Question

Interval of Convergence

Answer 1

Find, for any value of the constant a> 0, the interval of convergence for
\[\sum_{n=0}^{∞}\frac{ 1 }{ 1+a^n }x^n\]

Can someone help me?

Answer 2

determine the ratio of (n+1)/n to start with

Answer 3

My calculations say that it converges when x <1

Answer 4

I have used the ratio test.

Answer 5

\[\lim_{n\to inf}\frac{ x^{n+1}}{ 1+a^{n+1} }\frac{ 1+a^n }{ x^n}\]
\[|x|\lim_{n\to inf}\frac{1+a^n}{ 1+a^{n+1} }=0\]

Answer 6

pfft, i saw that as a poly for some reason

Answer 7

whats the limit you reach for the ratio ?

Answer 8

1/a by chance?

Answer 9

if so, then the radius of convergence is:

|x|/a < 1

|x| < a

with an interval of -a < x < a

Answer 10

the limit was 0 right?

Answer 11

so the series converges for all x

Answer 12

the limit wasnt 0

Answer 13

the limit is 0 the denominator grade is 1 more thatn the numerator grade

Answer 14

what am I doing wrong
\[\lim_{n \rightarrow ∞}\frac{ \frac{ 1 }{ 1+a^{(n+1)} }x^{n+1} }{ \frac{ 1 }{ 1+a^n }x^n}=\lim_{n \rightarrow ∞}\frac{ (a^n+1)*x }{ (a^{n+1}+1)*x^n}\]

Answer 15

a is not a variable, its a constant that is not determined yet

you exponential stuff at work

Answer 16

that x^n is not spose to remain in the denominator after simplifiaction

Answer 17

for now, we can just as well assume that a=5

Answer 18

for large values of n, the control is simply\[\frac{a^n}{a^{n-1}} \to\frac{1}{a}\]

Answer 19

except for a +1 not a -1 :)

Answer 20

ya
i got it

Answer 21

it is like u said @amistre64, and the interval of convergence too

Answer 22

but u must test for the endpoints too

Answer 23

i let the end points fend for themselves :)

Answer 24

:)

Answer 25

if 0<a<1 the limit gets tricker, seems to go to 1 if thats the case

Answer 26

not tricker, but the control moves from a to 1/1