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For which values of x does the power series converge or diverge?
Power series, derivatives, integrals, and different intervals of convergenceInterval of Converge for a Power SeriesFind the interval of convergence for these 3 power seriesDoes a terminating recurrence relation diverge?interval of convergence for power seriesFor which values of $x$ does these series converge absolutely, converge conditionally or divergePower series and intervals of convergencePower Series Convergence/DivergenceDoes an alternating sequence converge or diverge or none?How to find the smallest interval of convergence for a given power series that has unknowns?
$begingroup$
First, I know that the series converges when |x+2| < R and diverges when |x+2| > R. So now I have that the radius of convergence is somewhere between 2 and 3. However, this doesn't really give me too good of an indication of the interval of convergence. I would say that since 2 < R < 3 and a=2, then the interval of convergence would be 0 < x < 5, but please correct me on that if I am wrong. I'm kind of just taking a shot in the dark on that one.
So, I'm lead to believe that a and c would be divergent and b and d would be convergent. I just am in need of more insight on this because I'm trying to explain it to someone else but it has been years since I learned the topic so I'm very unsure of my knowledge.
calculus sequences-and-series convergence power-series
asked Nov 18 '14 at 1:02
mmmmmm
9391224
$endgroup$
add a comment |
$begingroup$
First, I know that the series converges when |x+2| < R and diverges when |x+2| > R. So now I have that the radius of convergence is somewhere between 2 and 3. However, this doesn't really give me too good of an indication of the interval of convergence. I would say that since 2 < R < 3 and a=2, then the interval of convergence would be 0 < x < 5, but please correct me on that if I am wrong. I'm kind of just taking a shot in the dark on that one.
So, I'm lead to believe that a and c would be divergent and b and d would be convergent. I just am in need of more insight on this because I'm trying to explain it to someone else but it has been years since I learned the topic so I'm very unsure of my knowledge.
calculus sequences-and-series convergence power-series
asked Nov 18 '14 at 1:02
mmmmmm
9391224
$endgroup$
add a comment |
$begingroup$
First, I know that the series converges when |x+2| < R and diverges when |x+2| > R. So now I have that the radius of convergence is somewhere between 2 and 3. However, this doesn't really give me too good of an indication of the interval of convergence. I would say that since 2 < R < 3 and a=2, then the interval of convergence would be 0 < x < 5, but please correct me on that if I am wrong. I'm kind of just taking a shot in the dark on that one.
So, I'm lead to believe that a and c would be divergent and b and d would be convergent. I just am in need of more insight on this because I'm trying to explain it to someone else but it has been years since I learned the topic so I'm very unsure of my knowledge.
calculus sequences-and-series convergence power-series
asked Nov 18 '14 at 1:02
mmmmmm
9391224
$endgroup$
First, I know that the series converges when |x+2| < R and diverges when |x+2| > R. So now I have that the radius of convergence is somewhere between 2 and 3. However, this doesn't really give me too good of an indication of the interval of convergence. I would say that since 2 < R < 3 and a=2, then the interval of convergence would be 0 < x < 5, but please correct me on that if I am wrong. I'm kind of just taking a shot in the dark on that one.
So, I'm lead to believe that a and c would be divergent and b and d would be convergent. I just am in need of more insight on this because I'm trying to explain it to someone else but it has been years since I learned the topic so I'm very unsure of my knowledge.
calculus sequences-and-series convergence power-series
calculus sequences-and-series convergence power-series
asked Nov 18 '14 at 1:02
mmmmmm
9391224
asked Nov 18 '14 at 1:02
mmmmmm
9391224
asked Nov 18 '14 at 1:02
mmmmmm
9391224
asked Nov 18 '14 at 1:02
mmmmmm
9391224
asked Nov 18 '14 at 1:02
mmmmmm
9391224
9391224
add a comment |
add a comment |
1 Answer
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$begingroup$
The power series must be convergent when $x$ satisfies that $-R <x+2< R$, and be divergent
when $x <-R$ and $x>R$ and be unkown when $x+2=-R$ or $x+2=R$, where $2le R le 3$.
So (a) is convergent; $b$ is divergent; $c$ and $d$ are unknown.
answered Nov 18 '14 at 1:22
PaulPaul
14k42465
$endgroup$
$begingroup$
Just a question to make sure I've got this correct. So when x=-3, x+2=-1 which falls between -2 and 2 which is the smallest possible interval of convergence so it converges. Then when x=2, x+2=4 which falls outside of -3 to 3 which is the largest possible interval of convergence so diverges. Then when x=-4, we have x+2=-2 which is a problem because R is possibly equal to -2. And lastly, if x=1, x+2=3 and since R is possibly equal to 3 we also have a problem.
$endgroup$
– mmm
Nov 18 '14 at 1:33
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The power series must be convergent when $x$ satisfies that $-R <x+2< R$, and be divergent
when $x <-R$ and $x>R$ and be unkown when $x+2=-R$ or $x+2=R$, where $2le R le 3$.
So (a) is convergent; $b$ is divergent; $c$ and $d$ are unknown.
answered Nov 18 '14 at 1:22
PaulPaul
14k42465
$endgroup$
$begingroup$
Just a question to make sure I've got this correct. So when x=-3, x+2=-1 which falls between -2 and 2 which is the smallest possible interval of convergence so it converges. Then when x=2, x+2=4 which falls outside of -3 to 3 which is the largest possible interval of convergence so diverges. Then when x=-4, we have x+2=-2 which is a problem because R is possibly equal to -2. And lastly, if x=1, x+2=3 and since R is possibly equal to 3 we also have a problem.
$endgroup$
– mmm
Nov 18 '14 at 1:33
add a comment |
$begingroup$
The power series must be convergent when $x$ satisfies that $-R <x+2< R$, and be divergent
when $x <-R$ and $x>R$ and be unkown when $x+2=-R$ or $x+2=R$, where $2le R le 3$.
So (a) is convergent; $b$ is divergent; $c$ and $d$ are unknown.
answered Nov 18 '14 at 1:22
PaulPaul
14k42465
$endgroup$
$begingroup$
Just a question to make sure I've got this correct. So when x=-3, x+2=-1 which falls between -2 and 2 which is the smallest possible interval of convergence so it converges. Then when x=2, x+2=4 which falls outside of -3 to 3 which is the largest possible interval of convergence so diverges. Then when x=-4, we have x+2=-2 which is a problem because R is possibly equal to -2. And lastly, if x=1, x+2=3 and since R is possibly equal to 3 we also have a problem.
$endgroup$
– mmm
Nov 18 '14 at 1:33
add a comment |
$begingroup$
The power series must be convergent when $x$ satisfies that $-R <x+2< R$, and be divergent
when $x <-R$ and $x>R$ and be unkown when $x+2=-R$ or $x+2=R$, where $2le R le 3$.
So (a) is convergent; $b$ is divergent; $c$ and $d$ are unknown.
answered Nov 18 '14 at 1:22
PaulPaul
14k42465
$endgroup$
The power series must be convergent when $x$ satisfies that $-R <x+2< R$, and be divergent
when $x <-R$ and $x>R$ and be unkown when $x+2=-R$ or $x+2=R$, where $2le R le 3$.
So (a) is convergent; $b$ is divergent; $c$ and $d$ are unknown.
answered Nov 18 '14 at 1:22
PaulPaul
14k42465
answered Nov 18 '14 at 1:22
PaulPaul
14k42465
answered Nov 18 '14 at 1:22
PaulPaul
14k42465
answered Nov 18 '14 at 1:22
PaulPaul
14k42465
14k42465
$begingroup$
Just a question to make sure I've got this correct. So when x=-3, x+2=-1 which falls between -2 and 2 which is the smallest possible interval of convergence so it converges. Then when x=2, x+2=4 which falls outside of -3 to 3 which is the largest possible interval of convergence so diverges. Then when x=-4, we have x+2=-2 which is a problem because R is possibly equal to -2. And lastly, if x=1, x+2=3 and since R is possibly equal to 3 we also have a problem.
$endgroup$
– mmm
Nov 18 '14 at 1:33
add a comment |
$begingroup$
Just a question to make sure I've got this correct. So when x=-3, x+2=-1 which falls between -2 and 2 which is the smallest possible interval of convergence so it converges. Then when x=2, x+2=4 which falls outside of -3 to 3 which is the largest possible interval of convergence so diverges. Then when x=-4, we have x+2=-2 which is a problem because R is possibly equal to -2. And lastly, if x=1, x+2=3 and since R is possibly equal to 3 we also have a problem.
$endgroup$
– mmm
Nov 18 '14 at 1:33
$begingroup$
Just a question to make sure I've got this correct. So when x=-3, x+2=-1 which falls between -2 and 2 which is the smallest possible interval of convergence so it converges. Then when x=2, x+2=4 which falls outside of -3 to 3 which is the largest possible interval of convergence so diverges. Then when x=-4, we have x+2=-2 which is a problem because R is possibly equal to -2. And lastly, if x=1, x+2=3 and since R is possibly equal to 3 we also have a problem.
$endgroup$
– mmm
Nov 18 '14 at 1:33
$begingroup$
Just a question to make sure I've got this correct. So when x=-3, x+2=-1 which falls between -2 and 2 which is the smallest possible interval of convergence so it converges. Then when x=2, x+2=4 which falls outside of -3 to 3 which is the largest possible interval of convergence so diverges. Then when x=-4, we have x+2=-2 which is a problem because R is possibly equal to -2. And lastly, if x=1, x+2=3 and since R is possibly equal to 3 we also have a problem.
$endgroup$
– mmm
Nov 18 '14 at 1:33
add a comment |
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