Question

If a series converges absolutely, does that impact its interval of convergence?

Answer 1

I have a question that is asking me to prove that a series converges for all x, but it has negative terms so I can only prove absolute convergence. So that is why I am asking this question

Answer 2

absolute convergence is stronger than conditional convergence

Answer 3

so yes

Answer 4

We didn't really get into all of the details of what conditional convergence was today

Answer 5

Is it possible for a series to converge for all x if it converges absolutely?

Answer 6

means not absolute convergence
simple example
\[\sum\frac{1}{n}\] does not converge but
\[\sum(-1)^n\frac{1}{n}\] does

Answer 7

not sure what you mean exactly
you got an example?

Answer 8

Yes. I am supposed to use a comparison test to show that \[\sum_{n=0}^{\infty}\frac{ x ^{3n} }{ 2n!+1 }\] converges for all x

Answer 9

the terms look positive to me

Answer 10

How?

Answer 11

x could be negative

Answer 12

not sure how you use a comparison test for series, usually use the ratio test

Answer 13

I think that counts

Answer 14

But my point is that x could be negative

Answer 15

in any case it is clear that the radius of convergence is \(\infty\)

Answer 16

if x is negative, the terms could be negative

Answer 17

oh of course x could be negative, but it is a fixed number, it does not change
say \(x=-10\) for example

Answer 18

If x is negative and n is odd, terms are negative

Answer 19

the idea is to compute \[\lim_{n\to \infty}|\frac{a_{n+1}}{a_n}|\]

Answer 20

then figure out which \(x\) values work, i.e make sure that ratio is less than one

Answer 21

Yes, but you can't use the ratio test in the first place if there are negative terms... or so I thought. Then you can only use it for absolute convergence

Answer 22

ok i see you are a bid confused here
n goes to infinity, x is a variable, but it is fixed as in \(f(x)=\frac{1}{1-x}\) which is equal to \(f(x)=\sum x^n\) if \(-1<x<1\)

Answer 23

in that example x could be negative, for example it could be \(-\frac{1}{2}\)

Answer 24

But 1/(1-x) didn't have a stipulation against negative terms

Answer 25

when you use the ratio test you are taking the absolute value of the terms, computing the limit, then deciding for which values of x the limit exists

Answer 26

But my teacher said that you can't use the ratio test in the first place if you can have negative terms T_T

Answer 27

ok lets take it slow
you are confusing the ratio test for seeing if a sum converges, like say \[\sum\frac{1}{n^2}\]

Answer 28

with checking for the radius of convergence of
\[\sum_{n=1}^{\infty}\frac{x^n}{n^2}\]

Answer 29

I have no idea

Answer 30

Although I can't find any part of my book saying anything against negative terms for the ratio test, but it does talk about absolute convergence

Answer 31

i see
to find the radius of convergence of \[\sum c_nx^n\] put \(a_n=c_nx^n\) and compute \[\lim_{n\to \infty}|\frac{a_{n+1}}{a_n}|\]

Answer 32

your answer will have an \(x\) in it

Answer 33

So there are NO stipulations on the interval of convergence if it converges absolutely?

Answer 34

then make sure \[\lim_{n\to \infty}|\frac{a_{n+1}}{a_n}|<1\]which will depend on \(x\) that is the radius of convergence

Answer 35

if what converges absolutely?

Answer 36

If the series does

Answer 37

It says that the ratio test is a tool to check for absolute convergence

Answer 38

i think i an unable to explain this, it seems you have confused a sum, a number, like
\[\sum_{n=1}^{\infty}\frac{1}{2^n}\] with a series that looks like
\[f(x)=\sum_{n=1}^{\infty}x^n\]

Answer 39

I just want to know that if it is possible for something to converge absolutely and converge for all x