Power series for $|z|$ outside radius of convergence diverges

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I have proved that if $|z| > R$ then the power series $sum_n = 0^infty |a_nz^n| = infty.$ However, I have trouble showing that the series $sum_n = 0^infty a_nz^n$ diverges as a result. I first tried assuming that $sum_n =0^infty a_nz^n = L$ converges. I'm not quite sure how to proceed. Any hints?

complex-analysis

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asked Jul 31 at 22:45

伽罗瓦

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  Start by showing that if the series converges for some $z in mathbbC$, then it converges whenever $|z_0| < |z|$.
  –  mheldman
  Jul 31 at 22:59
  

  
  
  
  

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up vote
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favorite

I have proved that if $|z| > R$ then the power series $sum_n = 0^infty |a_nz^n| = infty.$ However, I have trouble showing that the series $sum_n = 0^infty a_nz^n$ diverges as a result. I first tried assuming that $sum_n =0^infty a_nz^n = L$ converges. I'm not quite sure how to proceed. Any hints?

complex-analysis

share|cite|improve this question

asked Jul 31 at 22:45

伽罗瓦

781615

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Start by showing that if the series converges for some $z in mathbbC$, then it converges whenever $|z_0| < |z|$.
  –  mheldman
  Jul 31 at 22:59
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I have proved that if $|z| > R$ then the power series $sum_n = 0^infty |a_nz^n| = infty.$ However, I have trouble showing that the series $sum_n = 0^infty a_nz^n$ diverges as a result. I first tried assuming that $sum_n =0^infty a_nz^n = L$ converges. I'm not quite sure how to proceed. Any hints?

complex-analysis

share|cite|improve this question

asked Jul 31 at 22:45

伽罗瓦

781615

I have proved that if $|z| > R$ then the power series $sum_n = 0^infty |a_nz^n| = infty.$ However, I have trouble showing that the series $sum_n = 0^infty a_nz^n$ diverges as a result. I first tried assuming that $sum_n =0^infty a_nz^n = L$ converges. I'm not quite sure how to proceed. Any hints?

complex-analysis

share|cite|improve this question

asked Jul 31 at 22:45

伽罗瓦

781615

share|cite|improve this question

share|cite|improve this question

asked Jul 31 at 22:45

伽罗瓦

781615

asked Jul 31 at 22:45

伽罗瓦

781615

asked Jul 31 at 22:45

伽罗瓦

781615

781615

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Start by showing that if the series converges for some $z in mathbbC$, then it converges whenever $|z_0| < |z|$.
  –  mheldman
  Jul 31 at 22:59
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Start by showing that if the series converges for some $z in mathbbC$, then it converges whenever $|z_0| < |z|$.
  –  mheldman
  Jul 31 at 22:59
  

  
  
  
  

1

1

Start by showing that if the series converges for some $z in mathbbC$, then it converges whenever $|z_0| < |z|$.
–  mheldman
Jul 31 at 22:59

Start by showing that if the series converges for some $z in mathbbC$, then it converges whenever $|z_0| < |z|$.
–  mheldman
Jul 31 at 22:59

add a comment | 

2 Answers
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The radius of convergence is given by

$$R^-1 = limsup_n to infty left(|a_n| right)^1/n $$

If $|z| > R$ then

$$limsup_n to infty left(|a_nz^n| right)^1/n = limsup_n to infty left(|a_n| right)^1/n|z| = fracR >1, $$

and $|a_nz^n| > 1$ infinitely often. This implies that the series diverges since $a_nz^n notto 0$ as $n to infty$.

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answered Jul 31 at 23:16

RRL

43.4k42260

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If $|sqrt[n]a_nz| > 1$ your series diverges by the root test.
If $|sqrt[n]a_nz| < 1$ your series converges.

If $|frac a_n+1a_n z| > 1$ your series diverges by the ratio test.
If $|frac a_n+1a_n z| < 1$ your series converges.

If you haven't proven these, run a comparison test to a geometric series with a similar growth rate to prove divergence / convergence.

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answered Jul 31 at 23:01

Doug M

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

The radius of convergence is given by

$$R^-1 = limsup_n to infty left(|a_n| right)^1/n $$

If $|z| > R$ then

$$limsup_n to infty left(|a_nz^n| right)^1/n = limsup_n to infty left(|a_n| right)^1/n|z| = fracR >1, $$

and $|a_nz^n| > 1$ infinitely often. This implies that the series diverges since $a_nz^n notto 0$ as $n to infty$.

share|cite|improve this answer

answered Jul 31 at 23:16

RRL

43.4k42260

add a comment | 

up vote
1
down vote

The radius of convergence is given by

$$R^-1 = limsup_n to infty left(|a_n| right)^1/n $$

If $|z| > R$ then

$$limsup_n to infty left(|a_nz^n| right)^1/n = limsup_n to infty left(|a_n| right)^1/n|z| = fracR >1, $$

and $|a_nz^n| > 1$ infinitely often. This implies that the series diverges since $a_nz^n notto 0$ as $n to infty$.

share|cite|improve this answer

answered Jul 31 at 23:16

RRL

43.4k42260

add a comment | 

up vote
1
down vote

up vote
1
down vote

The radius of convergence is given by

$$R^-1 = limsup_n to infty left(|a_n| right)^1/n $$

If $|z| > R$ then

$$limsup_n to infty left(|a_nz^n| right)^1/n = limsup_n to infty left(|a_n| right)^1/n|z| = fracR >1, $$

and $|a_nz^n| > 1$ infinitely often. This implies that the series diverges since $a_nz^n notto 0$ as $n to infty$.

share|cite|improve this answer

answered Jul 31 at 23:16

RRL

43.4k42260

The radius of convergence is given by

$$R^-1 = limsup_n to infty left(|a_n| right)^1/n $$

If $|z| > R$ then

$$limsup_n to infty left(|a_nz^n| right)^1/n = limsup_n to infty left(|a_n| right)^1/n|z| = fracR >1, $$

and $|a_nz^n| > 1$ infinitely often. This implies that the series diverges since $a_nz^n notto 0$ as $n to infty$.

share|cite|improve this answer

answered Jul 31 at 23:16

RRL

43.4k42260

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 23:16

RRL

43.4k42260

answered Jul 31 at 23:16

RRL

43.4k42260

answered Jul 31 at 23:16

RRL

43.4k42260

43.4k42260

add a comment | 

add a comment | 

up vote
0
down vote

If $|sqrt[n]a_nz| > 1$ your series diverges by the root test.
If $|sqrt[n]a_nz| < 1$ your series converges.

If $|frac a_n+1a_n z| > 1$ your series diverges by the ratio test.
If $|frac a_n+1a_n z| < 1$ your series converges.

If you haven't proven these, run a comparison test to a geometric series with a similar growth rate to prove divergence / convergence.

share|cite|improve this answer

answered Jul 31 at 23:01

Doug M

39k31749

add a comment | 

up vote
0
down vote

If $|sqrt[n]a_nz| > 1$ your series diverges by the root test.
If $|sqrt[n]a_nz| < 1$ your series converges.

If $|frac a_n+1a_n z| > 1$ your series diverges by the ratio test.
If $|frac a_n+1a_n z| < 1$ your series converges.

If you haven't proven these, run a comparison test to a geometric series with a similar growth rate to prove divergence / convergence.

share|cite|improve this answer

answered Jul 31 at 23:01

Doug M

39k31749

add a comment | 

up vote
0
down vote

up vote
0
down vote

If $|sqrt[n]a_nz| > 1$ your series diverges by the root test.
If $|sqrt[n]a_nz| < 1$ your series converges.

If $|frac a_n+1a_n z| > 1$ your series diverges by the ratio test.
If $|frac a_n+1a_n z| < 1$ your series converges.

If you haven't proven these, run a comparison test to a geometric series with a similar growth rate to prove divergence / convergence.

share|cite|improve this answer

answered Jul 31 at 23:01

Doug M

39k31749

If $|sqrt[n]a_nz| > 1$ your series diverges by the root test.
If $|sqrt[n]a_nz| < 1$ your series converges.

If $|frac a_n+1a_n z| > 1$ your series diverges by the ratio test.
If $|frac a_n+1a_n z| < 1$ your series converges.

If you haven't proven these, run a comparison test to a geometric series with a similar growth rate to prove divergence / convergence.

share|cite|improve this answer

answered Jul 31 at 23:01

Doug M

39k31749

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 23:01

Doug M

39k31749

answered Jul 31 at 23:01

Doug M

39k31749

answered Jul 31 at 23:01

Doug M

39k31749

39k31749

add a comment | 

add a comment | 

 

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