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Absolute convergence of an infinite series and p-series test
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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
asked Dec 21 '14 at 21:34
user104221user104221
16618
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add a comment |
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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
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 |
$begingroup$
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
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
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
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
3
active
oldest
votes
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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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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 $|z|<1$.
answered Dec 21 '14 at 21:37
Adam HughesAdam Hughes
32.4k83770
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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.
answered Feb 3 at 11:08
Panchix RegenPanchix Regen
497
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add a comment |
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3 Answers
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active
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3 Answers
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$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered Dec 21 '14 at 21:40
user63181
add a comment |
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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 $|z|<1$.
answered Dec 21 '14 at 21:37
Adam HughesAdam Hughes
32.4k83770
$endgroup$
add a comment |
$begingroup$
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 $|z|<1$.
answered Dec 21 '14 at 21:37
Adam HughesAdam Hughes
32.4k83770
$endgroup$
add a comment |
$begingroup$
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 $|z|<1$.
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 $|z|<1$.
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
answered Dec 21 '14 at 21:37
Adam HughesAdam Hughes
32.4k83770
32.4k83770
add a comment |
add a comment |
$begingroup$
$$sum_{n=1}^inftyleft|frac{(-1)^nz^n}{sqrt{n}}right|=sum_{n=1}^inftyleft|frac{z^n}{sqrt{n}}right|geqsum_{n=1}^inftyleft|frac{z^n}{n}right|$$
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.
answered Feb 3 at 11:08
Panchix RegenPanchix Regen
497
$endgroup$
add a comment |
$begingroup$
$$sum_{n=1}^inftyleft|frac{(-1)^nz^n}{sqrt{n}}right|=sum_{n=1}^inftyleft|frac{z^n}{sqrt{n}}right|geqsum_{n=1}^inftyleft|frac{z^n}{n}right|$$
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.
answered Feb 3 at 11:08
Panchix RegenPanchix Regen
497
$endgroup$
add a comment |
$begingroup$
$$sum_{n=1}^inftyleft|frac{(-1)^nz^n}{sqrt{n}}right|=sum_{n=1}^inftyleft|frac{z^n}{sqrt{n}}right|geqsum_{n=1}^inftyleft|frac{z^n}{n}right|$$
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.
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.
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
answered Feb 3 at 11:08
Panchix RegenPanchix Regen
497
497
add a comment |
add a comment |
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$begingroup$
For $lvert zrvert < 1$, you have the domination by a convergent geometric series.
$endgroup$
– Daniel Fischer
Dec 21 '14 at 21:37