Question:Show that A=limn→∞√1+√12+√13+⋯+√1n

exists, and find the best estimate limit A.It is easy to show that

√1+√12+√13+⋯+√1n≤√1+√1+√1+⋯+√1

and it is well known that this limit

√1+√1+√1+⋯+√1 exists.So

A=limn→∞√1+√12+√13+⋯+√1nBut 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 10−8.

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