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[College] Compounded monthly problem. How do you do it?

Author : BlindDesire

Subreddit : learnmath

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Published : 7 hours ago

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[2nd year UG] Stuck on a Probability question

Author : ieuan_james

Subreddit : learnmath

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Published : 7 hours ago

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Can I get some help on this question?

Author : westlyroots

Subreddit : learnmath

Points : 1

Published : 6 hours ago

Comments : 2

Pacing

Author : matthias_j_doberman

Subreddit : learnmath

Points : 1

Published : 6 hours ago

Comments : 1

Prove with a contraposition

[Undergraduate Linear/Geometric Algebra] Avoiding overuse of coordinates?

Author : sleepingsquirrel

Subreddit : learnmath

Points : 1

Published : 6 hours ago

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Author : HopelesslyFamous

Subreddit : learnmath

Points : 1

Published : 6 hours ago

Comments : 2

Suggestions on how to teach someone 2-Step equations?

Author : Rebel_Scum83

Subreddit : learnmath

Points : 2

Published : 5 hours ago

Comments : 2

Double Integral Help

Author : SpiritGuide-

Subreddit : learnmath

Points : 1

Published : 5 hours ago

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Author : Mathywathy

Subreddit : learnmath

Points : 1

Published : 5 hours ago

Comments : 1

If the series n/n+1 diverges, multiplying it by -1 and 1 alternatingly won’t make it converge. Is there a theorem that says that?

Definitely not, because the claim is false. For example, the series 1/n diverges, but the series (-1)^{n}/n converges.

You can use any test you want on an alternating series, as long as it meets the hypotheses of the test. You can use ratio test, root test, nth term test, comparison test, whatever. The purpose of AST is that, in some cases (like (-1)^{n}/n), all of those tests are inconclusive, so we need something else.

Can use the integral test?

What does the integral test say?

It says that if the function equivalent of a(n) is positive, continuous, and decreasing for all x>0 and if the integral of f(x) with bounds 0 and infinity is a finite value, then the series converges, otherwise it diverges. I should have remembered the POSITIVE part.

Well, in this case, you'd quickly realize it doesn't work when you attempt to integrate (-1)^{x}.

Neither does the comparison test, right?

It's usable, but inconclusive for many alternating series.

It’s usable, as in I can compare the alternating series with another known ALTERNATING series? I can’t use it to compare alternating series with non alternating series?

As in you can use the test verbatim: if |An|<=Bn, and Bn converges as a series, then An converges absolutely. The absolute value signs are important. For example, you could compare (-1)^{n}/n^{2} to 1/n^{2}.

But if you try to do this with a conditionally convergent series, you won't be able to get anything out of it: for example, you can compare (-1)^{n}/n against 1/n, but 1/n is a divergent series. The comparison test can only tell you whether a series converges absolutely, not whether it converges conditionally.

If you mean comparing one alternating series to another alternating series, that does not work. For example, you might hope that if |An|<=|Bn|, where both are alternating, and if Bn is a convergent series, so is An. But this is false: for example, take Bn=(-1)^{n}/n and take An to be 1/n for n even, -1/n^{2} for n odd. Then both are alternating, and |An|<=|Bn|, but An diverges and Bn converges.

## arthur990807

Consider that the terms in your series don't approach 0.

## hadilbader812

The nth term test? But the alternating version of the series could approach zero.

## arthur990807

If a sequence {a_n} does not approach zero, then neither does (-1)

^{n}a_n.## hadilbader812

So the nth term can be used to prove that a series diverges?

## arthur990807

The n'th term test can only ever say your series diverges or be inconclusive.

Edit: massive derp.

## skullturf

No, the nth term test can say that a series diverges (if the terms don't approach 0) but the nth term test can never guarantee that a series converges.

## arthur990807

Bah. Typo. I meant "diverge", not "converge". Thanks for catching that.