Question

does anyone here know anything about real analysis?

Answer 1

ask!

Answer 2

ok heres the question

Answer 3

{sin(n(pi)/6)} of n conatined in real numbers
is it convergent and provide two subsequences

Answer 4

if n is going to be infinity there are kind of options.
n=6k
n=6k+1
....n=6k+5
and they make 6 different limit, so it is divergence.

Answer 5

it cannot converge because
what @mahmit2012 said,
it is \(\frac{1}{2}\) and \(-\frac{1}{2}\) infinitely often

Answer 6

how do i provide subsequences?

Answer 7

oh right, and also 0, \(\frac{\sqrt{3}}{2}\) etc etc

Answer 8

@mahmit2012 gave them to you in his/her reply

Answer 9

You just think about all options for n that you can build them.
In this question there are 6 options to show your sequence diverge.

Answer 10

why can i be sure that it has a convergent subsequence?

Answer 11

and how can i provide one?

Answer 12

?

Answer 13

for example sin(npi) is convergence
because you have two options n=2k and n=2k+1
but both of them gives you 0
So the sequence diverge to zero.

Answer 14

lets spell it out
if you take as a sub-sequence of \(n\) all natural numbers of the form \(6k\) then you get \[\sin(\frac{6k\pi}{6})=\sin(k\pi)=0\] so this subsequence is identically zero

Answer 15

now see if you can find one that is identically \(\frac{1}{2}\) for example