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Whats the order of starting.

Author : thoughtIlied

Subreddit : learnmath

Points : 2

Published : 7 hours ago

Comments : 0

Why when we want to get a linear approximation, we use x0=0 in a taylor polynomial?

What is the best way to learn math 112?

Author : ThEKiDBacK

Subreddit : learnmath

Points : 0

Published : 6 hours ago

Comments : 0

Area under x^x (calculus)

372-187 Using Regrouping

Author : BlueHorizon87

Subreddit : learnmath

Points : 1

Published : 3 hours ago

Comments : 2

Number of elements in shifted subsets

How to create a mathematical equation when there are more than two unknowns?

Author : Sibilisado

Subreddit : learnmath

Points : 2

Published : 4 hours ago

Comments : 2

Learning cryptography

Apparently A*sin(x) + B*cos(x) = C*sin(x+v). But why?

Author : reallyserious

Subreddit : learnmath

Points : 2

Published : 4 hours ago

Comments : 7

Natural Logarithms Help!

Author : falcaraz99

Subreddit : learnmath

Points : 1

Published : 4 hours ago

Comments : 5

For the love of god, can someone help me factor this trimonial?

Author : strangeburd

Subreddit : learnmath

Points : 1

Published : 4 hours ago

Comments : 8

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.