Question

help please

Answer 1

11.

Answer 2

Hmm, can you say anything about the `sequence` ?
What's going on with each individual term?
Limit?

Answer 3

hmm looks like the limit.. is this where we use different tests :O

Answer 4

No, this is before any fancy testing.
What is the limit of the `sequence` ?

Let's start by making sure it's zero,
otherwise this thing has no chance of converging.

Answer 5

multiplying by the greatest number?

Answer 6

I was trying to explain this to Sid earlier >.< Grr

The sequence is the list of terms,
\(\large\rm a_1, a_2,...,a_n\)

The series is the sum of those terms,
\(\large\rm a_1+a_2+...+a_n\)

Answer 7

Blah ignore that,
let's not go into more detail than we need.

Answer 8

\[\large\rm \lim_{n\to\infty}\frac{n}{n^3+1}\]Sequence converges yes?
That doesn't help us much, just something to keep in mind.

Answer 9

its convergent since you multiplied by 1/n^3 right ?

Answer 10

Sure, ya.
That's only the sequence though, not the series, so it doesn't tell us much.

Answer 11

So yes, right to one of our fancy tests.
I'm thinking in my head... Comparison Test.

This is my thought process...

Hmm n/(n^3+1) is very similar to n/n^3 right?

Answer 12

And n/n^3 can be rewritten as 1/n^2.
Hmm interesting.

Does that help us at all?
Anything clicking in your brain bucket? :)

Answer 13

hmm i would say by the p series test n comparison test

Answer 14

Good, yes, p-series tells you that \(\large\rm \sum\limits_{n=0}^{\infty} \frac{1}{n^2}\) converges.

Which means that \(\large\rm \sum\limits_{n=0}^{\infty} \frac{n}{n^3}\) converges.

Answer 15

question whats the difference between direct comparison test and limit comparison test ? i get them confuse when this stuff happens

Answer 16

Err I guess my index should start at 1, not 0.

Answer 17

Hmm, I'm not sure,
I'll have to look that up to refresh my memory :p

Answer 18

lool i have a formula sheet for me one sec

Answer 19

they look somewhat similar o.o

Answer 20

Ya they are pretty similar.

Answer 21

yeah D:

Answer 22

Direct Comparison doesn't use limits.
I can't think of a situation where we would use Limit Comparison,
I'd have to think about it some more.

Answer 23

Anyway, I picked n/n^3 because it work out nicely with our Direction Comparison.

Answer 24

By our P-series we were able to determine that \(\large\rm \sum\limits_{n=0}^{\infty} \frac{n}{n^3}\) is convergent.

So what do we need for this series \(\large\rm \sum\limits_{n=0}^{\infty} \frac{n}{n^3+1}\) to be convergent?
We need it to be "smaller" than a convergent series, right?
Does that make sense?

If the first series I listed is "bigger" than our series,
and it converges down,
then certainly our series will also converge because it's "smaller".

Answer 25

yeah

Answer 26

Crap I left the index wrong again :P

Answer 27

So this is ... maybe how you would want your answer to look:

\(\large\rm \frac{n}{n^3+1}<\frac{n}{n^3}\) for all n>0.

By P-Series Test \(\large\rm \sum\limits_{n=1}^{\infty} \frac{n}{n^3}\) converges.

Therefore Direct Comparison Test shows us that \(\large\rm \sum\limits_{n=1}^{\infty} \frac{n}{n^3+1}\) also converges.

Answer 28

ohhh

Answer 29

Oh, just so it looks like it does on your sheet,
maybe I should have written the first line like this,\[\large\rm 0\le\frac{n}{n^3+1}\le\frac{n}{n^3}\]for all n>0.

Answer 30

So now it's in this nice form,\[\large\rm 0\le a_n\le b_n\]

Answer 31

Does it make a little bit of sense? :O

Or was that just a blonde "Ohhhh" like you cracked a nail or something?

Answer 32

no haha it makes sense now ;p

Answer 33

about number 13

Answer 34

was thinking of geometric series

Answer 35

Sec, I gotta look at your cheat sheet again :d

Answer 36

Do you get to use the cheat sheet on your next test? :)

Answer 37

noo lool she gave us a short cheat sheet that we can use but its not specific though :o

Answer 38

I was trying to come up with a comparison we could use,
but can't think of anything nice and simple.

I think Ratio Test will work nicely for this one.

Answer 39

hmm so we use the ratio test when we have exponents ?

Answer 40

You had first mentioned geometric.
Notice that geometric really only works when we have both top and bottom being exponentials, not a mix of exponential and polynomial.\[\large\rm \frac{4^n}{7^n}\quad=\quad \left(\frac47\right)^n\]That's an example of when geometric would be very useful.

Answer 41

oh yeah

Answer 42

Sorry, I'm not exactly sure when to use what...
I just ran through the numbers real quick and worked out.

It was trial and error.

I guess I picked it because it's the most simple test in my opinion.
So I wanted to see if it worked before moving on to something a little more complex.

Answer 43

Keep in mind that there is not ONE WAY to do each of these.
This problem we're working on right now probably has 3 or 4 different tests that will work. I just like how the ratio test works out so that's the one I went with.

Do you know how to set up the ratio test for this one?
Draw it lady :O

Answer 44

Imma go make some food while you do that,
brb in like 7 mins >.<

DO ITTTTTTTTTTT

Answer 45

yesh lool one sec

Answer 46

Ok one small mistake.

Answer 47

Err actually a pretty big mistake I guess.

Answer 48

Because it would give you the WRONG answer.