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sqrt(1+3sqrt(1+3sqrt(1+(...)))) > y > sqrt(1+2sqrt(1+2sqrt(...))) Which means LHS is solution of y^23y1 and RHS is solution of y^22y1 => (3+sqrt(13))/2 > y > 1+sqrt(2) => 3.3 > y > 2.414 Siddhi, I think the left hand side of your inequality is incorrect  the factor is usually larger than 3. I don't know the answer to this question, but my guess is that the value is infinite. The square root of a very large number is still fairly large, and if you multiply that by anothr large number and take the square root you'll get an even larger number than the first square root. Perhaps you can program a computer to try this for some large, finite sequence. Hburgiel, No its a finite number, The factors become more and more insignificant as the proceed INTO the roots, So in reality the number is converging to a finite number. PS: I tried on google with increasing finite number and the number increases but with a decreasing slope, decreasing rate. (I felt it to be converging to the finite number 3) amogh, though the number increases but with decreasing slope, decreasing rate, it doesn't mean that the number is not infite  consider the harmony series \sum{\frac{1}{n}} Yes you are right luckyted, My explanation was not right but it is converging, Sorry I can't prove it but that's what I noticed! yes amogh,u r write...its a finite number n its 3 but how do u solve it or guess? I'm thinking on it, I'll post it here if I come up to anything good! Based on Siddhi's work, it follows that, there exists one x such that, and 2<x<3 y^2xy1=0 Also we can find a smaller range, by taking sqrt{1+2sqrt{1+3sqrt{1+3sqrt{...}}}} < y < sqrt{1+2sqrt{1+4sqrt{1+4sqrt{...}}}} (((y^21)/2)^21)/3 = (y^21)/2 y^4  8y^2 +3 =0 y=sqrt{4 + sqrt{13}} =2.757 y^410y^2+5=0 y=sqrt{5+2sqrt{5}} =3.08 2.76<y<3.08 Now this seems to be a limiting case but how can you justify it?.... We need to find a limiting equation or a second equation in terms of y and x. Anymore suggestions Welcome. Regards, Amogh.K Amogh, the two inequalities which you have used , do you have any proof that there exist 2<x<3 such that \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{...}}}} = \sqrt{1+x \sqrt{1+x \sqrt{1+x \sqrt{...}}}} and y < \sqrt{1+2\sqrt{1+4\sqrt{1+4\sqrt{...}}}} see the series is an infinite one, so if can't have any bounded value......but we can find the lower limit of the value of the expression. trivially it should be greater than zero. let y =sqrt(1+2sqrt(1+3sqrt(1+4sqrt(1+........)))) (y^2 1)/2 = sqrt(1+3sqrt(1+4sqrt(1+.....))) now sqrt(1+3sqrt(1+4sqrt(1+.....))) > sqrt(1+2sqrt(1+3sqrt(1+4sqrt(1+........)))) (y^21)/2 > y y^22y1>0 and we have y>0 these two simultaneous conditions leads us to a solution that y>2.414. See first of all, its obvious why y> sqrt(1+2 sqrt(1+ 2 sqrt(...))), isn't it? OK, I'm back from work. sqrt{3} > sqrt{sqrt{4}} > sqrt{sqrt{sqrt{5}}} ..... So what I'm trying to say is that the significance of the numbers goes on decreasing from left to right, ie. into the roots. sqrt{n}>sqrt{sqrt{n+1}} only when n is finite no, in case of any infinite no how can you say this!!!! the given series has no termination.
 3 years ago
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