Question

i need help with squeeze theorem how do i use squeeze theorem here . find whether the sequence converges or diverges {n^2cos(n)/1+n^2}

Answer 1

\[_{an}=\frac{ n^2\cos n }{1+n^2 }\]
find is the sequence diverges or converges using squeeze theorem

Answer 2

Our sequence is (I think?)$$a_n=\frac{n^2\cos n}{1+n^2}=\frac{n^2}{1+n^2}\cos n=\left(1-\frac{1}{n^2+1}\right)\cos n$$Now, the trick is to remember that \(-1\le\cos n\le1\), which means our sequence can easily be found to be bounded by multiplying by \(1-\frac1{1+n^2}\) to yield$$-\left(1-\frac1{1+n^2}\right)\le\left(1-\frac1{1+n^2}\right)\cos n\le\left(1-\frac1{1+n^2}\right)$$ and we can clearly see both sequences converge -- just not to the same value.$$-1\le\left(1-\frac1{n^2+1}\right)\cos n\le 1$$.

Answer 3

You should observe oscillatory behavior in the limit.

Answer 4

okay thanx a lot let me check if i will understand what you did when i write it down

Answer 5

Take a look at WolframAlpha's plot:
http://www.wolframalpha.com/input/?i=n%5E2%2F%281%2Bn%5E2%29+cos+n+for+n%3D-150+to+150

Answer 6

okay let me check it

Answer 7

why \[1-\frac{ 1 }{ 1+n^2 }\]
are we subtracting or multiplying both sides by \[1-\frac{ 1 }{ 1+n^2 }\]

Answer 8

i think seeing the graph i get it

Answer 9

@McCain $$\frac{n^2}{1+n^2}=\frac{n^2+1-1}{n^2+1}=\frac{n^2+1}{n^2+1}-\frac1{n^2+1}=1-\frac1{n^2+1}$$
and we're multiplying because \(\cos n\) alone is not our sequence, but \(\left(1-1\frac1{n^2+1}\right)\cos n\) is.

Answer 10

@McCain think of it like this. \(\frac34=1-\frac14, \frac89=1-\frac19, \text{etc}\dots\)

Answer 11

thanks i think i need more time working on squeeze theorem . so because or sequence is alternating an not approaching a single value should we conclude that it diverges?

Answer 12

@McCain indeed, since the sequence will continue to vary between \(-1\) and \(1\), never approaching a single value. You can think of it in terms of *inferior* and *superior* limits:$$\limsup_{n\to\infty}\left(\frac{n^2\cos n}{1+n^2}\right)=1\\\liminf_{n\to\infty}\left(\frac{n^2\cos n}{1+n^2}\right)=-1$$