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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    $begingroup$
    You are correct; it's likely a typo.
    $endgroup$
    – T. Bongers
    Dec 7 '18 at 1:16






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    $begingroup$
    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
















2












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














2












2








2





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













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edited Dec 7 '18 at 1:16









Eevee Trainer

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asked Dec 7 '18 at 1:15









AmarusAmarus

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457819








  • 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




    $begingroup$
    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










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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    $begingroup$

    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.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      0












      $begingroup$

      You are indeed correct.



      For coefficients $a_k:=(1+frac {1}{k})^{k^2}$, the Cauchy-Hadamard formula gives:



      $$R=frac {1}{limsup_{ktoinfty}(|a_k|)^{frac {1}{k}}}=frac {1}{limsup_{ktoinfty}(|(1+frac {1}{k})^{k^2}|)^{frac {1}{k}}}=frac {1}{limsup_{ktoinfty}((1+frac {1}{k})^{k^2})^{frac {1}{k}}}=frac {1}{limsup_{ktoinfty}(1+frac {1}{k})^k}=frac {1}{e}$$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        You are indeed correct.



        For coefficients $a_k:=(1+frac {1}{k})^{k^2}$, the Cauchy-Hadamard formula gives:



        $$R=frac {1}{limsup_{ktoinfty}(|a_k|)^{frac {1}{k}}}=frac {1}{limsup_{ktoinfty}(|(1+frac {1}{k})^{k^2}|)^{frac {1}{k}}}=frac {1}{limsup_{ktoinfty}((1+frac {1}{k})^{k^2})^{frac {1}{k}}}=frac {1}{limsup_{ktoinfty}(1+frac {1}{k})^k}=frac {1}{e}$$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          You are indeed correct.



          For coefficients $a_k:=(1+frac {1}{k})^{k^2}$, the Cauchy-Hadamard formula gives:



          $$R=frac {1}{limsup_{ktoinfty}(|a_k|)^{frac {1}{k}}}=frac {1}{limsup_{ktoinfty}(|(1+frac {1}{k})^{k^2}|)^{frac {1}{k}}}=frac {1}{limsup_{ktoinfty}((1+frac {1}{k})^{k^2})^{frac {1}{k}}}=frac {1}{limsup_{ktoinfty}(1+frac {1}{k})^k}=frac {1}{e}$$






          share|cite|improve this answer









          $endgroup$



          You are indeed correct.



          For coefficients $a_k:=(1+frac {1}{k})^{k^2}$, the Cauchy-Hadamard formula gives:



          $$R=frac {1}{limsup_{ktoinfty}(|a_k|)^{frac {1}{k}}}=frac {1}{limsup_{ktoinfty}(|(1+frac {1}{k})^{k^2}|)^{frac {1}{k}}}=frac {1}{limsup_{ktoinfty}((1+frac {1}{k})^{k^2})^{frac {1}{k}}}=frac {1}{limsup_{ktoinfty}(1+frac {1}{k})^k}=frac {1}{e}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 7 '18 at 1:30









          Mark PineauMark Pineau

          1,500414




          1,500414























              0












              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  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.






                  share|cite|improve this answer









                  $endgroup$



                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 7 '18 at 5:09









                  Claude LeiboviciClaude Leibovici

                  120k1157132




                  120k1157132






























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