<s>[INST] How to find this limit: $A=\lim_{n\to \infty}\sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{3}+\cdots+\sqrt{\frac{1}{n}}}}}$ Question:

Show that $$A=\lim_{n\to \infty}\sqrt{1+\sqrt{\dfrac{1}{2}+\sqrt{\dfrac{1}{3}+\cdots+\sqrt{\dfrac{1}{n}}}}}$$ 
  exists, and find the  best estimate limit $A$.

It is easy to show that

$$\sqrt{1+\sqrt{\dfrac{1}{2}+\sqrt{\dfrac{1}{3}+\cdots+\sqrt{\dfrac{1}{n}}}}}\le\sqrt{1+\sqrt{1+\sqrt{1+\cdots+\sqrt{1}}}}$$
  and it is well known that this limit
  $$\sqrt{1+\sqrt{1+\sqrt{1+\cdots+\sqrt{1}}}}$$ exists.

So
$$A=\lim_{n\to \infty}\sqrt{1+\sqrt{\dfrac{1}{2}+\sqrt{\dfrac{1}{3}+\cdots+\sqrt{\dfrac{1}{n}}}}}$$ 

But can use some math methods to find an approximation to this $A$ by hand?

and  I guess maybe this is true:
$$1<A\le (\pi)^{\frac{1}{e}}?$$
By the way:  we can prove  $A$ is a transcendental number?
Thank you very much!
 [/INST] Well, you can get very good approximations by computing:
$$\sqrt{a+\sqrt{a+\sqrt{a+...}}}=\frac{1+\sqrt{1+4a}}{2}$$

$$b_1=\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}...}}}=\frac{1+\sqrt{3}}{2}$$
$$b_2=\sqrt{\frac{1}{3}+\sqrt{\frac{1}{3}+\sqrt{\frac{1}{3}...}}}=\frac{1+\sqrt{7/3}}{2}$$
$$b_3=\sqrt{\frac{1}{4}+\sqrt{\frac{1}{4}+\sqrt{\frac{1}{4}...}}}=\frac{1+\sqrt{2}}{2}$$
Then you use the facts that:
$$A^2-1<b_1$$
$$(A^2-1)^2-\frac{1}{2}<b_2$$
And so on, to calculate upper bounds for $A$ with very good accuracy:
$$A<\sqrt{1+b_1}=1.53819$$
$$A<\sqrt{1+\sqrt{\frac{1}{2}+b_2}}=1.52580$$
$$A<\sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{3}+b_3}}}=1.52300$$

$$A<\sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{3}+\sqrt{\frac{1}{4}+b_4}}}}=1.52224$$

And so on. This is already very close to the accurate expression. It's a rather easy calculation and can be done by hand if needed.
For the lower boundary you can use $b=1$ (it's rather obvious):
$$A>\sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{3}+\sqrt{\frac{1}{4}+1}}}}=1.51844$$
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