21thank u..
what is the answer given in Polya and Szego Prob No.162 to this problem
?
19 Answers
66Well the problems seems to be difficult. I haven't been able so far to solve it entirely but this is what I have done. First thing to note is that the given sum exists. We consider a generalization:
f(x) = \sqrt{x+\sqrt{(x+1)+\sqrt{(x+2)+\ldots}}}
so that we are trying to find f(1). Its easy to see that f(x) satisfies a non-linear functional equation
f(x)^2 = x + f(x+1)
If some how we could solve this equation for f(x), then we can find the given sum.
Usually such kind of nested infinite radicals are solved this way. A famous sum of a similar kind was given by Ramanujan. He asked for the sum of the infinite nested radical
\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\ldots}}}}
(This equals 3)
p.s.: could you quote the source of the problem?
21
341after ricky's post, i saw that this prob is there in some books - Polya and Szego Prob No.162 for example. As to convergence the facts that the sequence monotonically increases and is bounded above by Ramanujan's sequence which converges to 3 are sufficient
1http://pballew.net/1935Herschfeld.pdf
Go to the above link and you will find what you need , Sir .
341That's really interesting. Could you give us some reference for this. I am unable to find any google results for this keyword
1This is a constant named as " Kasner Costant " . Its exact closed form expression hasn ' t been found out yet , but its approximate value , is 1 . 7533 .
66The sum exists for sure. The sum could be interpreted as the limit of a sequence of numbers of the kind
a_n = \sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+\ldots +\sqrt{n}}}}}
There is a general theorem regarding this. Check
http://mathworld.wolfram.com/NestedRadical.html
11sir, our 1st sem maths teacher gave this problem to us. however he neither told the method of evaluating this nor the answer..
the actual question was... 'whether the the sum existed and if it did, whats the answer??'
bt i don't know the actual source of this problem.. :(
66I hope you are talking about this:
\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+\ldots}}}}
11See, if you consider √∞ to be ∞ then the answer is ∞. But i dont think we can rightly say that so according to me √∞ is not defined. So, the answer is not defined. :P.
But maybe if you try limits you can get an accurate answer.
11@ rishabh, how can you approximate √1+√2+√3+...∞ = √2+√3+...∞ = x ??
we could have done that for √1+√1+√1+.....∞
so, i don't think it is right..
1i considered the answer to be some x.
then as this is an infinite series then we can approximate √1+x = x
=> on suaring x =1+√52
@fahad then 1 + 1/2 + 1/3 ................... should also not have any algebraic value but we still write it as 1/ (1-1/2) = 2
11@anant sir, yes sir i was talking about that..
but sir how to evaluate this one??
@rishabh, i don't know the answer.. :(
1It is imposibble that the answer is an algabric number, it should be a transcendental number.