Question

Interval of convergence of power series

Answer 1

Find, for any value of the constant a> 0, the interval of convergence of power series
\[\sum_{n=0}^{∞}\frac{ 1 }{ 1+a^n }x^n\]
With the radio test I find the series converges when \[\frac{ \left| x \right| }{ a }<1\]
Are you agree?

Answer 2

\[a_{n+1} = \frac{x^{n+1}}{1 + a^{n+1}}\]\[a_n = \frac{x^n}{1 + a^n}\]try the ratio test,\[\lim_{n\rightarrow \infty} \left|\frac{a_{n+1}}{a_n}\right|\]

Answer 3

I have used the ratio test..

Answer 4

well this is what i get\[\lim_{n\rightarrow \infty} |x|\frac{1 + a^n}{1+a^{n+1}}\]

Answer 5

Yes and \[\frac{ 1+a^n }{ 1+a^{n+1} }=\frac{ 1 }{ a }\] when n--> ∞

Answer 6

it depends on the value of a, if a < 1\[\lim_{n\rightarrow\infty} \frac{1 + a^n}{1 + a^{n+1}} = 1\]

Answer 7

Hmm...

Answer 8

notice that if |a| < 1\[\lim_{n\rightarrow \infty} \frac{1}{a^n} \neq 0\]

Answer 9

So when a<1 the converges interval is x. And when a≥1 converges interval is x/a

Answer 10

@exraven

Answer 11

if a < 1,\[|x| <1\]\[-1 < x < 1\]if a >= 1\[\frac{|x|}{a} < 1\]\[-a < x < a\]

Answer 12

Thank you.

Answer 13

is there something about that I also need to check the endpoints?

Answer 14

yes, you need to check endpoints

Answer 15

In both cases?

Answer 16

@satellite73 Do you now have I check the endpoints?

Answer 17

Or maybe @.Sam. or @ @phi can help me?

Answer 18

@Mertsj ?

Answer 19

sorry for the late reply, yes you should check both cases

Answer 20

@ash2326 can you help my find endpoints?

Answer 21

Or maybe @experimentX

Answer 22

seems okay .. what is the problem?

Answer 23

Do I not need to check the endpoints?

Answer 24

any info given about \( a \) ?

Answer 25

No... just a>0

Answer 26

there are two cases you need to check. a < 1, a > 0
if a<1, then |x| < 1 ... if x = 1, it diverges.

Answer 27

similarly if a=1, the same case as above.

if a>1, then ratio test yields, |x| < a, if x = a, then it diverges, as you can see
\[ \lim_{n \to \infty }\frac{a^n}{1 + a^{n+1}} = 1/a\]
if lim -> infinity a_n = 0 is a necessary case for convergence of series.

Answer 28

hence for all cases your series diverges at boundary

Answer 29

Thank you. I will use some time on response..

Answer 30

ok ok take your time.