How to find this limit: A=limn→∞√1+√12+√13+⋯+√1nA=\lim_{n\to \infty}\sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{3}+\cdots+\sqrt{\frac{1}{n}}}}}

Question:

Show that A=limn1+12+13++1n
exists, and find the best estimate limit A.

It is easy to show that

1+12+13++1n1+1+1++1
and it is well known that this limit
1+1+1++1 exists.

So
A=limn1+12+13++1n

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

and I guess maybe this is true:
1<A(π)1e?

By the way: we can prove A is a transcendental number?

Thank you very much!

Answer

The nature and closed form expression of these two related constants, i.e., the Nested Radical Constant and Somos's Quadratic Recurrence Constant, are (also) unknown. This would suggest that the same holds true for this one as well, meaning that we are dealing with an open question.

As far as numeric approximations are concerned, A(π+1)ln41+ln16 comes close, within an error of less than 108.

Attribution
Source : Link , Question Author : math110 , Answer Author : Lucian

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