How could it be that Sum(n^2/(n^3+1)) from n=0 to infinity is divergent?

Question

anonymous · Answer

For clarity:

$$
\sum\limits_{n=0}^\infty \frac{n^2}{n^3 + 1}
$$

anonymous · Answer

Wait sorry, completely forgot. Of course that series shouldn't converge! But it was alternating:

$$
\sum\limits_{n=0}^\infty (-1)^n \frac{n^2}{n^3 + 1}
$$

amistre64 · Answer

the limit of terms goes to 0

is the sequence decreasing?

those are the only criteria i know of to determine convergence with the alternating series thrm

amistre64 · Answer

http://www.wolframalpha.com/input/?i=sum_%7Bn%3D0%7D%5E%7Binf%7D+%28-1%29%5En+n%5E2%2F%28n%5E3%2B1%29 wolf says it converges

amistre64 · Answer

lucy! you have some splainin' to do ....

anonymous · Answer

@amistre64 I remember what the choices were now:

A) Absolutely converges
B) Conditionally converges
C) Diverges
D) Not enough information
E) None of the Above

The accepted answer was B, but since this series is just an expression, I really can't see what could be conditional about it.

Alas, my dispute was denied, so ¯\_(ツ)_/¯

amistre64 · Answer

you misspelt additionally

amistre64 · Answer

so we know its convergent, the question is to its absoluteness or conditionment ... love making new words

anonymous · Answer

Heh, indeed I did. The only places I've seen conditional convergence are in functions. e.g.:

$$f(x) = \sum_{n=1}^{\infty}x^n$$, which of course will only converge if the condition $|x| < 1$ is met.

amistre64 · Answer

A series is called A.C. if sum|an| is convergent.

A series is called C.C. if sum|an| is divergent, but sum an is convergent.

so the question is:  how does your material define these terms?

amistre64 · Answer

you agreed that sum n^2/(n^3+1) diverges  so sum |an| is divergent

but we showed that sum an is convergent by series test.

so it is CC  right?

amistre64 · Answer

'no function variables' is not a condition of CC by the way

anonymous · Answer

@amistre64 Ah, yes, our text definied it differently. I remember realizing that last year. Our teacher taught us that ac/cc was converge with or without absolute value, but our text said that ac/cc is "converges everywhere"/"converges on an interval".

amistre64 · Answer

you would do well to post the definition your course provides in order for us to assess it better :)

amistre64 · Answer

the actual courses wording; not your .. understanding .. ot it to prevent bias.

anonymous · Answer

@amistre64 Oh I don't have the text with me anymore. I took BC calc last year, but this year I'm not in any calc class. I do however remember that our teacher told us to ignore its for that chapter. The definitions of AC/CC were along the lines of:

$$ f(x) = \sum_{k=k_0}^n a_n$$ will absolutely converge iff the sum is finite over $\Re$, and will conditionally converge iff the sum is finite over a set $A \subset \Re $.

I'll see if any classmates can send me a copy of the definition though.