Question

IF series of a_n diverges, does the series of |a_n| also diverge?

Answer 1

I think yes because the series may still approach infinity or a nonzero number even when taking its absolute value.

Answer 2

Slight correction to your reasoning: if a series approaches a nonzero number, it can still be convergent. Take for instance a geometric series with \(|r|<1\). I think you meant sequence?

Anyway, are you supposed to write a proof, or just provide an example of a series \(\sum a_n\) and \(\sum |a_n|\) that diverge?

Answer 3

Also, if f(x) = 2x - x^2 + 1/3* x^3 - converges for all x, then f'''(0) is 2? I do not think we are able to find the derivative term by term so this would be false?

Answer 4

If it is true, provide an explanation... false then provide the example that disproves the statement

Answer 5

yes i slipped I meant sequence

Answer 6

Differentiation is a linear operation, and for that reason alone it's okay to differentiate term by term.

Answer 7

So is the statement above true?

Answer 8

HINT: \[a_n ~\le~|a_n|\]

Answer 9

By comparison test if we find that the series of an diverges then a n will diverge also! OHOHH

Answer 10

Gotcha :)

Answer 11

Also, if series of an and series of bn is divergent... would series of an*bn be divergent also? @geerky42

Answer 12

Not necessary. Saying \(0<a_n,~b_n<1\), then \(a_n,~b_n>a_nb_n\)

Answer 13

So in some cases, \(a_nb_n\) could be small enough for series to converges, right?

Answer 14

Simple example: Consider the series
\[\sum_{n=1}^\infty\frac{(-1)^n}{n}\]
where \(a_n=(-1)^n=\{-1,1,-1,1,\ldots\}\) and \(b_n=\dfrac{1}{n}=\left\{1,\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4},\ldots\right\}\). Both \(\sum a_n\) and \(\sum b_n\) are divergent, but \(\sum a_nb_n\) is not.