Hard math problem?

Question

Let sqrt denote the square-root operator. Find the following:
sqrt(1+2sqrt(3+4sqrt(5+6sqrt... where the iteration is taken to infinity.

Answer 1

The answer is e=2.718128.
The proof is n^2=e^n.logn, so the root of the highest number chosen whatever if is n, then other numbers will converge on to e as you reach the first number from the last. This will be something like 1+ 1/2!+1/3!+.........=e

Answer 2

Infinity

Answer 3

hint:
use a variable such as n
and define a recursive
function that gives the
same expression.

then find the limit of the
series as n approaches infinity

Answer 4

I don't have the answer but I gotta sqrt for ya & it may go to infinity....

Answer 5

That would be equal to 0/0