Confusion about radius of convergence of a power series
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I'm a bit confused about the following statement from a script. We look at the power series with coefficient $ a_k = (1+1/k)^{k^2}$ and want to compute the radius of convergence $R$ of the series. The solution given is $R=1$, applying the $k^2$ root to $a_k$. I don't get this: The Cauchy-Hadamard formula involves the $k$th root which would give $R=1/e$ in my opinion. Is this a mistake in the script or is it me who is mistaken?
real-analysis complex-analysis analysis power-series
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add a comment |
$begingroup$
I'm a bit confused about the following statement from a script. We look at the power series with coefficient $ a_k = (1+1/k)^{k^2}$ and want to compute the radius of convergence $R$ of the series. The solution given is $R=1$, applying the $k^2$ root to $a_k$. I don't get this: The Cauchy-Hadamard formula involves the $k$th root which would give $R=1/e$ in my opinion. Is this a mistake in the script or is it me who is mistaken?
real-analysis complex-analysis analysis power-series
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1
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You are correct; it's likely a typo.
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– T. Bongers
Dec 7 '18 at 1:16
1
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You do not say what power series actually is. If it is $$sum (1+1/k)^{k^2};x^{k^2}$$ then you should use the $k^2$ root.
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– GEdgar
Dec 7 '18 at 12:16
add a comment |
$begingroup$
I'm a bit confused about the following statement from a script. We look at the power series with coefficient $ a_k = (1+1/k)^{k^2}$ and want to compute the radius of convergence $R$ of the series. The solution given is $R=1$, applying the $k^2$ root to $a_k$. I don't get this: The Cauchy-Hadamard formula involves the $k$th root which would give $R=1/e$ in my opinion. Is this a mistake in the script or is it me who is mistaken?
real-analysis complex-analysis analysis power-series
$endgroup$
I'm a bit confused about the following statement from a script. We look at the power series with coefficient $ a_k = (1+1/k)^{k^2}$ and want to compute the radius of convergence $R$ of the series. The solution given is $R=1$, applying the $k^2$ root to $a_k$. I don't get this: The Cauchy-Hadamard formula involves the $k$th root which would give $R=1/e$ in my opinion. Is this a mistake in the script or is it me who is mistaken?
real-analysis complex-analysis analysis power-series
real-analysis complex-analysis analysis power-series
edited Dec 7 '18 at 1:16
Eevee Trainer
5,7961936
5,7961936
asked Dec 7 '18 at 1:15
AmarusAmarus
457819
457819
1
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You are correct; it's likely a typo.
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– T. Bongers
Dec 7 '18 at 1:16
1
$begingroup$
You do not say what power series actually is. If it is $$sum (1+1/k)^{k^2};x^{k^2}$$ then you should use the $k^2$ root.
$endgroup$
– GEdgar
Dec 7 '18 at 12:16
add a comment |
1
$begingroup$
You are correct; it's likely a typo.
$endgroup$
– T. Bongers
Dec 7 '18 at 1:16
1
$begingroup$
You do not say what power series actually is. If it is $$sum (1+1/k)^{k^2};x^{k^2}$$ then you should use the $k^2$ root.
$endgroup$
– GEdgar
Dec 7 '18 at 12:16
1
1
$begingroup$
You are correct; it's likely a typo.
$endgroup$
– T. Bongers
Dec 7 '18 at 1:16
$begingroup$
You are correct; it's likely a typo.
$endgroup$
– T. Bongers
Dec 7 '18 at 1:16
1
1
$begingroup$
You do not say what power series actually is. If it is $$sum (1+1/k)^{k^2};x^{k^2}$$ then you should use the $k^2$ root.
$endgroup$
– GEdgar
Dec 7 '18 at 12:16
$begingroup$
You do not say what power series actually is. If it is $$sum (1+1/k)^{k^2};x^{k^2}$$ then you should use the $k^2$ root.
$endgroup$
– GEdgar
Dec 7 '18 at 12:16
add a comment |
2 Answers
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You are indeed correct.
For coefficients $a_k:=(1+frac {1}{k})^{k^2}$, the Cauchy-Hadamard formula gives:
$$R=frac {1}{limsup_{ktoinfty}(|a_k|)^{frac {1}{k}}}=frac {1}{limsup_{ktoinfty}(|(1+frac {1}{k})^{k^2}|)^{frac {1}{k}}}=frac {1}{limsup_{ktoinfty}((1+frac {1}{k})^{k^2})^{frac {1}{k}}}=frac {1}{limsup_{ktoinfty}(1+frac {1}{k})^k}=frac {1}{e}$$
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Just another way to do it.
$$a_k=(1+frac {1}{k})^{k^2}implies log(a_k)=k^2log left(1+frac{1}{k}right)$$
$$log(a_k)-log(a_{k+1})=k^2log left(1+frac{1}{k}right)-(k+1)^2log left(1+frac{1}{k+1}right)$$ Now, use Taylor expansions to get
$$log(a_k)-log(a_{k+1})=-1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)$$
Continue with Taylor
$$frac{a_k} {a_{k+1}}=frac{1}{e}+frac{1}{3 e k^2}+Oleft(frac{1}{k^3}right)$$ making $R=frac{1}{e}$ as you found.
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2 Answers
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active
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2 Answers
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active
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$begingroup$
You are indeed correct.
For coefficients $a_k:=(1+frac {1}{k})^{k^2}$, the Cauchy-Hadamard formula gives:
$$R=frac {1}{limsup_{ktoinfty}(|a_k|)^{frac {1}{k}}}=frac {1}{limsup_{ktoinfty}(|(1+frac {1}{k})^{k^2}|)^{frac {1}{k}}}=frac {1}{limsup_{ktoinfty}((1+frac {1}{k})^{k^2})^{frac {1}{k}}}=frac {1}{limsup_{ktoinfty}(1+frac {1}{k})^k}=frac {1}{e}$$
$endgroup$
add a comment |
$begingroup$
You are indeed correct.
For coefficients $a_k:=(1+frac {1}{k})^{k^2}$, the Cauchy-Hadamard formula gives:
$$R=frac {1}{limsup_{ktoinfty}(|a_k|)^{frac {1}{k}}}=frac {1}{limsup_{ktoinfty}(|(1+frac {1}{k})^{k^2}|)^{frac {1}{k}}}=frac {1}{limsup_{ktoinfty}((1+frac {1}{k})^{k^2})^{frac {1}{k}}}=frac {1}{limsup_{ktoinfty}(1+frac {1}{k})^k}=frac {1}{e}$$
$endgroup$
add a comment |
$begingroup$
You are indeed correct.
For coefficients $a_k:=(1+frac {1}{k})^{k^2}$, the Cauchy-Hadamard formula gives:
$$R=frac {1}{limsup_{ktoinfty}(|a_k|)^{frac {1}{k}}}=frac {1}{limsup_{ktoinfty}(|(1+frac {1}{k})^{k^2}|)^{frac {1}{k}}}=frac {1}{limsup_{ktoinfty}((1+frac {1}{k})^{k^2})^{frac {1}{k}}}=frac {1}{limsup_{ktoinfty}(1+frac {1}{k})^k}=frac {1}{e}$$
$endgroup$
You are indeed correct.
For coefficients $a_k:=(1+frac {1}{k})^{k^2}$, the Cauchy-Hadamard formula gives:
$$R=frac {1}{limsup_{ktoinfty}(|a_k|)^{frac {1}{k}}}=frac {1}{limsup_{ktoinfty}(|(1+frac {1}{k})^{k^2}|)^{frac {1}{k}}}=frac {1}{limsup_{ktoinfty}((1+frac {1}{k})^{k^2})^{frac {1}{k}}}=frac {1}{limsup_{ktoinfty}(1+frac {1}{k})^k}=frac {1}{e}$$
answered Dec 7 '18 at 1:30
Mark PineauMark Pineau
1,500414
1,500414
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add a comment |
$begingroup$
Just another way to do it.
$$a_k=(1+frac {1}{k})^{k^2}implies log(a_k)=k^2log left(1+frac{1}{k}right)$$
$$log(a_k)-log(a_{k+1})=k^2log left(1+frac{1}{k}right)-(k+1)^2log left(1+frac{1}{k+1}right)$$ Now, use Taylor expansions to get
$$log(a_k)-log(a_{k+1})=-1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)$$
Continue with Taylor
$$frac{a_k} {a_{k+1}}=frac{1}{e}+frac{1}{3 e k^2}+Oleft(frac{1}{k^3}right)$$ making $R=frac{1}{e}$ as you found.
$endgroup$
add a comment |
$begingroup$
Just another way to do it.
$$a_k=(1+frac {1}{k})^{k^2}implies log(a_k)=k^2log left(1+frac{1}{k}right)$$
$$log(a_k)-log(a_{k+1})=k^2log left(1+frac{1}{k}right)-(k+1)^2log left(1+frac{1}{k+1}right)$$ Now, use Taylor expansions to get
$$log(a_k)-log(a_{k+1})=-1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)$$
Continue with Taylor
$$frac{a_k} {a_{k+1}}=frac{1}{e}+frac{1}{3 e k^2}+Oleft(frac{1}{k^3}right)$$ making $R=frac{1}{e}$ as you found.
$endgroup$
add a comment |
$begingroup$
Just another way to do it.
$$a_k=(1+frac {1}{k})^{k^2}implies log(a_k)=k^2log left(1+frac{1}{k}right)$$
$$log(a_k)-log(a_{k+1})=k^2log left(1+frac{1}{k}right)-(k+1)^2log left(1+frac{1}{k+1}right)$$ Now, use Taylor expansions to get
$$log(a_k)-log(a_{k+1})=-1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)$$
Continue with Taylor
$$frac{a_k} {a_{k+1}}=frac{1}{e}+frac{1}{3 e k^2}+Oleft(frac{1}{k^3}right)$$ making $R=frac{1}{e}$ as you found.
$endgroup$
Just another way to do it.
$$a_k=(1+frac {1}{k})^{k^2}implies log(a_k)=k^2log left(1+frac{1}{k}right)$$
$$log(a_k)-log(a_{k+1})=k^2log left(1+frac{1}{k}right)-(k+1)^2log left(1+frac{1}{k+1}right)$$ Now, use Taylor expansions to get
$$log(a_k)-log(a_{k+1})=-1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)$$
Continue with Taylor
$$frac{a_k} {a_{k+1}}=frac{1}{e}+frac{1}{3 e k^2}+Oleft(frac{1}{k^3}right)$$ making $R=frac{1}{e}$ as you found.
answered Dec 7 '18 at 5:09
Claude LeiboviciClaude Leibovici
120k1157132
120k1157132
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$begingroup$
You are correct; it's likely a typo.
$endgroup$
– T. Bongers
Dec 7 '18 at 1:16
1
$begingroup$
You do not say what power series actually is. If it is $$sum (1+1/k)^{k^2};x^{k^2}$$ then you should use the $k^2$ root.
$endgroup$
– GEdgar
Dec 7 '18 at 12:16