alex2100x
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I'm tackling some problems that give a specific series and then asks you to show whether it is absolutely convergent, conditionally convergent or divergent. I'm having a little trouble with doing this as I understand the theories of conditionally and absolutely convergence and the differences between them but I don't really see how they relate to the convergence tests I was taught before there was even mention of absolute and conditional convergence.

E.g.: Q1)

Is this series absolutely convergent, conditionally convergent or divergent:

\displaystyle \sum_{n=1}^{\infty}\frac{n^3+5n^  2+4}{2n^3\sqrt n+7n^2-1}

So what I did was to have a chain of elementary inequalities that showed here that my a_n (the quotient in the series) was always less than \displaystyle \frac{1}{9n^4} and then I said that since 0<a_n<=b_n and b_n converges (of the form 1/n^p with p>1) we have that a_n converges by the comparison test.

My problem is understanding what I have shown? The series converges according to my proof but that doesn't necessarily mean that it converges absolutely so would I be better of taking the absolute value of a_n before running the comparison test as then if it was shown to be absolutely convergent it would convergent in the normal sense.

I guess I don't quite understand what type of convergence the tests (ratio, root, comparison....etc) I've learnt actually prove?????

If anyone could clear this up for me and give me advice on how to tackle similar problems involving the ratio or root etc test instead of the comparison so I can have a better idea on how to tackle these I would be grateful.

Thanks!!!!!!!!
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Smaug123
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(Original post by alex2100x)
I'm tackling some problems that give a specific series and then asks you to show whether it is absolutely convergent, conditionally convergent or divergent. I'm having a little trouble with doing this as I understand the theories of conditionally and absolutely convergence and the differences between them but I don't really see how they relate to the convergence tests I was taught before there was even mention of absolute and conditional convergence.

E.g.: Q1)

Is this series absolutely convergent, conditionally convergent or divergent:

\displaystyle \sum_{n=1}^{\infty}\frac{n^3+5n^  2+4}{2n^3\sqrt n+7n^2-1}

So what I did was to have a chain of elementary inequalities that showed here that my a_n (the quotient in the series) was always less than \displaystyle \frac{1}{9n^4} and then I said that since 0<a_n<=b_n and b_n converges (of the form 1/n^p with p>1) we have that a_n converges by the comparison test.

My problem is understanding what I have shown? The series converges according to my proof but that doesn't necessarily mean that it converges absolutely so would I be better of taking the absolute value of a_n before running the comparison test as then if it was shown to be absolutely convergent it would convergent in the normal sense.

I guess I don't quite understand what type of convergence the tests (ratio, root, comparison....etc) I've learnt actually prove?????

If anyone could clear this up for me and give me advice on how to tackle similar problems involving the ratio or root etc test instead of the comparison so I can have a better idea on how to tackle these I would be grateful.

Thanks!!!!!!!!
You've shown absolute convergence. Indeed, if a sequence converges conditionally and is always positive, then it must converge absolutely. The comparison test as you stated it works on positive sequences only, so it shows absolute convergence.

The ratio test shows absolute convergence, because geometric sequences converge absolutely and the ratio test is fundamentally comparison with an absolutely-convergent sequence. (IIRC.)
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Ktulu666
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What the simple comparison test shows is absolute convergence, because you need the a_n to be >0 in the proof of the test (i.e: you sandwich them between 0 and b_n). If you're trying to work out what type of convergence the tests show, look at what conditions you need. Do you need a_n > 0? If so, you will be proving absolute convergence, either from the fact that a_n > 0 => |a_n| = a_n and hence sum(a_n) cvgt. => sum(|a_n|) cvgt. or by taking moduli to begin with. That's my limited understanding of the topic, would appreciate if someone could clarify what I've said if its unclear.
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james22
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I think you made a mistake in your calculations, a_n is not less than 1/(9n^4).
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alex2100x
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(Original post by james22)
I think you made a mistake in your calculations, a_n is not less than 1/(9n^4).
Thanks I just noticed a disgusting mistake from me.

Is this better?

\displaystyle a_n:=\frac{n^3+5n^2+4}{2n^3\sqrt n+7n^2-1} \geq \frac{n^3}{2n^4+7n^4}=\frac{1}{9  n}:=b_n which diverges since the infinite sum 1/n diverges.

Since:

\displaystyle 0&lt;b_n&lt;a_n~and~b_n~is~unbounded~t  hen~a_n~is~unbounded \rightarrow divergent.
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james22
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(Original post by alex2100x)
Thanks I just noticed a disgusting mistake from me.

Is this better?

\displaystyle a_n:=\frac{n^3+5n^2+4}{2n^3\sqrt n+7n^2-1} \geq \frac{n^3}{2n^4+7n^4}=\frac{1}{9  n}:=b_n which diverges since the infinite sum 1/n diverges.

Since:

\displaystyle 0&lt;b_n&lt;a_n~and~b_n~is~unbounded~t  hen~a_n~is~unbounded \rightarrow divergent.
Yes
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