Question

can the squeeze theorem be used here ?

Answer 1

@zepdrix

Answer 2

i was thinking that cos(3n) would be in between -1 < cos 3n < 1

Answer 3

Ugh I remember this one from a few days ago...
I still can't seem to figure it out.

Answer 4

oh right haha.. forgot to ask this to my teacher damn it ..

Answer 5

\[\large\rm \frac{-1}{1+(1.2)^n}\le \frac{\cos(3n)}{1+(1.2)^n}\le\frac{1}{1+(1.2)^n}\]
Limit on each side,\[\large\rm 0\le \frac{\cos(3n)}{1+(1.2)^n}\le0\]Ya doing that you can show that the `sequence` converges to 0.
I'm not sure that helps us with our series though.. hmm

Answer 6

hmm true

Answer 7

Wolfram is saying to use comparison test,
hmm thinking.

Answer 8

see page 3, problem 9

http://math.stanford.edu/~yjt/Math42A/class12_02112010ans.pdf

which is basically saying the same thing as what @zepdrix is saying

Answer 9

I thought you could only use comparison test on series that has all non-negative terms. How are you allowed to do that? Hmm weird :\

Answer 10

Oh oh,
"A series \(\large\rm \sum a_k\) is said to converge absolutely if the series of absolutes \(\large\rm \sum |a_k|\) converges."

Answer 11

So I guess if we can show that \(\large\rm \sum \left|\frac{cos(3n)}{1+(1.2)^n}\right|\) converges,
then we can show that our series converges, and not just conditionally.

Answer 12

\[\large\rm \left|\frac{\cos(3n)}{1+(1.2)^n}\right|\le \left|\frac{1}{1+(1.2)^n}\right|\]Cosine always gives us something between 0 and 1 in the numerator,
so this relationship holds true.
But we'd like to compare our terms directly to a P-series, ya?\[\large\rm \left|\frac{\cos(3n)}{1+(1.2)^n}\right|\le \left|\frac{1}{(1.2)^n}\right|\]
We know that \(\large\rm \sum\left|\frac{1}{(1.2)^n}\right|\) converges by P-series Test, ya?

Answer 13

I guess this one is a lot like all the comparisons we did,
just with a little twist.

Answer 14

What do you think lady? Confusing? :U

Answer 15

hmm so then we ignore the cos 3n? hmm

Answer 16

The cosine is kinda boring,
it's always giving us something between -1 and 1.
it's very predictable.

So we would like to make some type of comparison.
But our comparison test doesn't allow us to use a series which has negative terms.

Answer 17

So we're making use of this theorem:
"If \(\large\rm \sum|a_k|\) converges, then so does \(\large\rm \sum a_k\)."

My book calls it Theorem 10.6.4,
it doesn't have a nice title :\
your book is probably different.

Answer 18

So instead of looking at this series \(\large\rm \sum\frac{cos(3n)}{1+(1.2)^n}\),

we're looking at this series, \(\large\rm \sum\left|\frac{cos(3n)}{1+(1.2)^n}\right|\).

Since this new theorem has only positive terms,
yay! we can apply our comparison test.

Answer 19

Ok sorry I sort of said a lot without answering your question I think...
\(\large\rm 0\le |cos(3n)|\le1\)
The numerator is stuck between 0 and 1,
it makes it easy to ignore the numerator when apply a comparison, yes.

Answer 20

Since this new series* has only positive terms,

blah typo

Answer 21

Do you have that theorem in your book or cheat sheet anywhere?
The one about absolute convergence.

It's pretty handy!

Answer 22

oh i see now it makes sense . yeh i have it in my book but wording is complex :o lol

Answer 23

okay im struggling with this one too one sec.. almost done

Answer 24

Ohh sorry I forgot I had OpenStudy open >.<

Answer 25

Taylor Series? Mmmm I forgot how to do these,
One sec :d

Answer 26

Everything looks good besides this.
You forgot to plug in your n's here.