Question

Series convergence or diveregence

Answer 1

Both or just one? What's your best guess?

Answer 2

both. the problem is for qn 8, i havent dealt with trig yet so im not sure, but we suppose to do alternating test for tht right? for qn 9 im not sure how to approach tht

Answer 3

The first one should be simply an application of the divergence test. Take the limit as it approaches infinity. Which is also considered part of the alternating series test.

What does the divergence test say about number 8?

Answer 4

okay, since lim n->infinity cos pi/n = 0, no conclusion can be made?

Answer 5

No, cos(0)=1.

Answer 6

haha ya i just figured out gosh, so what shd i do with (-1)^n ?

Answer 7

Doesn't matter, -1 and +1 both make the divergence test pass, so it doesn't matter to us! =P Divergence!

Answer 8

i see, thank you thats clear indeed! :D what about qn 9?

Answer 9

Well what can you figure out from trying different p values? There are a few sort of observations of special cases that will obviously make it fail.

Answer 10

uhm, shd i apply divergence test also?

Answer 11

There are really just 3 things to show for the alt. series test.

Does the non alternating term approach 0 as n approaches infinity?

Is it larger than zero for all natural numbers you can pick for n? (infinity is not a number, so don't confuse this with the limit requirement above.)

Is the previous non alternating term greater than the next one?

So from here, what p's can you pick that will make this be violated?

Answer 12

ohh.. i thought p would be the natural value and n would be infinity.? is this a special case?

Answer 13

What do you mean? Explain more.

Answer 14

no coz u said, we sub natural numbers to n but isnt the limit n-->infinity?

Answer 15

I don't understand what you mean at all right now, sorry.

Answer 16

T^T okay.. i was just talking abt your point 2 in your previous reply. "is it larger thn 0 for all natual numbers..."

Answer 17

okay i have a guess, if p <1, series diverge am i correct?

Answer 18

Well by that I mean if you pick any natural number as "n" for:

\[\frac{ 1 }{ n+p }\]

will it be greater than 0? Yes.

But the limit as n approaches infinity is 0.

I think your guess is wrong, why did you choose it though?

Answer 19

A special case I was considering was what if you pick p to be any negative integer, what will happen?

Answer 20

I might be sort of skipping a step, what if p = -3.5? Then for a few FINITE steps the value of 1/(n+p) will be less than 1. But we can easily take them out and not consider them as part of the infinite sum, in which case we are fine. As long as we don't pick p to be an infinite number (which I've never heard of lol) we should be fine since we can always get rid of the finite number of negative terms and add it to our sum.

Answer 21

okay, if i choose the p values that would converge then, it will be, 0, <1, or >1, as long as its not inifinity?

Answer 22

to be honest i am so confused haha

Answer 23

i think its because the are 2 variables which is n and p.

Answer 24

haha it's ok. So why are you excluding 1? I'm trying to understand why you think 1 won't work here.

Answer 25

oh it shd be greater or equal to. so that means all real numbers except infinity?

Answer 26

Well infinity isn't a number, so you don't say you exclude infinity... That's sort of a nonsense statement.

Here, what if you pick p=-2. Then what happens?

Answer 27

Start writing out the sum of the first 4 terms of so and you'll see something.

Answer 28

nothing i guess? bcz when u sub in n-->infinity its still infinity right?

Answer 29

i mean the resulting answer is still 0.

Answer 30

Yes, but you're taking the LIMIT as you APPROACH infinity. Infinity is not a number. So you never "plug in" infinity. Infinity is just a shorthand for "increases without bound" and is not a number at all.

Answer 31

okay, so what shd i plug in for n then? if n is positive and p is negative, then the resulting is still positive (diverge) except when the resulting is 0. if n is negative, p is negative the resulting is negative which will mean the series will diverge?

Answer 32

No. Maybe just forget everything and start fresh?

Take 1/(n+p) and take the limit as n approaches infinity. You get 0. So we didn't fail the divergence test, which is a good sign for all p. So this doesn't give us any information on what p can or can't be. As far as we know it can be any real number right?

So let's continue what causes something to be divergent? If we have any term that is infinite. When might we have an infinite term? When we divide by 0, right? So when do we divide by zero? When n+p=0 yeah? Since n can be any natural number that means if we pick p to be any negative integer then eventually we will divide by zero when we get to that term!

Since that can't happen, we know that the numbers p can't be are just negative integers. All the other numbers are fine since we found out it passes the alt. series test otherwise!

Answer 33

okay that make sense, so all we need to do is to find p when the denominator is equal to 0.

Answer 34

Yeah basically, as long as it passes the alternating series test, you're good.

Answer 35

wait so which values exactly? since n can change so does p, im vague at what exact values they could be., can you give an example of a term?

Answer 36

or shd i just say as long as n+p =0 then that would be the anwer?

Answer 37

So let's look at p=-2.\[\sum_{n=1}^{\infty}\frac{ (-1)^n }{ n-2 }=\frac{ -1 }{ 1-2 }+\frac{ 1 }{ 2-2 }+\frac{ -1 }{ 3-2 }+...\]

What's wrong here?

Answer 38

the 2nd term is infinity. oh so u mean, when n=p?

Answer 39

p=-n?

Answer 40

Almost, but remember this comes from STOPPING when

n+p=0

So subtracting p from both sides you get

n=-p

Since n is eventually going to be all the natural numbers between 0 and infinity that means that p must not ever be any negative integer or else you'll get a crap term like this.

Answer 41

Ahh ok, you corrected yourself, but yeah it should make sense now.

Answer 42

i see i see, so the final answer would be, all real numbers except when p=-n ?

Answer 43

yep its all clear to me now thank you very much! :D