Calculus question

IIRC these tests can only be applied to series with positive terms.

Dammit.

thepedestrian (200)

I'm not sure if you found the answer but here is the explanation and apologies if this is vague. Just make another post. :)

I think your teacher is maybe trying to trick you but there are always absolute values around the ratio and root tests; and with these tests, the limit will always converge absolutely. You can probably find a derivation for these formulas online and these absolute values become essential when working with alternating series or finding a series for a function using Taylor series.

In both cases:
if limit < 1 -> Converges absolutely. There is no such thing as conditional convergene with Ratio and Root tests.
if limit > 1 -> Diverges.
if limit = 0 -> Test is inconclusive and you have to try a different test. A lot of times, you can use a comparison or limit test.

I'm not sure what you are asking in the latter question. Are you referring to a telescoping series?

I'm not sure what you are asking in the latter question.

Ignore it. I hadn't fully grasped the concepts of absolute and conditional convergence.

these absolute values become essential when working with alternating series

So then the tests can be used on alternating series?

Well, it doesn't matter, now. I've already taken the exam and I think I did pretty well. 3/4. Practically no one solved the series problem. If anyone would like to take a stab at it for fun, find p>0 such that ∑ n^5*(sqrt(n^p+7)-sqrt(n^p+3)) converges.

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ne555 (10692)

For alternating series the test is easier
a_n ---> 0
a_{n+1} <= a_n \forall n>N

However in your example a_n seems to diverge...

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For alternating series the test is easier
a_n ---> 0

What's the difference between this and the Leibniz test, other than it's not checking some propositions?

ne555 wrote:
However in your example a_n seems to diverge...

n^5 + ((n^p + 7)^.5 - (n^p + 3)^.5)
= n^5 * n^(p/2) * [sqrt(1 + 7/n^p) - sqrt(1 + 3/n^p)]

The thing in the brackets goes to zero, so it might not diverge. Also helios' series wasn't alternating. I tried and failed to use the ratio test - I couldn't even find the limit of the sequence.

@helios, ne555

ne555 wrote:
a_n ---> 0 a_{n+1} <= a_n \forall n>N

I think helios is correct. You don't need the second condition. Intuitively one would expect the first one to imply the second: the sequence surely can't converge if it doesn't reach a point after which its terms get consecutively smaller. I may be wrong though - it's years since I did this stuff.
Looks like I was

EDIT: If you look at the proof on this Wikipedia Page:
http://en.wikipedia.org/wiki/Alternating_series_test
I can't see that the monotonicity is used in the proof of convergence, just the proof of the condition of the error. Have a look and see if you agree with me.

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helios wrote:
So then the tests can be used on alternating series?

Yes, but only to test for absolute convergence.
And, of course, you'll have to use abs(An+1/An).

Does this -> ∑ n^5*(sqrt(n^p+7)-sqrt(n^p+3)) converge for p = 2011?

EDIT: Yes, it does \(^o^)/

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Ok, here's the proof.

∑ n⁵ * [ sqrt(n^p + 7) - sqrt(n^p + 3) ] =

∑ (n⁵ / n⁷) * n⁷ * [ sqrt(n^p + 7) - sqrt(n^p + 3) ] =

∑ (1 / n²) * n^{7 + p/2} * [ sqrt(1 + 7/n^p) - sqrt(1 + 3/n^p) ] =

∑ (1 / n²) * A / B, where

A = sqrt(1 + 7/n^p) - sqrt(1 + 3/n^p) and

B = n^{-7 - p/2}

Now, if I pick a p such that p > 7 + p/2 => ... => p > 14, lim (n -> oo) of (A / B) is < oo.

This means that if p >14, there is a κ such that the terms of my series
become smaller than the respective terms of ∑ κ / n² after a certain point.

Since that last one converges, my series converges too.

Yes, but only to test for absolute convergence.

Uh-huh... Then there is possible conditional convergence at the values of x where the test returns 1, and at other values there's either absolute convergence or divergence, correct?

I can't see that the monotonicity is used in the proof of convergence, just the proof of the condition of the error. Have a look and see if you agree with me.

It's used in the assumption that a₁ + (-a₂+a₃) + (-a₄+a₅) + ... is smaller than a₁.

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Oh yeah - I knew I'd read it too fast ^^

EDIT: I'm fairly rusty on all this stuff. The problem is that (at my uni and for my course at least), they make you do all this rigorous stuff and get you in the right mindset and then they gradually remove it again and you start being sloppy just as you were at A level (e.g. differentiating vectors with the gradient operator as if they were normal scalar derivatives and just hoping for the best :P).

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Engineering? I'm asking because treating mathematics as a tool rather than a logical system seems to fit engineers.

It was quantum mechanics actually ^^ I never really got past the mental barrier with that course though.

But yeah it does sound like an engineer thing - I guess our lecturers indoctrinated us in the same way. Are you on a pure maths course, then?

By the way, a little anecdote from my awesome analysis teacher:

There's an engineer, a physicist and a mathematician who are asked to prove that all odd numbers are prime.
The mathematician says "Three's prime, five's prime, seven's prime. I'm sure I can prove the rest by induction".
The physicist says "Three's prime, five's prime, seven's prime, eleven's prime. I'm sure I can put the rest down to experimental error"
The engineer says "Three's prime, five's prime, seven's prime, nine's prime, eleven's prime."

The moral of this story is that mathematicians are lazy, physicists believe what they want to believe and engineers are careless. XD

Oh, and there was the one about McTaggot being the pope... Do you have any 'fun' lecturers?

helios wrote:
Then there is possible conditional convergence at the values of x where the test returns 1, and at other values there's either absolute convergence or divergence, correct?

Hmmm... I don't seem to be able to find an example where the ratio
test returns > 1 and the series converges conditionally, so, I'll say yes.

Hmmm... I don't seem to be able to find an example where the ratio
test returns > 1 and the series converges conditionally, so, I'll say yes.

Great.

By the way, a little anecdote from my awesome analysis teacher:

There's an engineer, a physicist and a mathematician who are asked to prove that all odd numbers are prime.
The mathematician says "Three's prime, five's prime, seven's prime. I'm sure I can prove the rest by induction".
The physicist says "Three's prime, five's prime, seven's prime, eleven's prime. I'm sure I can put the rest down to experimental error"
The engineer says "Three's prime, five's prime, seven's prime, nine's prime, eleven's prime."

The moral of this story is that mathematicians are lazy, physicists believe what they want to believe and engineers are careless. XD

Nice.

Are you on a pure maths course, then?

I'm in computer science, but I'm also interested in mathematics in general.

Do you have any 'fun' lecturers?

Not really. At least, not yet. I do have a funny anecdote, though. One of the calculus professors this semester was teaching mathematics for medicine (read: watered down calculus) the previous one. See if you can guess why some of the students there thought that this was the natural logarithm's graph:
f(x)=(x>0)?log(x):e
Hint: it has something to do with iteratively evaluating the function.

ne555 (10692)

@m4ster r0shi: great, p>10 -> converge

I like this:
An astronomer, a physicist and a mathematician are on a train in Scotland. The astronomer looks out of the window, sees a black sheep standing in a field, and remarks, "How odd. Scottish sheep are black." "No, no, no!" says the physicist. "Only some Scottish sheep are black." The mathematician rolls his eyes at his companions' muddled thinking and says, "In Scotland, there is at least one sheep, at least one side of which appears to be black from here."

@ne555 Haha. Looks like all our lecturers do tell the same stories. As they go, that one's pretty good.

@helios Sorry, can't see it. Logarithms are undefined for x <= 0 so f(x) = (x>0)?log(x):e surely becomes f(x) = log(x) as far as logarithms are concerned, in which case I can't see how it could be mistaken for ln(x). You'll have to enlighten me ;)