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        riddles >> putnam exam (pure math) >> A sequence
        
(Message started by: wmat1234 on Aug 31st, 2007, 7:18am)
Title: A sequence
Post by wmat1234 on Aug 31st, 2007, 7:18am
Sequence:

x_1 = 1

x_n = x_{n-1} + sqrt(x_{n-1})

what is lim x_n/n^2
Title: Re: A sequence
Post by Eigenray on Aug 31st, 2007, 9:21am
Let yn = 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gifxn; then

yn+12 = yn2 + 2yn.

On the one hand,

[hide]yn+12 < (yn+1)2,

so yn < n, and limsup yn/n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif 1.[/hide]

On the other hand,

[hide]Clearly xn http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif xn-1 + 1, so xn (hence yn) goes to infinity.  Now, fix 0<a<1.  Then there exists an N such that for n>N, yn > C = a2/[2(1-a)], and therefore

yn+12 = yn2 + 2yn > yn2 + 2ayn + a2 = (yn + a)2,

so for n>N, yn > C + a(n-N).

Therefore liminf yn/n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif a.  Since this holds for all a<1, we have liminf yn/n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 1.  Since also yn < n, it follows lim yn/n = 1, and thus lim xn/n2 = 1/4.[/hide]


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