Question

Is this following sequence bounded?
And does it converge? If so find the limit:
Xn=( n^2cos(n*pi))/(2n^2 + n)

Answer 1

\[x _{n}= (n ^{2} \cos(n \Pi))\div(2n ^{2} + n)\]

Answer 2

cos(n*pi) = (-1)^n

for integers, so might be helpful, I don't know.

Answer 3

I'm not sure what to do thats why im a little confused

Answer 4

I honestly never remembered the tests for these, I would sometimes just try to look at it in different ways and be able to more or less figure it out through reckoning.

Answer 5

Like, obviously the top is oscillating between + and - and the limit of a single term will approach 1/2. Seems like it would be divergent.

Answer 6

But maybe I'm missing something, it's kind of tricky looking. Good luck. Try looking at Paul's online notes for calculus 2.

Answer 7

It's a practice for an upcoming test .. Thankss anyways I'll check Paul' online notes

Answer 8

Hmmm... Wolfram Alpha says the root and ratio test are inconclusive.

http://www.wolframalpha.com/input/?i=infinite%20sum%20(%20n%5E2cos(n*pi))%2F(2n%5E2%20%2B%20n)%20&t=crmtb01

Just an extra tool to help you practice for your test.

Answer 9

So its means that the sequence isnt bounded ?

Answer 10

Yeah. As you approach infinity you will either get a -1 or +1 times your term because of the cosine term, right?

Answer 11

So remove that cosine part and just take the limit as it approaches infinity and the divergence test will show you that if it was just a positive (or negative) term it would fail, since the limit will be 1/2.

So if you're "at" infinity and have 1/2 with every other number oscillating between +1 and -1 then eventually you'll just be at:

+1/2-1/2+1/2-1/2+1/2-1/2+... which doesn't converge for the same reason that
\[\lim_{x \rightarrow \infty} \cos(x)\] doesn't converge.

Answer 12

In fact, it'll look almost exactly like a more discrete form of this limit. \[\lim_{x \rightarrow \infty} \frac{ 1 }{ 2} \cos(x)\]

Answer 13

I think I'm starting to get it .. So that means that the fucntion will oscillate between -.5 and . 5 AND DIVERGE TO INFINITY ? Sorry for the caps loll

Answer 14

Well, not necessarily diverge TO infinity but doesn't converge AT infinity because at infinity you don't know if it's at +1 or -1, but it's not infinity itself.

Answer 15

I meant when n goes to infinity it oscilllates between -1and 1 but never converges to a limit

Answer 16

Yeah.

Answer 17

Cool Thanks :)