Can you prove that
sqrt(1+2sqrt(1+3sqrt(1+4sqrt...............
is equal to three?

Is there a generalised method of tackling series or sums similar to this for, say nth roots? What if those nth roots were complex numbers?

You can imagine the above series as being represented by an infinitude of square root signs. What if, instead of square root signs, I had brackets which would place their contents to the power of n, for example:
(x+1(x+2(x+3(x+4(x+5(x+6....)² )² )² )² )² )² )² ?
Obviously, for some x, the above might tend to infinity, but when might it converge and what will it's value then be?

Also, instead of going up in consecutive numbers, what if I went up in squares or cubes?:
sqrt(1+1sqrt(1+4sqrt(1+9sqrt(1+.........

Thanks for any answers given for these series!

Has anyone got any ideas on the things that Anon is asking about? I remember reading that Ramanujan posted the question to some mathematical journal and that no-one could answer it for quite a while, so I imagine that it's not an easy problem, although very interesting.

How can we show that
sqrt(1+x x sqrt(1+(x+1) x sqrt(1+...))) = x+1?

I know this discussion is quite old now, but I'm a new member and just came across this about two weeks ago while looking around. I found it very interesting and so tried to work it out. I was able to find a way to get to the solution and so I just couldn't resist posting it.
Considering that the solution for Ramanujan's series was already given in the message as : sqrt(1+x x sqrt(1+(x+1) x sqrt(1+...))) = x+1; I was able to formulate a way to prove/find it. I did this by forming a "difference equation" or "recurrence relation". Here's how
:-
Consider that F_x =sqrt(1+x x sqrt(1+(x+1) x sqrt(.....))). So we shall have F_x ² =1+xF_x+1 . Now the problem has been reduced to the solution of a difference equation. Unfortunately this is a non-linear difference equation with variable coefficients. But considering that the solution is known to us, or even if the form of the solution is known, we can assume it to be F_x =ax+b, where a and b are constants to be determined.

Substituting this value of F_x in the above equation, then equating the coefficients of various powers of x, we find that a=b=1 are the only possible values. Thus leading to F_x =x+1 - the required solution!! Note that even if we "guess" F_x in a more general manner as F_x =a.p^x .x + b.q^x , although it becomes a bit more complicated F_x still comes out to be (x+1). I guess, a similar line of attack will lead to solutions for most of the series mentioned by Anon above, even if the order of equations or solutions may vary.

There's a problem with the above. What Ashish has shown is that if F_x is of the form ax + b then F_x = x + 1 as required. It is certainly true that F_x = x + 1 does satisfy his difference equation so it is plausible. But F_x = x + 1 is certainly not the only solution. For example, set F₁ = 2. The recurrence relationship now gives us values of F for x = 2,3,4,... However if we were to plot a graph of F against x then we could join up the curve between x = 1 and x = 2 with any continuous curve which starts at (1,2) and (2,3). This would then automatically induce a continuous curve between x = 2 and x = 3 (from the recurrence relationship) and between x = 3 and x = 4, etc. But there are so many different ways of joining up (1,2) and (2,3) so there are loads and loads of other solutions to the recurrence relationship and we need somehow to prove that our one is the one that corresponds to the original expression. It seems very tricky.

Hello Michael,
I do see your point. But I had even tried it with a more general guess of a.x.p^x +b.q^x , since in linear difference equations we have a complementary function of the type c1 x a^x +c2 x b^x ..etc; and the solution turned out to be x+1 again.
Also, I haven't been able to solve the other 2 equations, since I didn't have the right "guesses" available to me. I'll try to find some reference that helps in solving non-linear difference equations. In the mean time I think, we should assume a solution of the form F_x = a₀ xx⁰ + a₁ xx¹ + a₂ x x² + a₃ x x³ + .... Here a₀ ,a₁ ,a₂ etc are constants. Most functions can be expressed this way (from Taylor's expansion). And if we then replace this in the difference equation and solve for a₀ ,a₁ ,a₂ ... using the initial conditions my guess is we'll end up with x+1. But I haven't tried it yet, since I'm quite busy with my exams right now. But I'll try to do it soon and post whatever I get out of it. Maybe we'll need to include negative coefficients as well !? I'll see. But my guess is that x+1 is the only solution.
In the mean time here's something curious that I did. I made a program that assumes a value of F_n , n> > x and calculates F_x from it. It turned out that for n= 50, 100 ... etc and values of F_n assumed over a very wide range, the value of F₁ turns out to be very close to 2 (of the order 10^-10 or smaller)!! Similarly for other F_x that I tried. Amazingly stabilizing! That makes me tend to believe that x+1 is the right and distinct solution.
I do agree that for an infinite number of initial conditions there would be infinite different solutions to the equation. But I think that there's always a unique solution, given an initial condition with sufficient constants. Here, I believe that one constant is sufficient since F_x is defined in terms of F_x+1 and some coefficients.
Ashish

Michael is not saying the answer is not x+1. He just said that you have NOT proved the answer is x+1 for all x. You have guessed what the solution form might be, but there may be other solutions which are not in your form which also satisfies the recurrence. In other words, you have to prove that your solution is the unique solution of the (non-linear) difference equation. However, this is not true, and you probably have to go back to the original definition of F_x and show that F_x , as a limit of a sequence, converges to the limit x+1, for any x> -1.

Kerwin

I don't have any idea how to go about solving such equations without making initial guesses. I think that's how linear difference equations are also solved, although it's not quite that obvious. Maybe there exists a solution with one (or more) arbitrary constants that always satisifies the equation. But I just can't seem to prove that or even otherwise. .
Still, I'll keep trying.
Ashish

Hi Ashish,

There's certainly nothing wrong with making guesses at the solution to difference equations. However we do need to prove that we have the correct solution.

If you are dealing with a linear discrete recurrence relationship, e.g.

f(n+2) = 5f(n+1) - 6f(n)

where f is from Z -> R you can solve this by guessing solutions. Specifically you guess solutions of the form aⁿ and you come up with 2ⁿ and 3ⁿ . Then you can show that A 2ⁿ + B 3ⁿ also satisfies this recurrence relationship.

Setting g(n) = (3f(0) - f(1)) 2ⁿ + (f(1) - 2f(0)) 3ⁿ - f(n) you obtain:

g(0) = 0
g(1) = 0
g(n) = 5g(n-1) - 6g(n-2)

Therefore g(n) = 0 for all integers n

Hence it is true for all integers n that f(n) = (3f(0) - f(1)) 2ⁿ + (f(1) - 2f(0)) 3ⁿ so we have proven that our choice of functional form for f is correct - it is the only possible form.

So essentially we have proved uniqueness. But it is clear no that uniqueness theorem exists for your recurrence relationship. Your equation only relates F_x to F_x+1 but F is from R to R! Therefore essentially, f can do anything it wants when x is in [1,2] provided it is continuous and hits the right values at the endpoints.