Find the radius of convergence of $sum_{n=0}^{infty} n!x^{n^{2}}$ [duplicate]












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  • Radius of convergence of power series $sum_0^{infty} n!x^{n^2}$

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I have to find find the radius of convergence of the series $$sum_{n=0}^{infty} n!x^{n^{2}}$$.



Here $$lim_{ntoinfty} frac{(n+1)!}{n!} = infty$$. Then radius of convergence is 0. Where I'm doing wrong?
Please help






real-analysis sequences-and-series analysis power-series







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marked as duplicate by Key Flex, José Carlos Santos complex-analysis
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 Jan 11 at 7:08


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    $begingroup$
    Wrong formula. What values do $a_n$ take?
    $endgroup$
    – xbh
    Jan 11 at 5:25
















2












$begingroup$



This question already has an answer here:




  • Radius of convergence of power series $sum_0^{infty} n!x^{n^2}$

    3 answers




I have to find find the radius of convergence of the series $$sum_{n=0}^{infty} n!x^{n^{2}}$$.



Here $$lim_{ntoinfty} frac{(n+1)!}{n!} = infty$$. Then radius of convergence is 0. Where I'm doing wrong?
Please help






real-analysis sequences-and-series analysis power-series







share|cite|improve this question











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marked as duplicate by Key Flex, José Carlos Santos complex-analysis
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 Jan 11 at 7:08


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 1




    $begingroup$
    Wrong formula. What values do $a_n$ take?
    $endgroup$
    – xbh
    Jan 11 at 5:25














2












2








2


1



$begingroup$



This question already has an answer here:




  • Radius of convergence of power series $sum_0^{infty} n!x^{n^2}$

    3 answers




I have to find find the radius of convergence of the series $$sum_{n=0}^{infty} n!x^{n^{2}}$$.



Here $$lim_{ntoinfty} frac{(n+1)!}{n!} = infty$$. Then radius of convergence is 0. Where I'm doing wrong?
Please help






real-analysis sequences-and-series analysis power-series







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Radius of convergence of power series $sum_0^{infty} n!x^{n^2}$

    3 answers




I have to find find the radius of convergence of the series $$sum_{n=0}^{infty} n!x^{n^{2}}$$.



Here $$lim_{ntoinfty} frac{(n+1)!}{n!} = infty$$. Then radius of convergence is 0. Where I'm doing wrong?
Please help





This question already has an answer here:




  • Radius of convergence of power series $sum_0^{infty} n!x^{n^2}$

    3 answers







real-analysis sequences-and-series analysis power-series




real-analysis sequences-and-series analysis power-series






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share|cite|improve this question













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edited Jan 11 at 7:51









Omojola Micheal

1,831324




1,831324










asked Jan 11 at 5:18









user633319user633319

384




384




marked as duplicate by Key Flex, José Carlos Santos complex-analysis
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 Jan 11 at 7:08


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 Jan 11 at 7:08


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    $begingroup$
    Wrong formula. What values do $a_n$ take?
    $endgroup$
    – xbh
    Jan 11 at 5:25














  • 1




    $begingroup$
    Wrong formula. What values do $a_n$ take?
    $endgroup$
    – xbh
    Jan 11 at 5:25








1




1




$begingroup$
Wrong formula. What values do $a_n$ take?
$endgroup$
– xbh
Jan 11 at 5:25




$begingroup$
Wrong formula. What values do $a_n$ take?
$endgroup$
– xbh
Jan 11 at 5:25










2 Answers
2






active

oldest

votes


















1












$begingroup$

You have found the radius of convergence of $sum n! x^{n}$ not that of $sum n! x^{n^{2}}$. Here $a_n=0$ of $n$ is not a square and $a_n=k!$ if $n=k^{2}$. To show that the series $sum n! |x^{n^{2}}|$converges only for $|x|<1$ use Stirling's approximation. You have to observe that $e^{-n} e^{n^{2} ln, x} e^{(n+frac 1 2) ln, n} to infty$ for $|x|geq 1$ whereas the series is dominated by a series of the type $sum Ce^{-delta |x|}$ for $|x| <1$. Hence the radius of convergence is $1$.






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$endgroup$





















    -1












    $begingroup$

    By the Cauchy-Hadamard theorem, $r=frac1{limsup_{ntoinfty}sqrt[n^2]{n!}}=1$.






    share|cite|improve this answer











    $endgroup$




















      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      You have found the radius of convergence of $sum n! x^{n}$ not that of $sum n! x^{n^{2}}$. Here $a_n=0$ of $n$ is not a square and $a_n=k!$ if $n=k^{2}$. To show that the series $sum n! |x^{n^{2}}|$converges only for $|x|<1$ use Stirling's approximation. You have to observe that $e^{-n} e^{n^{2} ln, x} e^{(n+frac 1 2) ln, n} to infty$ for $|x|geq 1$ whereas the series is dominated by a series of the type $sum Ce^{-delta |x|}$ for $|x| <1$. Hence the radius of convergence is $1$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        You have found the radius of convergence of $sum n! x^{n}$ not that of $sum n! x^{n^{2}}$. Here $a_n=0$ of $n$ is not a square and $a_n=k!$ if $n=k^{2}$. To show that the series $sum n! |x^{n^{2}}|$converges only for $|x|<1$ use Stirling's approximation. You have to observe that $e^{-n} e^{n^{2} ln, x} e^{(n+frac 1 2) ln, n} to infty$ for $|x|geq 1$ whereas the series is dominated by a series of the type $sum Ce^{-delta |x|}$ for $|x| <1$. Hence the radius of convergence is $1$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          You have found the radius of convergence of $sum n! x^{n}$ not that of $sum n! x^{n^{2}}$. Here $a_n=0$ of $n$ is not a square and $a_n=k!$ if $n=k^{2}$. To show that the series $sum n! |x^{n^{2}}|$converges only for $|x|<1$ use Stirling's approximation. You have to observe that $e^{-n} e^{n^{2} ln, x} e^{(n+frac 1 2) ln, n} to infty$ for $|x|geq 1$ whereas the series is dominated by a series of the type $sum Ce^{-delta |x|}$ for $|x| <1$. Hence the radius of convergence is $1$.






          share|cite|improve this answer









          $endgroup$



          You have found the radius of convergence of $sum n! x^{n}$ not that of $sum n! x^{n^{2}}$. Here $a_n=0$ of $n$ is not a square and $a_n=k!$ if $n=k^{2}$. To show that the series $sum n! |x^{n^{2}}|$converges only for $|x|<1$ use Stirling's approximation. You have to observe that $e^{-n} e^{n^{2} ln, x} e^{(n+frac 1 2) ln, n} to infty$ for $|x|geq 1$ whereas the series is dominated by a series of the type $sum Ce^{-delta |x|}$ for $|x| <1$. Hence the radius of convergence is $1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 11 at 5:46









          Kavi Rama MurthyKavi Rama Murthy

          55.4k42057




          55.4k42057























              -1












              $begingroup$

              By the Cauchy-Hadamard theorem, $r=frac1{limsup_{ntoinfty}sqrt[n^2]{n!}}=1$.






              share|cite|improve this answer











              $endgroup$


















                -1












                $begingroup$

                By the Cauchy-Hadamard theorem, $r=frac1{limsup_{ntoinfty}sqrt[n^2]{n!}}=1$.






                share|cite|improve this answer











                $endgroup$
















                  -1












                  -1








                  -1





                  $begingroup$

                  By the Cauchy-Hadamard theorem, $r=frac1{limsup_{ntoinfty}sqrt[n^2]{n!}}=1$.






                  share|cite|improve this answer











                  $endgroup$



                  By the Cauchy-Hadamard theorem, $r=frac1{limsup_{ntoinfty}sqrt[n^2]{n!}}=1$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 11 at 16:18

























                  answered Jan 11 at 6:18









                  Chris CusterChris Custer

                  11.6k3824




                  11.6k3824















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