
Re: Sequence limit
Posted:
Oct 7, 2013 1:46 PM


On 07/10/2013 15:34, Mohan Pawar wrote:
> I won't attempt to attain a low level to match you in ranting and > displaying vitriol. At times it is better to keep shut and let others > wonder that you are stupid than to open your mouth and remove their all > doubts. Bart Goddard, when others critically read the solution under > "somehow explained to Bart Goddard" you will have removed their doubts > about you.
Bart might be a bit blunt at times, but he is one of the most knowledgable contributors remaining in sci.math.
> Let y = sin (1/m) ^(m) Taking log on both sides => ln(y) = m > ln(sin (1/m)) Take limit on both sides as m>0 and evaluating it > =>lim m>0 ln(y) = lim m>0 m ln(sin (1/m))= 0 (at the time of > evaluating limit, m=0 is the multiplier and one doesn?t need to care > about value of ln(sin (1/m)). Also, the limit is determinate.)
You do need to care about the value of log sin(1/m). Note that here m = 1/n in the original problem, so that m is a reciprocal of a natural number. With this class of m, 1/m can be arbitrarily close to an integer multiple of pi, and so sin(1/m) can be arbitarily close to zero and logsin(1/m) can be the negative of an arbitrarily large number. The question is, whether then multiplying logsin(1/m) will give you a sequence tending to zero or not. A priori, logsin(1/m) might get large enough so that even mlogsin(1/m) doesn't tend to zero. (It doesn't but that needs some nontrivial mathematics).
You should realize that if you have two sequences (a_n) converging to zero and (b_n) then it is not automatic that the sequence (a_n b_n) converges to zero. To determine it limiting behaviour, you need more information about both the original sequences. (Your botched argument is the case a_n = 1/n and b_n = logsin(1/n); certainly a_n > 0 but that is not enough by itself to conlcude a_n b_n > 0. Note that (b_n) is here an *unbounded* sequence.)
> =>lim m>0 ln(y) = 0 => lim m>0 y = e^0=1 => lim m>0 sin (1/m) > ^(m)= 1 as before. (ALSO, VERIFIABLE ON WOLFRAM ALPHA)
Well, if Wolfie Alpha says it's true, it must be true :)
Returning to the original and less confusing notation then (i) the sequence (c_n) defined for natural numbers n by c_n = sin(n)^{1/n} converges to 1, and (ii) the function f(x) defined for real x>0 by f(x) = sin(x)^{1/x} does not have a limit as x > infinity.
As we have seen, (ii) is easy, but (i) is relatively difficult to prove.

