Alternative examples, alternating series test

[Abscissas]

Hey guys, this one is just for funnsies. So when dealing with an alternating series test, 3 requirements must be met, :
Alternating
u(sub n) ≥ u(sub n+1) for all n ≥ N, for some integer N
u(sub n) → 0 as n → ∞.

So I have been coming up with examples where of these are true, and one isnt. A little bit further, I am not sure if this makes sense, but is possible to find an example for each where like, one diverges and one converges, or does that not make sense?

 

[micromass]

Come on, come up with some examples yourself first! It's not hard to find some counterexamples! Tell us what you've got already, and I'll help you fill in the parts you don't have.

(By the way, there are not always examples possible where one diverges and one converges. I can see one obvious condition where it wouldn't be the case).

 

[Abscissas]

Oh right! forgot about that, havnt slept in waaayy to long. Alright so i think the sum of ((-1)^2n)/n wouldn't alternate (rule 1) but seems to follow the rest, sum of cos(pi)/n I think would alternate and go to zero. I am thinking the one that your thinking of, which I've had a hunch on, is u(sub n) ≥ u(sub n+1) for all n ≥ N, for some integer N for diverging, but i can't really put it into words or make sense of it,its just a hunch , probably due to my lack of sleep

 

[micromass]

The one I'm thinking of is ##u_n \rightarrow 0##. If this is false, can the series ever converge?

 

[Abscissas]

Well, couldn't it converge to someone other than 0?

 

[Abscissas]

also, don't forget that u (Sub n) is in absolute values, sorry forgot to put that

 

[micromass]

Do you know the "divergence test"?

 

[Abscissas]

Right, if it doesn't go to zero then it diverges because of the infinite sums. forgot about that. Yeah so i guess for that one it is impossible

 

[micromass]

OK, so you have ##\sum 1/n## which does not alternate but does satisfy the other conditions. This series does not converge. Do you have one which does converge?

 

[Abscissas]

∑sin(n)/n I think would converge

 

[Abscissas]

never mind it alternates haha

 

[Abscissas]

∑Sin(npi)/n right? because it would keep going to 0s and not alternating

 

[micromass]

∑sin(n)/n I think would converge

Does it alternate? I'm not so sure? Alternating means it gets positive and negative at each turn. Does this do this?

 

[micromass]

∑Sin(npi)/n right? because it would keep going to 0s and not alternating

You could probably write this easier.

 

[Abscissas]

I just tried to plug it into my calculator and it denied me. My instincts tell me it doesnt. but I put down sin(3pi/4)= negative and sin(3pi/4)= still negative, and before pi its positive

 

[Abscissas]

∑cos/n?

 

[micromass]

Again, doesn't seem alternating to me...

 

[Abscissas]

Hang on, would ∑1/n work? because the absolute value of the next term has to be greater then the last right?

 

[micromass]

Here is one ##\sum_n 2^{-n}##.

 

[micromass]

Hang on, would ∑1/n work? because the absolute value of the next term has to be greater then the last right?

No, the sequence needs to be decreasing. Every next term needs to have smaller absolute value.

 

[Abscissas]

Ohhhh, ∑((-1)^2n)/n would work too right?

 

[Abscissas]

That makes a lot more sense, and destroys my hunch about that rule not working

 

[micromass]

Ohhhh, ∑((-1)^2n)/n would work too right?

OK, but that is just ##\sum 1/n##.

 

[Abscissas]

Ohh okay, i see what you did there now. I feel dumb :p. Well hey, thanks for helping, its a lot of fun , and I feel like I am getting a better understanding of this test now. Would you like to help me further?

 

[micromass]

Of course, what else would you like help on?

 

[Abscissas]

Im thinking a divergence example for the decreasing rule where both others are true, I am trying to come up with something, ill post it once i have something respectable

 

[micromass]

[tex]1 - (1 + 1) + 1/2 - (1/2 + 1/2) + 1/3 - (1/3 + 1/3) + ...[/tex]

 

[Abscissas]

Thats a really cool one, written it would be ∑n-(n+n) but wouldn't it violate the a (sub n)→ 0. I think it would go to negative infinity right? That was a cool one though and I think to fix that you could put it over n, but nvm then it gets smaller. Oh by the way, i totally couldn't come up with anything, I am assuming this was a clue to send me in the right direction, so I am going to play around with it a little

 

[micromass]

Thats a really cool one, written it would be ∑n-(n+n)

Write it a bit differently (and use ##1/n## instead of ##n##).

 

[Abscissas]

So when dealing with these, should I avoid simplifying? because i got ∑-1/n but before simplification i got ∑1/n-(1/n+1/n)

 

[micromass]

So when dealing with these, should I avoid simplifying? because i got ∑-1/n but before simplification i got ∑1/n-(1/n+1/n)

Sure, but write it even differently. Write it so that the odd term is ##1/n## and the even term is ##-2/n##.

 

[Abscissas]

OHHHHHHHHH I see what you did now ∑1/n-2/n which would diverge because to goes to 2 and your a genius haha thanks man

 

[Abscissas]

Idk why, but i was trying to make it converge

 

[Abscissas]

But wait, it would then violate the rule that it goes to 0, and i feel like I am wrong on this one, but its not alternating

 

[Abscissas]

Because each sum, is negative, or when we say its alternating do we just mean each term?