Question

Calc II question please...
Can someone please explain if the sequence cosn/n^4 converges or diverges? I know the formal definition of Squeeze but I came up with ---1 13 years ago

Answer 1

cosine is stuck between -1 and 1, so it converges for sure

Answer 2

\[a_n = \frac{ cosn }{ n^4 }\]
\[-1\le(cosn/n^4)\le1\]

Answer 3

but I have to show the squeeze theorem for the convergence but I am not getting a limit that =L

Answer 4

try
\[-\frac{1}{n^4}\leq \frac{\cos(n)}{n^4}\leq \frac{1}{n^4}\]

Answer 5

why is it over n^4

Answer 6

since the left and right both go to zero, so does the middle one

Answer 7

Are the left and right parts (x1/n^4)? Is that why

Answer 8

you wrote \(\frac{\cos(n)}{n^4}\) right?

Answer 9

yes

Answer 10

that is why i wrote \[-\frac{1}{n^4}\leq \frac{\cos(n)}{n^4}\leq \frac{1}{n^4}\]

Answer 11

\(-1\leq \cos(n)\leq 1\) divide both sides by \(n^4\)

Answer 12

why wouldn't I multiply each side, like to get rid of it in the middle

Answer 13

you are trying to show \(\frac{\cos(n)}{n^4}\) converges

Answer 14

the numerator is stuck between -1 and 1, so the whole thing is stuck between \(-\frac{1}{n^4}\) and \(\frac{1}{n^4}\)

Answer 15

yes, I am but I have to show what you did in terms of the definition of squeeze

Answer 16

it is the squeeze, it is squeezed between to things that go to zero

Answer 17

okay so can something converge to zero?

Answer 18

do sin and cos always converge to 0? I thought they are alternating sequences

Answer 19

\[\lim_{n\to \infty}\frac{-1}{n^4}=0\] and
\[\lim_{n\to \infty}\frac{1}{n^4}=0\] and
\[\frac{\cos(n)}{n^4}\] is squeezed between them

Answer 20

I see that now for sure.

Answer 21

it is true that \(\cos(n)\) has no limit as \(n\to \infty\) but you have
\[\frac{\cos(n)}{n^4}\]

Answer 22

so what would be convergence of \[\sum_{n=1}^{\infty} \cos n/n^4\]

Answer 23

again it converges

Answer 24

but wouldn't it be comparable to a p series with p=4>1

Answer 25

you can compare it to
\[\sum\frac{1}{n^4}\]which converges like mad

Answer 26

uh oh, p>1 converges?

Answer 27

in fact it is absolutely convergent

Answer 28

yes if \(p>1\) then \(\sum\frac{1}{n^p}\) converges

Answer 29

actually, I am getting this because I know abs convergence is true if absolute value of a_n converges, correct?

Answer 30

so if theres no negative one to cos(n), then it would have to converge....is that right?

Answer 31

yes

Answer 32

yayyyy!! thank you. you are awesome!!

Answer 33

although in this case i think you have to use the comparison test
ratio or root won't work

Answer 34

yw

Answer 35

in fact not only does it converge, but it is rather small i think
you can check here
http://www.wolframalpha.com/input/?i=sum+1+to+infinity+cos%28n%29%2Fn^4

Answer 36

in the case of cos n x 1/n^4, is a_n less than b_n for all n?

Answer 37

and how can you tell?

Answer 38

\[|\frac{\cos(n)}{n^4}|\leq \frac{1}{n^4}\]

Answer 39

gotta run, good luck

Answer 40

thanks tons of help