Mixing
Determining the radius of convergence of $sum_{n=0}^ infty (2/z)^n$. [closed]
-1
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I've read that the power series $sum_{n=0}^ infty (2/z)^n$ converges for $ vert z vert >2$ however it's been so long since I did radius of convergence that I can't seem to see how we got there! Please can someone help?
real-analysis sequences-and-series analysis convergence power-series
asked Jan 8 at 11:56
JanittJanitt
102
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closed as off-topic by RRL, egreg, Shailesh, mrtaurho, KReiser Jan 9 at 9:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, egreg, Shailesh, mrtaurho, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
-1
$begingroup$
I've read that the power series $sum_{n=0}^ infty (2/z)^n$ converges for $ vert z vert >2$ however it's been so long since I did radius of convergence that I can't seem to see how we got there! Please can someone help?
real-analysis sequences-and-series analysis convergence power-series
asked Jan 8 at 11:56
JanittJanitt
102
$endgroup$
closed as off-topic by RRL, egreg, Shailesh, mrtaurho, KReiser Jan 9 at 9:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, egreg, Shailesh, mrtaurho, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
en.wikipedia.org/wiki/Geometric_series#Proof_of_convergence
$endgroup$
– lab bhattacharjee
Jan 8 at 11:57
2
$begingroup$
That is NOT, strictly speaking, a "power series" because the powers or z are negative and does not have a "radius of convergence" in the usual sense. Thinking of it as a geometric series it converges for $left|frac{2}{z}right|< 1$ so for |z|> 2. It converges outside the circle of radius 2.
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– user247327
Jan 8 at 12:07
$begingroup$
I'm perfectly aware of this but as I stated above I read that this was the case hence was just asking for an elaboration. thank you anyway.
$endgroup$
– Janitt
Jan 9 at 15:28
add a comment |
-1
-1
-1
$begingroup$
I've read that the power series $sum_{n=0}^ infty (2/z)^n$ converges for $ vert z vert >2$ however it's been so long since I did radius of convergence that I can't seem to see how we got there! Please can someone help?
real-analysis sequences-and-series analysis convergence power-series
asked Jan 8 at 11:56
JanittJanitt
102
$endgroup$
I've read that the power series $sum_{n=0}^ infty (2/z)^n$ converges for $ vert z vert >2$ however it's been so long since I did radius of convergence that I can't seem to see how we got there! Please can someone help?
real-analysis sequences-and-series analysis convergence power-series
real-analysis sequences-and-series analysis convergence power-series
asked Jan 8 at 11:56
JanittJanitt
102
asked Jan 8 at 11:56
JanittJanitt
102
asked Jan 8 at 11:56
JanittJanitt
102
asked Jan 8 at 11:56
JanittJanitt
102
asked Jan 8 at 11:56
JanittJanitt
102
102
closed as off-topic by RRL, egreg, Shailesh, mrtaurho, KReiser Jan 9 at 9:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, egreg, Shailesh, mrtaurho, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by RRL, egreg, Shailesh, mrtaurho, KReiser Jan 9 at 9:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, egreg, Shailesh, mrtaurho, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
en.wikipedia.org/wiki/Geometric_series#Proof_of_convergence
$endgroup$
– lab bhattacharjee
Jan 8 at 11:57
2
$begingroup$
That is NOT, strictly speaking, a "power series" because the powers or z are negative and does not have a "radius of convergence" in the usual sense. Thinking of it as a geometric series it converges for $left|frac{2}{z}right|< 1$ so for |z|> 2. It converges outside the circle of radius 2.
$endgroup$
– user247327
Jan 8 at 12:07
$begingroup$
I'm perfectly aware of this but as I stated above I read that this was the case hence was just asking for an elaboration. thank you anyway.
$endgroup$
– Janitt
Jan 9 at 15:28
add a comment |
$begingroup$
en.wikipedia.org/wiki/Geometric_series#Proof_of_convergence
$endgroup$
– lab bhattacharjee
Jan 8 at 11:57
2
$begingroup$
That is NOT, strictly speaking, a "power series" because the powers or z are negative and does not have a "radius of convergence" in the usual sense. Thinking of it as a geometric series it converges for $left|frac{2}{z}right|< 1$ so for |z|> 2. It converges outside the circle of radius 2.
$endgroup$
– user247327
Jan 8 at 12:07
$begingroup$
I'm perfectly aware of this but as I stated above I read that this was the case hence was just asking for an elaboration. thank you anyway.
$endgroup$
– Janitt
Jan 9 at 15:28
$begingroup$
en.wikipedia.org/wiki/Geometric_series#Proof_of_convergence
$endgroup$
– lab bhattacharjee
Jan 8 at 11:57
$begingroup$
en.wikipedia.org/wiki/Geometric_series#Proof_of_convergence
$endgroup$
– lab bhattacharjee
Jan 8 at 11:57
2
2
$begingroup$
That is NOT, strictly speaking, a "power series" because the powers or z are negative and does not have a "radius of convergence" in the usual sense. Thinking of it as a geometric series it converges for $left|frac{2}{z}right|< 1$ so for |z|> 2. It converges outside the circle of radius 2.
$endgroup$
– user247327
Jan 8 at 12:07
$begingroup$
That is NOT, strictly speaking, a "power series" because the powers or z are negative and does not have a "radius of convergence" in the usual sense. Thinking of it as a geometric series it converges for $left|frac{2}{z}right|< 1$ so for |z|> 2. It converges outside the circle of radius 2.
$endgroup$
– user247327
Jan 8 at 12:07
$begingroup$
I'm perfectly aware of this but as I stated above I read that this was the case hence was just asking for an elaboration. thank you anyway.
$endgroup$
– Janitt
Jan 9 at 15:28
$begingroup$
I'm perfectly aware of this but as I stated above I read that this was the case hence was just asking for an elaboration. thank you anyway.
$endgroup$
– Janitt
Jan 9 at 15:28
add a comment |
1 Answer
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1
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$sum_{n=0}^ infty (2/z)^n$ is a geometric series. Hence $sum_{n=0}^ infty (2/z)^n$ converges $ iff frac{2}{|z|} <1 iff |z|>2.$
answered Jan 8 at 12:17
FredFred
45.3k1847
$endgroup$
$begingroup$
Brilliant, thank you!
$endgroup$
– Janitt
Jan 9 at 15:24
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
$sum_{n=0}^ infty (2/z)^n$ is a geometric series. Hence $sum_{n=0}^ infty (2/z)^n$ converges $ iff frac{2}{|z|} <1 iff |z|>2.$
answered Jan 8 at 12:17
FredFred
45.3k1847
$endgroup$
$begingroup$
Brilliant, thank you!
$endgroup$
– Janitt
Jan 9 at 15:24
add a comment |
1
$begingroup$
$sum_{n=0}^ infty (2/z)^n$ is a geometric series. Hence $sum_{n=0}^ infty (2/z)^n$ converges $ iff frac{2}{|z|} <1 iff |z|>2.$
answered Jan 8 at 12:17
FredFred
45.3k1847
$endgroup$
$begingroup$
Brilliant, thank you!
$endgroup$
– Janitt
Jan 9 at 15:24
add a comment |
1
1
1
$begingroup$
$sum_{n=0}^ infty (2/z)^n$ is a geometric series. Hence $sum_{n=0}^ infty (2/z)^n$ converges $ iff frac{2}{|z|} <1 iff |z|>2.$
answered Jan 8 at 12:17
FredFred
45.3k1847
$endgroup$
$sum_{n=0}^ infty (2/z)^n$ is a geometric series. Hence $sum_{n=0}^ infty (2/z)^n$ converges $ iff frac{2}{|z|} <1 iff |z|>2.$
answered Jan 8 at 12:17
FredFred
45.3k1847
answered Jan 8 at 12:17
FredFred
45.3k1847
answered Jan 8 at 12:17
FredFred
45.3k1847
answered Jan 8 at 12:17
FredFred
45.3k1847
45.3k1847
$begingroup$
Brilliant, thank you!
$endgroup$
– Janitt
Jan 9 at 15:24
add a comment |
$begingroup$
Brilliant, thank you!
$endgroup$
– Janitt
Jan 9 at 15:24
$begingroup$
Brilliant, thank you!
$endgroup$
– Janitt
Jan 9 at 15:24
$begingroup$
Brilliant, thank you!
$endgroup$
– Janitt
Jan 9 at 15:24
add a comment |
$begingroup$
en.wikipedia.org/wiki/Geometric_series#Proof_of_convergence
$endgroup$
– lab bhattacharjee
Jan 8 at 11:57
2
$begingroup$
That is NOT, strictly speaking, a "power series" because the powers or z are negative and does not have a "radius of convergence" in the usual sense. Thinking of it as a geometric series it converges for $left|frac{2}{z}right|< 1$ so for |z|> 2. It converges outside the circle of radius 2.
$endgroup$
– user247327
Jan 8 at 12:07
$begingroup$
I'm perfectly aware of this but as I stated above I read that this was the case hence was just asking for an elaboration. thank you anyway.
$endgroup$
– Janitt
Jan 9 at 15:28