Absolute convergence of an infinite series and p-series test

0

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Why does the infinite series $sum_{n=1}^{infty}frac{(-1)^n}{sqrt n}z^n$ where $zin mathbb{C}$ converge absolutely for $|z|<1$. Doesn't the series diverge because if we apply the absolute values, we can use p-series test and since $p<1$, the series diverges?
Also, I am trying to find a value of $z$ with $|z|=1$ such that the series converges but I am stuck on this part too.

real-analysis sequences-and-series complex-analysis divergent-series absolute-convergence

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asked Dec 21 '14 at 21:34

user104221user104221

16618

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  For $lvert zrvert < 1$, you have the domination by a convergent geometric series.
  $endgroup$
  – Daniel Fischer
  Dec 21 '14 at 21:37
  

  
  
  
  

add a comment | 

0

$begingroup$

Why does the infinite series $sum_{n=1}^{infty}frac{(-1)^n}{sqrt n}z^n$ where $zin mathbb{C}$ converge absolutely for $|z|<1$. Doesn't the series diverge because if we apply the absolute values, we can use p-series test and since $p<1$, the series diverges?
Also, I am trying to find a value of $z$ with $|z|=1$ such that the series converges but I am stuck on this part too.

real-analysis sequences-and-series complex-analysis divergent-series absolute-convergence

share|cite|improve this question

asked Dec 21 '14 at 21:34

user104221user104221

16618

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For $lvert zrvert < 1$, you have the domination by a convergent geometric series.
  $endgroup$
  – Daniel Fischer
  Dec 21 '14 at 21:37
  

  
  
  
  

add a comment | 

0

0

0

$begingroup$

Why does the infinite series $sum_{n=1}^{infty}frac{(-1)^n}{sqrt n}z^n$ where $zin mathbb{C}$ converge absolutely for $|z|<1$. Doesn't the series diverge because if we apply the absolute values, we can use p-series test and since $p<1$, the series diverges?
Also, I am trying to find a value of $z$ with $|z|=1$ such that the series converges but I am stuck on this part too.

real-analysis sequences-and-series complex-analysis divergent-series absolute-convergence

share|cite|improve this question

asked Dec 21 '14 at 21:34

user104221user104221

16618

$endgroup$

Why does the infinite series $sum_{n=1}^{infty}frac{(-1)^n}{sqrt n}z^n$ where $zin mathbb{C}$ converge absolutely for $|z|<1$. Doesn't the series diverge because if we apply the absolute values, we can use p-series test and since $p<1$, the series diverges?
Also, I am trying to find a value of $z$ with $|z|=1$ such that the series converges but I am stuck on this part too.

real-analysis sequences-and-series complex-analysis divergent-series absolute-convergence

real-analysis sequences-and-series complex-analysis divergent-series absolute-convergence

share|cite|improve this question

asked Dec 21 '14 at 21:34

user104221user104221

16618

share|cite|improve this question

asked Dec 21 '14 at 21:34

user104221user104221

16618

share|cite|improve this question

share|cite|improve this question

asked Dec 21 '14 at 21:34

user104221user104221

16618

asked Dec 21 '14 at 21:34

user104221user104221

16618

asked Dec 21 '14 at 21:34

user104221user104221

16618

16618

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For $lvert zrvert < 1$, you have the domination by a convergent geometric series.
  $endgroup$
  – Daniel Fischer
  Dec 21 '14 at 21:37
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For $lvert zrvert < 1$, you have the domination by a convergent geometric series.
  $endgroup$
  – Daniel Fischer
  Dec 21 '14 at 21:37
  

  
  
  
  

$begingroup$
For $lvert zrvert < 1$, you have the domination by a convergent geometric series.
$endgroup$
– Daniel Fischer
Dec 21 '14 at 21:37

$begingroup$
For $lvert zrvert < 1$, you have the domination by a convergent geometric series.
$endgroup$
– Daniel Fischer
Dec 21 '14 at 21:37

add a comment | 

3 Answers
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2

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Hint

• Use the ratio test to prove that the radius of convergence is $R=1$.


• Use the Dirichlet's test to prove the convergence for $|z|=1$ and $zne1$, and for $z=1$ use the Leibniz criterion to prove the convergence.

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answered Dec 21 '14 at 21:40

user63181

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1

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No, that's only for ordinary series. You can see by the root test that

$$lim_{ntoinfty}left|sqrt{n}z^nright|^{1/n}=lim_{ntoinfty}n^{1/2n}|z|=|z|$$

converges absolutely when this limit is $<1$, i.e. when $|z|<1$.

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answered Dec 21 '14 at 21:37

Adam HughesAdam Hughes

32.4k83770

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add a comment | 

0

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$$sum_{n=1}^inftyleft|frac{(-1)^nz^n}{sqrt{n}}right|=sum_{n=1}^inftyleft|frac{z^n}{sqrt{n}}right|geqsum_{n=1}^inftyleft|frac{z^n}{n}right|$$
If $|z|=1$, the rightmost part becomes the harmonic series:
$$sum_{n=1}^inftyfrac{1}{n}=1+frac{1}{2}+frac{1}{3}+...$$ for which we know diverges. If $|z|=1$ diverges, $|z|>1$ should diverge as well. It follows that the LHS of the inequality diverges for these values as well.

Now, if $|z|<1$, the LHS of the inequality will become a sum of infinite geometric series with common ratio |z|<1 and converges.

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answered Feb 3 at 11:08

Panchix RegenPanchix Regen

497

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

2

$begingroup$

Hint

• Use the ratio test to prove that the radius of convergence is $R=1$.


• Use the Dirichlet's test to prove the convergence for $|z|=1$ and $zne1$, and for $z=1$ use the Leibniz criterion to prove the convergence.

share|cite|improve this answer

answered Dec 21 '14 at 21:40

user63181

$endgroup$

add a comment | 

2

$begingroup$

Hint

• Use the ratio test to prove that the radius of convergence is $R=1$.


• Use the Dirichlet's test to prove the convergence for $|z|=1$ and $zne1$, and for $z=1$ use the Leibniz criterion to prove the convergence.

share|cite|improve this answer

answered Dec 21 '14 at 21:40

user63181

$endgroup$

add a comment | 

2

2

2

$begingroup$

Hint

• Use the ratio test to prove that the radius of convergence is $R=1$.


• Use the Dirichlet's test to prove the convergence for $|z|=1$ and $zne1$, and for $z=1$ use the Leibniz criterion to prove the convergence.

share|cite|improve this answer

answered Dec 21 '14 at 21:40

user63181

$endgroup$

Hint

• Use the ratio test to prove that the radius of convergence is $R=1$.


• Use the Dirichlet's test to prove the convergence for $|z|=1$ and $zne1$, and for $z=1$ use the Leibniz criterion to prove the convergence.

share|cite|improve this answer

answered Dec 21 '14 at 21:40

user63181

share|cite|improve this answer

share|cite|improve this answer

answered Dec 21 '14 at 21:40

user63181

answered Dec 21 '14 at 21:40

user63181

answered Dec 21 '14 at 21:40

user63181

add a comment | 

add a comment | 

1

$begingroup$

No, that's only for ordinary series. You can see by the root test that

$$lim_{ntoinfty}left|sqrt{n}z^nright|^{1/n}=lim_{ntoinfty}n^{1/2n}|z|=|z|$$

converges absolutely when this limit is $<1$, i.e. when $|z|<1$.

share|cite|improve this answer

answered Dec 21 '14 at 21:37

Adam HughesAdam Hughes

32.4k83770

$endgroup$

add a comment | 

1

$begingroup$

No, that's only for ordinary series. You can see by the root test that

$$lim_{ntoinfty}left|sqrt{n}z^nright|^{1/n}=lim_{ntoinfty}n^{1/2n}|z|=|z|$$

converges absolutely when this limit is $<1$, i.e. when $|z|<1$.

share|cite|improve this answer

answered Dec 21 '14 at 21:37

Adam HughesAdam Hughes

32.4k83770

$endgroup$

add a comment | 

1

1

1

$begingroup$

No, that's only for ordinary series. You can see by the root test that

$$lim_{ntoinfty}left|sqrt{n}z^nright|^{1/n}=lim_{ntoinfty}n^{1/2n}|z|=|z|$$

converges absolutely when this limit is $<1$, i.e. when $|z|<1$.

share|cite|improve this answer

answered Dec 21 '14 at 21:37

Adam HughesAdam Hughes

32.4k83770

$endgroup$

No, that's only for ordinary series. You can see by the root test that

$$lim_{ntoinfty}left|sqrt{n}z^nright|^{1/n}=lim_{ntoinfty}n^{1/2n}|z|=|z|$$

converges absolutely when this limit is $<1$, i.e. when $|z|<1$.

share|cite|improve this answer

answered Dec 21 '14 at 21:37

Adam HughesAdam Hughes

32.4k83770

share|cite|improve this answer

share|cite|improve this answer

answered Dec 21 '14 at 21:37

Adam HughesAdam Hughes

32.4k83770

answered Dec 21 '14 at 21:37

Adam HughesAdam Hughes

32.4k83770

answered Dec 21 '14 at 21:37

Adam HughesAdam Hughes

32.4k83770

32.4k83770

add a comment | 

add a comment | 

0

$begingroup$

$$sum_{n=1}^inftyleft|frac{(-1)^nz^n}{sqrt{n}}right|=sum_{n=1}^inftyleft|frac{z^n}{sqrt{n}}right|geqsum_{n=1}^inftyleft|frac{z^n}{n}right|$$
If $|z|=1$, the rightmost part becomes the harmonic series:
$$sum_{n=1}^inftyfrac{1}{n}=1+frac{1}{2}+frac{1}{3}+...$$ for which we know diverges. If $|z|=1$ diverges, $|z|>1$ should diverge as well. It follows that the LHS of the inequality diverges for these values as well.

Now, if $|z|<1$, the LHS of the inequality will become a sum of infinite geometric series with common ratio |z|<1 and converges.

share|cite|improve this answer

answered Feb 3 at 11:08

Panchix RegenPanchix Regen

497

$endgroup$

add a comment | 

0

$begingroup$

$$sum_{n=1}^inftyleft|frac{(-1)^nz^n}{sqrt{n}}right|=sum_{n=1}^inftyleft|frac{z^n}{sqrt{n}}right|geqsum_{n=1}^inftyleft|frac{z^n}{n}right|$$
If $|z|=1$, the rightmost part becomes the harmonic series:
$$sum_{n=1}^inftyfrac{1}{n}=1+frac{1}{2}+frac{1}{3}+...$$ for which we know diverges. If $|z|=1$ diverges, $|z|>1$ should diverge as well. It follows that the LHS of the inequality diverges for these values as well.

Now, if $|z|<1$, the LHS of the inequality will become a sum of infinite geometric series with common ratio |z|<1 and converges.

share|cite|improve this answer

answered Feb 3 at 11:08

Panchix RegenPanchix Regen

497

$endgroup$

add a comment | 

0

0

0

$begingroup$

$$sum_{n=1}^inftyleft|frac{(-1)^nz^n}{sqrt{n}}right|=sum_{n=1}^inftyleft|frac{z^n}{sqrt{n}}right|geqsum_{n=1}^inftyleft|frac{z^n}{n}right|$$
If $|z|=1$, the rightmost part becomes the harmonic series:
$$sum_{n=1}^inftyfrac{1}{n}=1+frac{1}{2}+frac{1}{3}+...$$ for which we know diverges. If $|z|=1$ diverges, $|z|>1$ should diverge as well. It follows that the LHS of the inequality diverges for these values as well.

Now, if $|z|<1$, the LHS of the inequality will become a sum of infinite geometric series with common ratio |z|<1 and converges.

share|cite|improve this answer

answered Feb 3 at 11:08

Panchix RegenPanchix Regen

497

$endgroup$

$$sum_{n=1}^inftyleft|frac{(-1)^nz^n}{sqrt{n}}right|=sum_{n=1}^inftyleft|frac{z^n}{sqrt{n}}right|geqsum_{n=1}^inftyleft|frac{z^n}{n}right|$$
If $|z|=1$, the rightmost part becomes the harmonic series:
$$sum_{n=1}^inftyfrac{1}{n}=1+frac{1}{2}+frac{1}{3}+...$$ for which we know diverges. If $|z|=1$ diverges, $|z|>1$ should diverge as well. It follows that the LHS of the inequality diverges for these values as well.

Now, if $|z|<1$, the LHS of the inequality will become a sum of infinite geometric series with common ratio |z|<1 and converges.

share|cite|improve this answer

answered Feb 3 at 11:08

Panchix RegenPanchix Regen

497

share|cite|improve this answer

share|cite|improve this answer

answered Feb 3 at 11:08

Panchix RegenPanchix Regen

497

answered Feb 3 at 11:08

Panchix RegenPanchix Regen

497

answered Feb 3 at 11:08

Panchix RegenPanchix Regen

497

497

add a comment | 

add a comment | 

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