Why do $sum na_n x^{n-1}$ and $sum na_n x^n$ have the same radius of convergence?
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Why is it that $sum na_n x^{n-1}$ and $sum na_n x^n$ have the same radius of convergence $R$?
How can I show that
$$lim sup|na_n|^{1/n} = lim sup |na_n x|^{1/n}$$ since this would mean that both series have the same radius of convergence $R$?
real-analysis sequences-and-series convergence
asked Dec 18 '18 at 3:45
The BoscoThe Bosco
541212
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add a comment |
$begingroup$
Why is it that $sum na_n x^{n-1}$ and $sum na_n x^n$ have the same radius of convergence $R$?
How can I show that
$$lim sup|na_n|^{1/n} = lim sup |na_n x|^{1/n}$$ since this would mean that both series have the same radius of convergence $R$?
real-analysis sequences-and-series convergence
asked Dec 18 '18 at 3:45
The BoscoThe Bosco
541212
$endgroup$
2
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You can directly note they have the same radius of converge because one is just x times the other
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– mathworker21
Dec 18 '18 at 3:54
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That is exactly what I want to prove
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– The Bosco
Dec 18 '18 at 11:51
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can you be more specific?
$endgroup$
– mathworker21
Dec 18 '18 at 11:55
add a comment |
$begingroup$
Why is it that $sum na_n x^{n-1}$ and $sum na_n x^n$ have the same radius of convergence $R$?
How can I show that
$$lim sup|na_n|^{1/n} = lim sup |na_n x|^{1/n}$$ since this would mean that both series have the same radius of convergence $R$?
real-analysis sequences-and-series convergence
asked Dec 18 '18 at 3:45
The BoscoThe Bosco
541212
$endgroup$
Why is it that $sum na_n x^{n-1}$ and $sum na_n x^n$ have the same radius of convergence $R$?
How can I show that
$$lim sup|na_n|^{1/n} = lim sup |na_n x|^{1/n}$$ since this would mean that both series have the same radius of convergence $R$?
real-analysis sequences-and-series convergence
real-analysis sequences-and-series convergence
asked Dec 18 '18 at 3:45
The BoscoThe Bosco
541212
asked Dec 18 '18 at 3:45
The BoscoThe Bosco
541212
asked Dec 18 '18 at 3:45
The BoscoThe Bosco
541212
asked Dec 18 '18 at 3:45
The BoscoThe Bosco
541212
asked Dec 18 '18 at 3:45
The BoscoThe Bosco
541212
541212
2
$begingroup$
You can directly note they have the same radius of converge because one is just x times the other
$endgroup$
– mathworker21
Dec 18 '18 at 3:54
$begingroup$
That is exactly what I want to prove
$endgroup$
– The Bosco
Dec 18 '18 at 11:51
$begingroup$
can you be more specific?
$endgroup$
– mathworker21
Dec 18 '18 at 11:55
add a comment |
2
$begingroup$
You can directly note they have the same radius of converge because one is just x times the other
$endgroup$
– mathworker21
Dec 18 '18 at 3:54
$begingroup$
That is exactly what I want to prove
$endgroup$
– The Bosco
Dec 18 '18 at 11:51
$begingroup$
can you be more specific?
$endgroup$
– mathworker21
Dec 18 '18 at 11:55
2
2
$begingroup$
You can directly note they have the same radius of converge because one is just x times the other
$endgroup$
– mathworker21
Dec 18 '18 at 3:54
$begingroup$
You can directly note they have the same radius of converge because one is just x times the other
$endgroup$
– mathworker21
Dec 18 '18 at 3:54
$begingroup$
That is exactly what I want to prove
$endgroup$
– The Bosco
Dec 18 '18 at 11:51
$begingroup$
That is exactly what I want to prove
$endgroup$
– The Bosco
Dec 18 '18 at 11:51
$begingroup$
can you be more specific?
$endgroup$
– mathworker21
Dec 18 '18 at 11:55
$begingroup$
can you be more specific?
$endgroup$
– mathworker21
Dec 18 '18 at 11:55
add a comment |
1 Answer
1
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oldest
votes
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Instead of using any formula for radius of convergence simply use the fact that $sum na_nx^{n-1}$ converges iff $sum na_nx^n$ does, for any $x neq 0$. Obviously, this implies that they have the same radius of convergence.
answered Dec 18 '18 at 5:59
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
$endgroup$
$begingroup$
But proving that part that one converges iff the other one does is what I don't get.
$endgroup$
– The Bosco
Dec 18 '18 at 11:09
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If $sum a_n $ converges then $sum c a_n$ also converges for any constant $c$. Apply this with $c=x$ in one direction and $c =frac 1 x$ in the other direction.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 11:43
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
active
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votes
$begingroup$
Instead of using any formula for radius of convergence simply use the fact that $sum na_nx^{n-1}$ converges iff $sum na_nx^n$ does, for any $x neq 0$. Obviously, this implies that they have the same radius of convergence.
answered Dec 18 '18 at 5:59
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
$endgroup$
$begingroup$
But proving that part that one converges iff the other one does is what I don't get.
$endgroup$
– The Bosco
Dec 18 '18 at 11:09
$begingroup$
If $sum a_n $ converges then $sum c a_n$ also converges for any constant $c$. Apply this with $c=x$ in one direction and $c =frac 1 x$ in the other direction.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 11:43
add a comment |
$begingroup$
Instead of using any formula for radius of convergence simply use the fact that $sum na_nx^{n-1}$ converges iff $sum na_nx^n$ does, for any $x neq 0$. Obviously, this implies that they have the same radius of convergence.
answered Dec 18 '18 at 5:59
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
$endgroup$
$begingroup$
But proving that part that one converges iff the other one does is what I don't get.
$endgroup$
– The Bosco
Dec 18 '18 at 11:09
$begingroup$
If $sum a_n $ converges then $sum c a_n$ also converges for any constant $c$. Apply this with $c=x$ in one direction and $c =frac 1 x$ in the other direction.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 11:43
add a comment |
$begingroup$
Instead of using any formula for radius of convergence simply use the fact that $sum na_nx^{n-1}$ converges iff $sum na_nx^n$ does, for any $x neq 0$. Obviously, this implies that they have the same radius of convergence.
answered Dec 18 '18 at 5:59
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
$endgroup$
Instead of using any formula for radius of convergence simply use the fact that $sum na_nx^{n-1}$ converges iff $sum na_nx^n$ does, for any $x neq 0$. Obviously, this implies that they have the same radius of convergence.
answered Dec 18 '18 at 5:59
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
answered Dec 18 '18 at 5:59
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
answered Dec 18 '18 at 5:59
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
answered Dec 18 '18 at 5:59
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
55.4k42057
$begingroup$
But proving that part that one converges iff the other one does is what I don't get.
$endgroup$
– The Bosco
Dec 18 '18 at 11:09
$begingroup$
If $sum a_n $ converges then $sum c a_n$ also converges for any constant $c$. Apply this with $c=x$ in one direction and $c =frac 1 x$ in the other direction.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 11:43
add a comment |
$begingroup$
But proving that part that one converges iff the other one does is what I don't get.
$endgroup$
– The Bosco
Dec 18 '18 at 11:09
$begingroup$
If $sum a_n $ converges then $sum c a_n$ also converges for any constant $c$. Apply this with $c=x$ in one direction and $c =frac 1 x$ in the other direction.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 11:43
$begingroup$
But proving that part that one converges iff the other one does is what I don't get.
$endgroup$
– The Bosco
Dec 18 '18 at 11:09
$begingroup$
But proving that part that one converges iff the other one does is what I don't get.
$endgroup$
– The Bosco
Dec 18 '18 at 11:09
$begingroup$
If $sum a_n $ converges then $sum c a_n$ also converges for any constant $c$. Apply this with $c=x$ in one direction and $c =frac 1 x$ in the other direction.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 11:43
$begingroup$
If $sum a_n $ converges then $sum c a_n$ also converges for any constant $c$. Apply this with $c=x$ in one direction and $c =frac 1 x$ in the other direction.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 11:43
add a comment |
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2
$begingroup$
You can directly note they have the same radius of converge because one is just x times the other
$endgroup$
– mathworker21
Dec 18 '18 at 3:54
$begingroup$
That is exactly what I want to prove
$endgroup$
– The Bosco
Dec 18 '18 at 11:51
$begingroup$
can you be more specific?
$endgroup$
– mathworker21
Dec 18 '18 at 11:55