Shmuel (Seymour J.) Metz <firstname.lastname@example.org> wrote in news:email@example.com:
> In <XnsA24E846DE2C77goddardbenetscapenet@188.8.131.52>, on 10/03/2013 > at 06:01 PM, Bart Goddard <firstname.lastname@example.org> said: > >>Calculus techniques imply the answer is 1. >
I mean that if a calculus student works the problem, he probably elides over the difference between a continuous and a discrete variable, sees sign, assumes the numerator is bounded and decides the limit of the log is zero and then concludes, almost correctly, that the limit is 1. Since in the discrete case, all the zeros are missed, it's likely that the limit really is 1.
>>Anyone wrestled with the subtlies of this problem? > > So far you haven't shown a problem. Show your purported proof and > maybe someone will take the time to point out the error in it.
There is no purported proof. Who knows what you're talking about now? I purported nothing.
>>E.g., can you construct a subsequence n_k such >>that sin (n_k) goes to zero so fast that the >>exponent can't pull it up to 1? > > For all k, n_k=0. Lim k->oo |sin(n_k)|^(1/k) = 0.
Not a subsequence. But thanks for playing.
> Is this homework?
It was someone's homework. I stated the source of the problem.