Question

Is this series correct? (last one on the page)
File attached.

Answer 1

Is it \(\sum_{n=1}^{\infty} \frac{n^2+1}{3n^2-n+2}?\)

Answer 2

If so, then it diverges by the divergence test, since its limit is different than zero.

Answer 3

Yeah, that's the equation, and Divergence is TFD right?

Answer 4

What's TFD?

Answer 5

Test For Divergence? I think?

Answer 6

last term test; if not 0 diverges

Answer 7

I think so! :D

Answer 8

Thats what I get for taking an online class...

Answer 9

The test states that every convergent series \(an\) has a limit zero as n goes to infinity.

Answer 10

it states no such thing ...

it states if it doesNOT go to zero, it diverges.

if it goes to zero, another test has to be done

Answer 11

That's exactly what I said @amistre.

Answer 12

if it converges, it will have to be zero yes lol

Answer 13

its your british accent; hard to read

Answer 14

LOL!

Answer 15

Wait, what if the problem is \[n^{2}+1{/}3n^3+2\]

Answer 16

then the test for divergence fails and we have to do another test

Answer 17

This is without any doubt diverges. I don't need to study math to tell :P

Answer 18

divergent*

Answer 19

And what psychic powers did you use to get that?

Answer 20

Haha. No powers, just the theorem I just mentioned above. Can you tell what's the limit of the expression you gave above as \(n\to \infty\)?

Answer 21

Can that be done just by inspection of the limit?

Answer 22

i think he meant:\[\frac{n^{2}+1}{3n^3+2}\]

Answer 23

yes, how did you get it to do that?

Answer 24

\frac{top}{bottom}

Answer 25

ahh, thanks

Answer 26

I'm studying here with a friend, and he seems to think its done by using LCT not TFD

Answer 27

with bn = 1/n

Answer 28

tfd comes up zero, so it fails that one

Answer 29

youd have to pick another test to determine c or d

Answer 30

compare it to 1/n seems most likely

Answer 31

can someone help me with my problem?

Answer 32

Oh I see! Well, it's still divergent. As for the limit for rational functions f(n) as n goes to infinity, you have three cases:
1) The degree of the top is greater than the bottom, the limit is \(\infty\).
2) The degree of the top is less than the bottom, the limit is 0.
3) The degree of the top of the bottom and the the top are the same, the limit is the coefficient of the term of the higher degree on the top over the coefficient of the term with the highest degree in the bottom.

Answer 33

its not a good candidate for the integral; unless you can get the wolf to do it :)

Answer 34

So \(\frac{n^2+1}{3n^2+2} \to \frac{1}{3} \text{ as } n \to \infty.\)

Answer 35

right, but the next one has a degree 3 on the underside

Answer 36

Haha! You've already got a next one?!

Answer 37

I don't understand why you would need to use the comparison test.

Answer 38

because you can't tell by inspection of the limit as far as I know

Answer 39

\[\frac{n^{2}+1}{3n^3-n+2}\gt\frac{1}{n}\]

\[n^{3}+n \gt 3n^3-n+2\]

\[0 \gt 2n^3-2n+2\]

doesnt look to be greater than 1/n

Answer 40

Mk, so that means it diverges, correct?

Answer 41

i/n diverges, and anything greater than 1/n would diverge; but this is less than it so we have to do another test if we can

Answer 42

ratio and roots are inconclusive ...

Answer 43

Mk, that works

Answer 44

what is "Mk" ?

Answer 45

Translating my version of "ok" in real life into text. Phonetically its more or less mmkay

Answer 46

oooh, got it :)

Answer 47

:P

Answer 48

And do either of you two do Diff Equations?

Answer 49

i dabble, nothing fancy tho

Answer 50

If I past this Calc finals in 3 hours, I'll be taking Diffy next semester