Hello
I am trying to figure out if the following sequence converges:
Pn={(1/n)*sin(npi/2))} from n=1 to infinity.
This is how I am approaching it. I am using the fact that a sequence converges iff all of its subsequences converge to the same limit. So, I assume that Pn converges to p.
Then, I look at the subsequence {sin(npi/2)}. This subsequence doesn't converge because it fluctuates between -1,0 and 1 depending on the value of n. So, since this sequence doesn't have a limit and it doesn't converge, then Pn diverges. Am I right?
Thanks
so, if the lim 1/n = 0 and the lim sin(npi/2) doesn't exist, then can i use the fact that lim(1/n)(sin(npi/2)) = lim1/n * lim sin(npi/2)? or is this property only used for sequences that converge?
how do i go about proving if the sequence converges?
SHould I find a p and an n0 such that |pn - p| < e for all n>= n0?
Thanks!
i have another question for you:
how do i show that pn={(ncos(npi))/(2n+3)} converges or diverges.
THis is what i did so far:
[ncos(npi)][/(2n+3)]=[cos(npi)][/(2+(3/n)]
as n --> infinity, the numerator fluctuates from -1 and 1. The denominator, goes to 2. So, as n goes to infinity, the sequences takes the values -1/2 or 1/2 depending on the cos(npi).
Does this sound good to prove that it diverges?
THanks