Mixing
Complex Infinite Series
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Having trouble with this infinite series and deciding whether it converges or diverges.
The series:
$$sum_{n=1}^infty n(frac{1}{2i})^n$$
My thoughts are that you take the modulus of the fraction and get $frac{1}{2}$ to the exponent $n$ makes it go to $0$ and then multiplied by $n$ make it
$$infty*0$$ which is always divergent right, making the series diverge? Can someone also clarify that this is the case?
sequences-and-series complex-numbers
asked Feb 2 at 16:42
MathstatsstudentMathstatsstudent
1076
$endgroup$
add a comment |
$begingroup$
Having trouble with this infinite series and deciding whether it converges or diverges.
The series:
$$sum_{n=1}^infty n(frac{1}{2i})^n$$
My thoughts are that you take the modulus of the fraction and get $frac{1}{2}$ to the exponent $n$ makes it go to $0$ and then multiplied by $n$ make it
$$infty*0$$ which is always divergent right, making the series diverge? Can someone also clarify that this is the case?
sequences-and-series complex-numbers
asked Feb 2 at 16:42
MathstatsstudentMathstatsstudent
1076
$endgroup$
2
$begingroup$
No this is not true, the summation of $n cdot r^n$ converges where $|r| lt 1$
$endgroup$
– Peter Foreman
Feb 2 at 16:45
add a comment |
$begingroup$
Having trouble with this infinite series and deciding whether it converges or diverges.
The series:
$$sum_{n=1}^infty n(frac{1}{2i})^n$$
My thoughts are that you take the modulus of the fraction and get $frac{1}{2}$ to the exponent $n$ makes it go to $0$ and then multiplied by $n$ make it
$$infty*0$$ which is always divergent right, making the series diverge? Can someone also clarify that this is the case?
sequences-and-series complex-numbers
asked Feb 2 at 16:42
MathstatsstudentMathstatsstudent
1076
$endgroup$
Having trouble with this infinite series and deciding whether it converges or diverges.
The series:
$$sum_{n=1}^infty n(frac{1}{2i})^n$$
My thoughts are that you take the modulus of the fraction and get $frac{1}{2}$ to the exponent $n$ makes it go to $0$ and then multiplied by $n$ make it
$$infty*0$$ which is always divergent right, making the series diverge? Can someone also clarify that this is the case?
sequences-and-series complex-numbers
sequences-and-series complex-numbers
asked Feb 2 at 16:42
MathstatsstudentMathstatsstudent
1076
asked Feb 2 at 16:42
MathstatsstudentMathstatsstudent
1076
asked Feb 2 at 16:42
MathstatsstudentMathstatsstudent
1076
asked Feb 2 at 16:42
MathstatsstudentMathstatsstudent
1076
asked Feb 2 at 16:42
MathstatsstudentMathstatsstudent
1076
1076
2
$begingroup$
No this is not true, the summation of $n cdot r^n$ converges where $|r| lt 1$
$endgroup$
– Peter Foreman
Feb 2 at 16:45
add a comment |
2
$begingroup$
No this is not true, the summation of $n cdot r^n$ converges where $|r| lt 1$
$endgroup$
– Peter Foreman
Feb 2 at 16:45
2
2
$begingroup$
No this is not true, the summation of $n cdot r^n$ converges where $|r| lt 1$
$endgroup$
– Peter Foreman
Feb 2 at 16:45
$begingroup$
No this is not true, the summation of $n cdot r^n$ converges where $|r| lt 1$
$endgroup$
– Peter Foreman
Feb 2 at 16:45
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
First let’s look if the series converges absolutely.
For this, we need to see if $sum b_n = sum frac{n}{2^n}$ converges. And this is immediate using the ratio test
as $limlimits_{nto infty}frac{b_{n+1}}{b_n} =1/2<1$.
Conclusion: the given series converges absolutely hence converges
answered Feb 2 at 16:47
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
$begingroup$
Limit 0 in the ratio test, sure about that?
$endgroup$
– Did
Feb 2 at 16:51
$begingroup$
@Did Thanks for asking the question!
$endgroup$
– mathcounterexamples.net
Feb 2 at 16:53
add a comment |
$begingroup$
Hint. Your first thought is correct: look at the modulus.
Your reasoning about $infty * 0$ is wrong.
Try the ratio test.
If you know about the geometric series
$$
1 + x + x^2 + cdots
$$
you can differentiate, multiply by $x$ and actually find out what your series converges to.
answered Feb 2 at 16:47
Ethan BolkerEthan Bolker
46.2k553121
$endgroup$
add a comment |
$begingroup$
Consider $sum_{n=0}^infty nz^n$. The radius of convergence is $r=limsup_{ntoinfty}frac1{n^{frac1n}}=1$. Since $midfrac1{2i}mid=frac12 $, the series converges.
answered Feb 2 at 16:50
Chris CusterChris Custer
14.4k3827
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
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oldest
votes
$begingroup$
First let’s look if the series converges absolutely.
For this, we need to see if $sum b_n = sum frac{n}{2^n}$ converges. And this is immediate using the ratio test
as $limlimits_{nto infty}frac{b_{n+1}}{b_n} =1/2<1$.
Conclusion: the given series converges absolutely hence converges
answered Feb 2 at 16:47
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
$begingroup$
Limit 0 in the ratio test, sure about that?
$endgroup$
– Did
Feb 2 at 16:51
$begingroup$
@Did Thanks for asking the question!
$endgroup$
– mathcounterexamples.net
Feb 2 at 16:53
add a comment |
$begingroup$
First let’s look if the series converges absolutely.
For this, we need to see if $sum b_n = sum frac{n}{2^n}$ converges. And this is immediate using the ratio test
as $limlimits_{nto infty}frac{b_{n+1}}{b_n} =1/2<1$.
Conclusion: the given series converges absolutely hence converges
answered Feb 2 at 16:47
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
$begingroup$
Limit 0 in the ratio test, sure about that?
$endgroup$
– Did
Feb 2 at 16:51
$begingroup$
@Did Thanks for asking the question!
$endgroup$
– mathcounterexamples.net
Feb 2 at 16:53
add a comment |
$begingroup$
First let’s look if the series converges absolutely.
For this, we need to see if $sum b_n = sum frac{n}{2^n}$ converges. And this is immediate using the ratio test
as $limlimits_{nto infty}frac{b_{n+1}}{b_n} =1/2<1$.
Conclusion: the given series converges absolutely hence converges
answered Feb 2 at 16:47
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
First let’s look if the series converges absolutely.
For this, we need to see if $sum b_n = sum frac{n}{2^n}$ converges. And this is immediate using the ratio test
as $limlimits_{nto infty}frac{b_{n+1}}{b_n} =1/2<1$.
Conclusion: the given series converges absolutely hence converges
answered Feb 2 at 16:47
mathcounterexamples.netmathcounterexamples.net
26.9k22158
edited Feb 2 at 16:53
answered Feb 2 at 16:47
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Feb 2 at 16:47
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Feb 2 at 16:47
mathcounterexamples.netmathcounterexamples.net
26.9k22158
26.9k22158
$begingroup$
Limit 0 in the ratio test, sure about that?
$endgroup$
– Did
Feb 2 at 16:51
$begingroup$
@Did Thanks for asking the question!
$endgroup$
– mathcounterexamples.net
Feb 2 at 16:53
add a comment |
$begingroup$
Limit 0 in the ratio test, sure about that?
$endgroup$
– Did
Feb 2 at 16:51
$begingroup$
@Did Thanks for asking the question!
$endgroup$
– mathcounterexamples.net
Feb 2 at 16:53
$begingroup$
Limit 0 in the ratio test, sure about that?
$endgroup$
– Did
Feb 2 at 16:51
$begingroup$
Limit 0 in the ratio test, sure about that?
$endgroup$
– Did
Feb 2 at 16:51
$begingroup$
@Did Thanks for asking the question!
$endgroup$
– mathcounterexamples.net
Feb 2 at 16:53
$begingroup$
@Did Thanks for asking the question!
$endgroup$
– mathcounterexamples.net
Feb 2 at 16:53
add a comment |
$begingroup$
Hint. Your first thought is correct: look at the modulus.
Your reasoning about $infty * 0$ is wrong.
Try the ratio test.
If you know about the geometric series
$$
1 + x + x^2 + cdots
$$
you can differentiate, multiply by $x$ and actually find out what your series converges to.
answered Feb 2 at 16:47
Ethan BolkerEthan Bolker
46.2k553121
$endgroup$
add a comment |
$begingroup$
Hint. Your first thought is correct: look at the modulus.
Your reasoning about $infty * 0$ is wrong.
Try the ratio test.
If you know about the geometric series
$$
1 + x + x^2 + cdots
$$
you can differentiate, multiply by $x$ and actually find out what your series converges to.
answered Feb 2 at 16:47
Ethan BolkerEthan Bolker
46.2k553121
$endgroup$
add a comment |
$begingroup$
Hint. Your first thought is correct: look at the modulus.
Your reasoning about $infty * 0$ is wrong.
Try the ratio test.
If you know about the geometric series
$$
1 + x + x^2 + cdots
$$
you can differentiate, multiply by $x$ and actually find out what your series converges to.
answered Feb 2 at 16:47
Ethan BolkerEthan Bolker
46.2k553121
$endgroup$
Hint. Your first thought is correct: look at the modulus.
Your reasoning about $infty * 0$ is wrong.
Try the ratio test.
If you know about the geometric series
$$
1 + x + x^2 + cdots
$$
you can differentiate, multiply by $x$ and actually find out what your series converges to.
answered Feb 2 at 16:47
Ethan BolkerEthan Bolker
46.2k553121
answered Feb 2 at 16:47
Ethan BolkerEthan Bolker
46.2k553121
answered Feb 2 at 16:47
Ethan BolkerEthan Bolker
46.2k553121
answered Feb 2 at 16:47
Ethan BolkerEthan Bolker
46.2k553121
46.2k553121
add a comment |
add a comment |
$begingroup$
Consider $sum_{n=0}^infty nz^n$. The radius of convergence is $r=limsup_{ntoinfty}frac1{n^{frac1n}}=1$. Since $midfrac1{2i}mid=frac12 $, the series converges.
answered Feb 2 at 16:50
Chris CusterChris Custer
14.4k3827
$endgroup$
add a comment |
$begingroup$
Consider $sum_{n=0}^infty nz^n$. The radius of convergence is $r=limsup_{ntoinfty}frac1{n^{frac1n}}=1$. Since $midfrac1{2i}mid=frac12 $, the series converges.
answered Feb 2 at 16:50
Chris CusterChris Custer
14.4k3827
$endgroup$
add a comment |
$begingroup$
Consider $sum_{n=0}^infty nz^n$. The radius of convergence is $r=limsup_{ntoinfty}frac1{n^{frac1n}}=1$. Since $midfrac1{2i}mid=frac12 $, the series converges.
answered Feb 2 at 16:50
Chris CusterChris Custer
14.4k3827
$endgroup$
Consider $sum_{n=0}^infty nz^n$. The radius of convergence is $r=limsup_{ntoinfty}frac1{n^{frac1n}}=1$. Since $midfrac1{2i}mid=frac12 $, the series converges.
answered Feb 2 at 16:50
Chris CusterChris Custer
14.4k3827
answered Feb 2 at 16:50
Chris CusterChris Custer
14.4k3827
answered Feb 2 at 16:50
Chris CusterChris Custer
14.4k3827
answered Feb 2 at 16:50
Chris CusterChris Custer
14.4k3827
14.4k3827
add a comment |
add a comment |
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$begingroup$
No this is not true, the summation of $n cdot r^n$ converges where $|r| lt 1$
$endgroup$
– Peter Foreman
Feb 2 at 16:45