Question

What is the value of this expression ...... ?

Answer 1

\(\sqrt{1+2\sqrt{1+3\sqrt{1+4..}}}\)

Answer 2

@mukushla

Answer 3

isnt it infinite...

Answer 4

Yes the sequence is infinite

Answer 5

and the value of expression isnt infinite

Answer 6

Don't think so.........

Answer 7

@eliassaab

Answer 8

We can prove that

\[x+1=\sqrt{1+x\sqrt{1+(x+1)\sqrt{...}}}\]

but this is a hard work....

Answer 9

this problem is related to functional equations...

Answer 10

@mukushla
Please solve it............

Answer 11

let

\( f(x)=\sqrt{1+x\sqrt{1+(x+1)\sqrt{...}}}\)

then

\( f^2(x)=1+x\sqrt{1+(x+1)\sqrt{...}}=1+x f(x+1)\) so problem changes to find the answer of this functional equation...

\( f^2(x)=1+x f(x+1)\) and \(f(x) \ge 0 \)

Answer 12

ok

Answer 13

this is not a exact solution....

Let us look for a polynomial \(f(x)\) which satisfies our equation. If \(f(x)\) were a polynomial of degree \(n\) then the left-hand side of equation would be of degree \(2n\) and the right hand side of degree \(n + \)1. it means \(n+1=2n\) ----> \(n=1\)
So our polynomial is of degree one. Let \(f(x) = ax + b\). put this in the equation We get

\((ax + b)^2 = 1+x (ax + a + b) \) and it gives \(a=b=1\) so \(f(x)=x+1\).

Using Advanced methods of solving functional equations gives \(f(x)=x+1\) too.

Answer 14

letting \(x=1\) gives the answer for that radical...

Answer 15

I think letting x=2 gives the value

Answer 16

sorry .... x=2

Answer 17

original Q was posted by
http://en.wikipedia.org/wiki/Srinivasa_Ramanujan

Answer 18

yeah and i learned that way from a book that is not my solution...

Answer 19

lol ... it wasn't soled for six months ... after it got posted!!

Answer 20

lol ... it wasn't solved for six months ... after it got posted!! thank you for posting solution!!

Answer 21

:)