How do I find the least upper bound / greatest lower bound of

>>8806057
Babby's first calc homework?

>>8806063
uhh don't think so nice try though

>>8806057
Casually, it's the largest and smallest number in the range of the sequence for all n.

Look up the formal definition and follow it for a proof, but it's -1 and 1 respectively.

>>8806071
But senpai there are no values of n (integer, guess I could have specified) for which sin(n)=1 or -1. I get the LUB's and GLB's don't have to necessarily be in the set, but how do I know there's no number less than 1 that also bounds the set from above?

>>8806082
You could try reading the course you should have been following more closely. Also I'm pretty sure there are answers to this all over the internet.

Now why don't you fuck off and follow the board's rules ?

>>8806082
Sin is periodic with 2pi.
As n gets large, eventually you get something that's arbitrarily close to pi and -pi times a constant.

As n goes to infinity, there's some power of 10 such that 10^k*pi such that pi becomes a natural number.

How you answer the question depends on the level of understanding implied with the reading

>>8806098
Pi/2 sorry.

There's also a theorem about the derivative of a function increasing/decreasing on an interval implying that the sequence f (xn) is bounded above/below so you could use that

>>8806082
Of course there are no integer solutions. You are not looking for solutions, you are looking for a proof that there are integers which get you arbitrarily close to 1 or -1.

>>8806082
https://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(number_theory)

>>8806168
ty for a useful response senpai

>>8806190
That doesn't actually help you.

http://artofproblemsolving.com/community/c7h363309p1996432

>>8806209
you're right; it's not as immediately relevant as i first thought. an alternative might be to use weyl's equidistribution theorem to show that the sequence 1,2,3,... mod 2pi is equidistributed on the interval [0,2pi). assume for the sake of contradiction there is no subsequence that converges to pi/2. then there is an open interval containing pi/2 which does not contain any elements of the sequence. now consider a smooth bump function which vanishes outside of this interval and use the definition of equidistribution to get a contradiction.