# Thread: sequence- diverges/converges and limits

1. ## sequence- diverges/converges and limits

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

2. Originally Posted by inthequestofproofs
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
You need to be careful what happend to the factor $\displaystyle \frac{1}{n}$.

So you have the product of two sequences one of which is bounded and the other is going to zero...

3. 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!

4. Originally Posted by inthequestofproofs
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!
You should have a theorem about the first hint that I gave you.
As an alternative you could use the squeeze theorem.

$\displaystyle \frac{-1}{n}\le P_n \le \frac{1}{n}$

Or just write out some terms and see if you can find a pattern. This have an equivilent form that is much simpler.

Good luck.

5. I got it!
using the squeeze theorem, i know lim -1/n = 0 and lim1/n =0 thus, the lim pn =0 by squeeze theorem.

Thanks

6. 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

7. Why make so difficult?
$\displaystyle \frac{{ - 1}} {n} \leqslant \frac{1} {n}\sin \left( {\frac{{n\pi }} {2}} \right) \leqslant \frac{1} {n}$

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### sin(npi/2) converge or diverge

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