Why does the criterion for convergence of a power series not imply every series with bounded terms converges?
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I am reading Complex Made Simple by David C. Ullrich. There is a result from which I am deducing bogus conclusions, so I must be misunderstanding it somehow:
Lemma 1.0. Suppose $(c_n)_{n = 0}^{infty}$ is a sequence of complex numbers, and define $R in [0, infty]$ by
$$R = sup {r ge 0: text{the sequence } (c_nr^n) text{ is bounded}}.$$
Then the power series $sum_{n=0}^{infty}c_n(z-z_0)^n$ converges absolutely and uniformly on every compact subset of the disk $D(z_0, R)$ and diverges at every point $z$ with $|z-z_0|>R$.
My bogus conclusion:
Let $c_n$ be a sequence of complex numbers and suppose that $c_n r^n$ is bounded. Then $sum_{n=0}^{infty} c_n r^n$ converges.
My reasoning:
Let $c_n$ be any sequence of complex numbers. The series $sum_{n=0}^{infty}c_n(z-z_0)^n$ converges absolutely whenever $|z - z_0|<R$, so $sum_{n=0}^{infty}c_nr^n$ converges whenever $r < R$, so $sum_{n=0}^{infty}c_nr^n$ converges whenever $c_n r^n$ is bounded.
complex-analysis convergence power-series
edited Dec 22 '18 at 23:09
Eric Wofsey
182k12209337
asked Dec 22 '18 at 22:51
OviOvi
12.4k1038112
$endgroup$
add a comment |
$begingroup$
I am reading Complex Made Simple by David C. Ullrich. There is a result from which I am deducing bogus conclusions, so I must be misunderstanding it somehow:
Lemma 1.0. Suppose $(c_n)_{n = 0}^{infty}$ is a sequence of complex numbers, and define $R in [0, infty]$ by
$$R = sup {r ge 0: text{the sequence } (c_nr^n) text{ is bounded}}.$$
Then the power series $sum_{n=0}^{infty}c_n(z-z_0)^n$ converges absolutely and uniformly on every compact subset of the disk $D(z_0, R)$ and diverges at every point $z$ with $|z-z_0|>R$.
My bogus conclusion:
Let $c_n$ be a sequence of complex numbers and suppose that $c_n r^n$ is bounded. Then $sum_{n=0}^{infty} c_n r^n$ converges.
My reasoning:
Let $c_n$ be any sequence of complex numbers. The series $sum_{n=0}^{infty}c_n(z-z_0)^n$ converges absolutely whenever $|z - z_0|<R$, so $sum_{n=0}^{infty}c_nr^n$ converges whenever $r < R$, so $sum_{n=0}^{infty}c_nr^n$ converges whenever $c_n r^n$ is bounded.
complex-analysis convergence power-series
edited Dec 22 '18 at 23:09
Eric Wofsey
182k12209337
asked Dec 22 '18 at 22:51
OviOvi
12.4k1038112
$endgroup$
add a comment |
$begingroup$
I am reading Complex Made Simple by David C. Ullrich. There is a result from which I am deducing bogus conclusions, so I must be misunderstanding it somehow:
Lemma 1.0. Suppose $(c_n)_{n = 0}^{infty}$ is a sequence of complex numbers, and define $R in [0, infty]$ by
$$R = sup {r ge 0: text{the sequence } (c_nr^n) text{ is bounded}}.$$
Then the power series $sum_{n=0}^{infty}c_n(z-z_0)^n$ converges absolutely and uniformly on every compact subset of the disk $D(z_0, R)$ and diverges at every point $z$ with $|z-z_0|>R$.
My bogus conclusion:
Let $c_n$ be a sequence of complex numbers and suppose that $c_n r^n$ is bounded. Then $sum_{n=0}^{infty} c_n r^n$ converges.
My reasoning:
Let $c_n$ be any sequence of complex numbers. The series $sum_{n=0}^{infty}c_n(z-z_0)^n$ converges absolutely whenever $|z - z_0|<R$, so $sum_{n=0}^{infty}c_nr^n$ converges whenever $r < R$, so $sum_{n=0}^{infty}c_nr^n$ converges whenever $c_n r^n$ is bounded.
complex-analysis convergence power-series
edited Dec 22 '18 at 23:09
Eric Wofsey
182k12209337
asked Dec 22 '18 at 22:51
OviOvi
12.4k1038112
$endgroup$
I am reading Complex Made Simple by David C. Ullrich. There is a result from which I am deducing bogus conclusions, so I must be misunderstanding it somehow:
Lemma 1.0. Suppose $(c_n)_{n = 0}^{infty}$ is a sequence of complex numbers, and define $R in [0, infty]$ by
$$R = sup {r ge 0: text{the sequence } (c_nr^n) text{ is bounded}}.$$
Then the power series $sum_{n=0}^{infty}c_n(z-z_0)^n$ converges absolutely and uniformly on every compact subset of the disk $D(z_0, R)$ and diverges at every point $z$ with $|z-z_0|>R$.
My bogus conclusion:
Let $c_n$ be a sequence of complex numbers and suppose that $c_n r^n$ is bounded. Then $sum_{n=0}^{infty} c_n r^n$ converges.
My reasoning:
Let $c_n$ be any sequence of complex numbers. The series $sum_{n=0}^{infty}c_n(z-z_0)^n$ converges absolutely whenever $|z - z_0|<R$, so $sum_{n=0}^{infty}c_nr^n$ converges whenever $r < R$, so $sum_{n=0}^{infty}c_nr^n$ converges whenever $c_n r^n$ is bounded.
complex-analysis convergence power-series
complex-analysis convergence power-series
edited Dec 22 '18 at 23:09
Eric Wofsey
182k12209337
asked Dec 22 '18 at 22:51
OviOvi
12.4k1038112
edited Dec 22 '18 at 23:09
Eric Wofsey
182k12209337
asked Dec 22 '18 at 22:51
OviOvi
12.4k1038112
edited Dec 22 '18 at 23:09
Eric Wofsey
182k12209337
edited Dec 22 '18 at 23:09
Eric Wofsey
182k12209337
edited Dec 22 '18 at 23:09
Eric Wofsey
182k12209337
182k12209337
asked Dec 22 '18 at 22:51
OviOvi
12.4k1038112
asked Dec 22 '18 at 22:51
OviOvi
12.4k1038112
asked Dec 22 '18 at 22:51
OviOvi
12.4k1038112
12.4k1038112
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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The problem comes in the last step. Just because $sum_{n=1}^infty c_nr^n$ converges with $r lt R$ you cannot conclude that $sum_{n=1}^infty c_nR^n$ converges. As an example, let $c_n=1$ for all $n$. We note that $R=1$ here. $c_nR^n=1$, so is bounded. For any $r lt 1$, $sum_{n=1}^infty c_nr^n$ converges absolutely, but $sum_{n=1}^infty c_nR^n$ does not converge.
answered Dec 22 '18 at 23:02
Ross MillikanRoss Millikan
293k23197371
$endgroup$
$begingroup$
Hi! I $think$ I understand, but could I please run my whole reasoning by you to check? My reasoning is this: Let $c_n$ be a sequence and let's say $R=1$. Take $r=0.5$. Then $c_n (0.5)^n$ is bounded. Let $z_0=0$. The theorem tells us that $sum_{n=0}^{infty} c_nz^n$ converges absolutely for any $z$ with $|z|<1$; pick $z = 0.5$, Then $sum_{n=0}^{infty} c_n (0.5)^n$ converges. But here I made the mistake; the conclusion we can draw from here is that $sum_{n=0}^{infty} c_nr^n$ converges for "almost every" $r$ for which $c_nr^n$ is bounded (for any $r$ with $|r|<1$). (continued)
$endgroup$
– Ovi
Dec 29 '18 at 14:45
$begingroup$
(continued) But we cannot draw the conclusion that "If $c_n r^n$ is bounded, then $sum_{n=0}^{infty} c_n r^n$ converges, because it may be the case that $c_n R^n$ is bounded, but $sum_{n=0}^{infty} c_n R^n$ does not converge.
$endgroup$
– Ovi
Dec 29 '18 at 14:45
$begingroup$
Yes, that is correct. The sum will converge for every $r$ with modulus strictly less than $R$, but not necessarily on the circle $|r|=R$.
$endgroup$
– Ross Millikan
Dec 29 '18 at 15:26
$begingroup$
Thank you! ${}{}{}$
$endgroup$
– Ovi
Dec 29 '18 at 16:06
add a comment |
$begingroup$
The fact that $c_n r^n$ is bounded, means that $r$ is in the set we're taking the sup of, so $r le R$. But the convergence of $sum_{n=0}^infty c_n s^n$ is only guaranteed for $s < R$. It could very well be that $r=R$;
a simple example is $c_n = (-1)^n$ and $r=1$.
answered Dec 22 '18 at 22:59
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
add a comment |
$begingroup$
The condition $r<R$ is not equivalent to $c_nr^n$ being bounded. It is possible that $c_nr^n$ is bounded for $r=R$ as well, and in that case we cannot conclude that the series converges.
answered Dec 22 '18 at 22:58
Eric WofseyEric Wofsey
182k12209337
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The problem comes in the last step. Just because $sum_{n=1}^infty c_nr^n$ converges with $r lt R$ you cannot conclude that $sum_{n=1}^infty c_nR^n$ converges. As an example, let $c_n=1$ for all $n$. We note that $R=1$ here. $c_nR^n=1$, so is bounded. For any $r lt 1$, $sum_{n=1}^infty c_nr^n$ converges absolutely, but $sum_{n=1}^infty c_nR^n$ does not converge.
answered Dec 22 '18 at 23:02
Ross MillikanRoss Millikan
293k23197371
$endgroup$
$begingroup$
Hi! I $think$ I understand, but could I please run my whole reasoning by you to check? My reasoning is this: Let $c_n$ be a sequence and let's say $R=1$. Take $r=0.5$. Then $c_n (0.5)^n$ is bounded. Let $z_0=0$. The theorem tells us that $sum_{n=0}^{infty} c_nz^n$ converges absolutely for any $z$ with $|z|<1$; pick $z = 0.5$, Then $sum_{n=0}^{infty} c_n (0.5)^n$ converges. But here I made the mistake; the conclusion we can draw from here is that $sum_{n=0}^{infty} c_nr^n$ converges for "almost every" $r$ for which $c_nr^n$ is bounded (for any $r$ with $|r|<1$). (continued)
$endgroup$
– Ovi
Dec 29 '18 at 14:45
$begingroup$
(continued) But we cannot draw the conclusion that "If $c_n r^n$ is bounded, then $sum_{n=0}^{infty} c_n r^n$ converges, because it may be the case that $c_n R^n$ is bounded, but $sum_{n=0}^{infty} c_n R^n$ does not converge.
$endgroup$
– Ovi
Dec 29 '18 at 14:45
$begingroup$
Yes, that is correct. The sum will converge for every $r$ with modulus strictly less than $R$, but not necessarily on the circle $|r|=R$.
$endgroup$
– Ross Millikan
Dec 29 '18 at 15:26
$begingroup$
Thank you! ${}{}{}$
$endgroup$
– Ovi
Dec 29 '18 at 16:06
add a comment |
$begingroup$
The problem comes in the last step. Just because $sum_{n=1}^infty c_nr^n$ converges with $r lt R$ you cannot conclude that $sum_{n=1}^infty c_nR^n$ converges. As an example, let $c_n=1$ for all $n$. We note that $R=1$ here. $c_nR^n=1$, so is bounded. For any $r lt 1$, $sum_{n=1}^infty c_nr^n$ converges absolutely, but $sum_{n=1}^infty c_nR^n$ does not converge.
answered Dec 22 '18 at 23:02
Ross MillikanRoss Millikan
293k23197371
$endgroup$
$begingroup$
Hi! I $think$ I understand, but could I please run my whole reasoning by you to check? My reasoning is this: Let $c_n$ be a sequence and let's say $R=1$. Take $r=0.5$. Then $c_n (0.5)^n$ is bounded. Let $z_0=0$. The theorem tells us that $sum_{n=0}^{infty} c_nz^n$ converges absolutely for any $z$ with $|z|<1$; pick $z = 0.5$, Then $sum_{n=0}^{infty} c_n (0.5)^n$ converges. But here I made the mistake; the conclusion we can draw from here is that $sum_{n=0}^{infty} c_nr^n$ converges for "almost every" $r$ for which $c_nr^n$ is bounded (for any $r$ with $|r|<1$). (continued)
$endgroup$
– Ovi
Dec 29 '18 at 14:45
$begingroup$
(continued) But we cannot draw the conclusion that "If $c_n r^n$ is bounded, then $sum_{n=0}^{infty} c_n r^n$ converges, because it may be the case that $c_n R^n$ is bounded, but $sum_{n=0}^{infty} c_n R^n$ does not converge.
$endgroup$
– Ovi
Dec 29 '18 at 14:45
$begingroup$
Yes, that is correct. The sum will converge for every $r$ with modulus strictly less than $R$, but not necessarily on the circle $|r|=R$.
$endgroup$
– Ross Millikan
Dec 29 '18 at 15:26
$begingroup$
Thank you! ${}{}{}$
$endgroup$
– Ovi
Dec 29 '18 at 16:06
add a comment |
$begingroup$
The problem comes in the last step. Just because $sum_{n=1}^infty c_nr^n$ converges with $r lt R$ you cannot conclude that $sum_{n=1}^infty c_nR^n$ converges. As an example, let $c_n=1$ for all $n$. We note that $R=1$ here. $c_nR^n=1$, so is bounded. For any $r lt 1$, $sum_{n=1}^infty c_nr^n$ converges absolutely, but $sum_{n=1}^infty c_nR^n$ does not converge.
answered Dec 22 '18 at 23:02
Ross MillikanRoss Millikan
293k23197371
$endgroup$
The problem comes in the last step. Just because $sum_{n=1}^infty c_nr^n$ converges with $r lt R$ you cannot conclude that $sum_{n=1}^infty c_nR^n$ converges. As an example, let $c_n=1$ for all $n$. We note that $R=1$ here. $c_nR^n=1$, so is bounded. For any $r lt 1$, $sum_{n=1}^infty c_nr^n$ converges absolutely, but $sum_{n=1}^infty c_nR^n$ does not converge.
answered Dec 22 '18 at 23:02
Ross MillikanRoss Millikan
293k23197371
answered Dec 22 '18 at 23:02
Ross MillikanRoss Millikan
293k23197371
answered Dec 22 '18 at 23:02
Ross MillikanRoss Millikan
293k23197371
answered Dec 22 '18 at 23:02
Ross MillikanRoss Millikan
293k23197371
293k23197371
$begingroup$
Hi! I $think$ I understand, but could I please run my whole reasoning by you to check? My reasoning is this: Let $c_n$ be a sequence and let's say $R=1$. Take $r=0.5$. Then $c_n (0.5)^n$ is bounded. Let $z_0=0$. The theorem tells us that $sum_{n=0}^{infty} c_nz^n$ converges absolutely for any $z$ with $|z|<1$; pick $z = 0.5$, Then $sum_{n=0}^{infty} c_n (0.5)^n$ converges. But here I made the mistake; the conclusion we can draw from here is that $sum_{n=0}^{infty} c_nr^n$ converges for "almost every" $r$ for which $c_nr^n$ is bounded (for any $r$ with $|r|<1$). (continued)
$endgroup$
– Ovi
Dec 29 '18 at 14:45
$begingroup$
(continued) But we cannot draw the conclusion that "If $c_n r^n$ is bounded, then $sum_{n=0}^{infty} c_n r^n$ converges, because it may be the case that $c_n R^n$ is bounded, but $sum_{n=0}^{infty} c_n R^n$ does not converge.
$endgroup$
– Ovi
Dec 29 '18 at 14:45
$begingroup$
Yes, that is correct. The sum will converge for every $r$ with modulus strictly less than $R$, but not necessarily on the circle $|r|=R$.
$endgroup$
– Ross Millikan
Dec 29 '18 at 15:26
$begingroup$
Thank you! ${}{}{}$
$endgroup$
– Ovi
Dec 29 '18 at 16:06
add a comment |
$begingroup$
Hi! I $think$ I understand, but could I please run my whole reasoning by you to check? My reasoning is this: Let $c_n$ be a sequence and let's say $R=1$. Take $r=0.5$. Then $c_n (0.5)^n$ is bounded. Let $z_0=0$. The theorem tells us that $sum_{n=0}^{infty} c_nz^n$ converges absolutely for any $z$ with $|z|<1$; pick $z = 0.5$, Then $sum_{n=0}^{infty} c_n (0.5)^n$ converges. But here I made the mistake; the conclusion we can draw from here is that $sum_{n=0}^{infty} c_nr^n$ converges for "almost every" $r$ for which $c_nr^n$ is bounded (for any $r$ with $|r|<1$). (continued)
$endgroup$
– Ovi
Dec 29 '18 at 14:45
$begingroup$
(continued) But we cannot draw the conclusion that "If $c_n r^n$ is bounded, then $sum_{n=0}^{infty} c_n r^n$ converges, because it may be the case that $c_n R^n$ is bounded, but $sum_{n=0}^{infty} c_n R^n$ does not converge.
$endgroup$
– Ovi
Dec 29 '18 at 14:45
$begingroup$
Yes, that is correct. The sum will converge for every $r$ with modulus strictly less than $R$, but not necessarily on the circle $|r|=R$.
$endgroup$
– Ross Millikan
Dec 29 '18 at 15:26
$begingroup$
Thank you! ${}{}{}$
$endgroup$
– Ovi
Dec 29 '18 at 16:06
$begingroup$
Hi! I $think$ I understand, but could I please run my whole reasoning by you to check? My reasoning is this: Let $c_n$ be a sequence and let's say $R=1$. Take $r=0.5$. Then $c_n (0.5)^n$ is bounded. Let $z_0=0$. The theorem tells us that $sum_{n=0}^{infty} c_nz^n$ converges absolutely for any $z$ with $|z|<1$; pick $z = 0.5$, Then $sum_{n=0}^{infty} c_n (0.5)^n$ converges. But here I made the mistake; the conclusion we can draw from here is that $sum_{n=0}^{infty} c_nr^n$ converges for "almost every" $r$ for which $c_nr^n$ is bounded (for any $r$ with $|r|<1$). (continued)
$endgroup$
– Ovi
Dec 29 '18 at 14:45
$begingroup$
Hi! I $think$ I understand, but could I please run my whole reasoning by you to check? My reasoning is this: Let $c_n$ be a sequence and let's say $R=1$. Take $r=0.5$. Then $c_n (0.5)^n$ is bounded. Let $z_0=0$. The theorem tells us that $sum_{n=0}^{infty} c_nz^n$ converges absolutely for any $z$ with $|z|<1$; pick $z = 0.5$, Then $sum_{n=0}^{infty} c_n (0.5)^n$ converges. But here I made the mistake; the conclusion we can draw from here is that $sum_{n=0}^{infty} c_nr^n$ converges for "almost every" $r$ for which $c_nr^n$ is bounded (for any $r$ with $|r|<1$). (continued)
$endgroup$
– Ovi
Dec 29 '18 at 14:45
$begingroup$
(continued) But we cannot draw the conclusion that "If $c_n r^n$ is bounded, then $sum_{n=0}^{infty} c_n r^n$ converges, because it may be the case that $c_n R^n$ is bounded, but $sum_{n=0}^{infty} c_n R^n$ does not converge.
$endgroup$
– Ovi
Dec 29 '18 at 14:45
$begingroup$
(continued) But we cannot draw the conclusion that "If $c_n r^n$ is bounded, then $sum_{n=0}^{infty} c_n r^n$ converges, because it may be the case that $c_n R^n$ is bounded, but $sum_{n=0}^{infty} c_n R^n$ does not converge.
$endgroup$
– Ovi
Dec 29 '18 at 14:45
$begingroup$
Yes, that is correct. The sum will converge for every $r$ with modulus strictly less than $R$, but not necessarily on the circle $|r|=R$.
$endgroup$
– Ross Millikan
Dec 29 '18 at 15:26
$begingroup$
Yes, that is correct. The sum will converge for every $r$ with modulus strictly less than $R$, but not necessarily on the circle $|r|=R$.
$endgroup$
– Ross Millikan
Dec 29 '18 at 15:26
$begingroup$
Thank you! ${}{}{}$
$endgroup$
– Ovi
Dec 29 '18 at 16:06
$begingroup$
Thank you! ${}{}{}$
$endgroup$
– Ovi
Dec 29 '18 at 16:06
add a comment |
$begingroup$
The fact that $c_n r^n$ is bounded, means that $r$ is in the set we're taking the sup of, so $r le R$. But the convergence of $sum_{n=0}^infty c_n s^n$ is only guaranteed for $s < R$. It could very well be that $r=R$;
a simple example is $c_n = (-1)^n$ and $r=1$.
answered Dec 22 '18 at 22:59
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
add a comment |
$begingroup$
The fact that $c_n r^n$ is bounded, means that $r$ is in the set we're taking the sup of, so $r le R$. But the convergence of $sum_{n=0}^infty c_n s^n$ is only guaranteed for $s < R$. It could very well be that $r=R$;
a simple example is $c_n = (-1)^n$ and $r=1$.
answered Dec 22 '18 at 22:59
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
add a comment |
$begingroup$
The fact that $c_n r^n$ is bounded, means that $r$ is in the set we're taking the sup of, so $r le R$. But the convergence of $sum_{n=0}^infty c_n s^n$ is only guaranteed for $s < R$. It could very well be that $r=R$;
a simple example is $c_n = (-1)^n$ and $r=1$.
answered Dec 22 '18 at 22:59
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
The fact that $c_n r^n$ is bounded, means that $r$ is in the set we're taking the sup of, so $r le R$. But the convergence of $sum_{n=0}^infty c_n s^n$ is only guaranteed for $s < R$. It could very well be that $r=R$;
a simple example is $c_n = (-1)^n$ and $r=1$.
answered Dec 22 '18 at 22:59
Henno BrandsmaHenno Brandsma
106k347114
answered Dec 22 '18 at 22:59
Henno BrandsmaHenno Brandsma
106k347114
answered Dec 22 '18 at 22:59
Henno BrandsmaHenno Brandsma
106k347114
answered Dec 22 '18 at 22:59
Henno BrandsmaHenno Brandsma
106k347114
106k347114
add a comment |
add a comment |
$begingroup$
The condition $r<R$ is not equivalent to $c_nr^n$ being bounded. It is possible that $c_nr^n$ is bounded for $r=R$ as well, and in that case we cannot conclude that the series converges.
answered Dec 22 '18 at 22:58
Eric WofseyEric Wofsey
182k12209337
$endgroup$
add a comment |
$begingroup$
The condition $r<R$ is not equivalent to $c_nr^n$ being bounded. It is possible that $c_nr^n$ is bounded for $r=R$ as well, and in that case we cannot conclude that the series converges.
answered Dec 22 '18 at 22:58
Eric WofseyEric Wofsey
182k12209337
$endgroup$
add a comment |
$begingroup$
The condition $r<R$ is not equivalent to $c_nr^n$ being bounded. It is possible that $c_nr^n$ is bounded for $r=R$ as well, and in that case we cannot conclude that the series converges.
answered Dec 22 '18 at 22:58
Eric WofseyEric Wofsey
182k12209337
$endgroup$
The condition $r<R$ is not equivalent to $c_nr^n$ being bounded. It is possible that $c_nr^n$ is bounded for $r=R$ as well, and in that case we cannot conclude that the series converges.
answered Dec 22 '18 at 22:58
Eric WofseyEric Wofsey
182k12209337
answered Dec 22 '18 at 22:58
Eric WofseyEric Wofsey
182k12209337
answered Dec 22 '18 at 22:58
Eric WofseyEric Wofsey
182k12209337
answered Dec 22 '18 at 22:58
Eric WofseyEric Wofsey
182k12209337
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