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












2












$begingroup$


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



I applied ratio test, it went inconclusive and in root test i got $|x|^n n!^{1/n}$ but $n!^{1/n} $ diverges as n gets bigger.



So please help me to proceed






real-analysis sequences-and-series power-series







share|cite|improve this question









$endgroup$












  • $begingroup$
    I realize there is already an accepted answer, but just a a note: the ratio test is not inconclusive. Here we have $$a_n = n! x^{n^2}$$ and so $$lvert a_{n+1}/a_n rvert = (n+1) lvert x rvert ^{2n+1} = (n+1) e^{(2n+1)log(lvert x rvert)}.$$ Now if $lvert x rvert < 1$, so that $log(lvert x vert) < 0$, then this is a decaying exponential and the limit is zero as $n to infty$. If $ lvert x rvert ge 1$, then this limit is $+infty$. Hence the ratio test tells us that we have convergence iff $lvert x rvert < 1$.
    $endgroup$
    – User8128
    Jan 11 at 16:29










  • $begingroup$
    @User8128 thanks
    $endgroup$
    – Cloud JR
    Jan 11 at 17:15
















2












$begingroup$


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



I applied ratio test, it went inconclusive and in root test i got $|x|^n n!^{1/n}$ but $n!^{1/n} $ diverges as n gets bigger.



So please help me to proceed






real-analysis sequences-and-series power-series







share|cite|improve this question









$endgroup$












  • $begingroup$
    I realize there is already an accepted answer, but just a a note: the ratio test is not inconclusive. Here we have $$a_n = n! x^{n^2}$$ and so $$lvert a_{n+1}/a_n rvert = (n+1) lvert x rvert ^{2n+1} = (n+1) e^{(2n+1)log(lvert x rvert)}.$$ Now if $lvert x rvert < 1$, so that $log(lvert x vert) < 0$, then this is a decaying exponential and the limit is zero as $n to infty$. If $ lvert x rvert ge 1$, then this limit is $+infty$. Hence the ratio test tells us that we have convergence iff $lvert x rvert < 1$.
    $endgroup$
    – User8128
    Jan 11 at 16:29










  • $begingroup$
    @User8128 thanks
    $endgroup$
    – Cloud JR
    Jan 11 at 17:15














2












2








2





$begingroup$


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



I applied ratio test, it went inconclusive and in root test i got $|x|^n n!^{1/n}$ but $n!^{1/n} $ diverges as n gets bigger.



So please help me to proceed






real-analysis sequences-and-series power-series







share|cite|improve this question









$endgroup$




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



I applied ratio test, it went inconclusive and in root test i got $|x|^n n!^{1/n}$ but $n!^{1/n} $ diverges as n gets bigger.



So please help me to proceed






real-analysis sequences-and-series power-series




real-analysis sequences-and-series power-series






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 11 at 3:56









Cloud JRCloud JR

920518




920518












  • $begingroup$
    I realize there is already an accepted answer, but just a a note: the ratio test is not inconclusive. Here we have $$a_n = n! x^{n^2}$$ and so $$lvert a_{n+1}/a_n rvert = (n+1) lvert x rvert ^{2n+1} = (n+1) e^{(2n+1)log(lvert x rvert)}.$$ Now if $lvert x rvert < 1$, so that $log(lvert x vert) < 0$, then this is a decaying exponential and the limit is zero as $n to infty$. If $ lvert x rvert ge 1$, then this limit is $+infty$. Hence the ratio test tells us that we have convergence iff $lvert x rvert < 1$.
    $endgroup$
    – User8128
    Jan 11 at 16:29










  • $begingroup$
    @User8128 thanks
    $endgroup$
    – Cloud JR
    Jan 11 at 17:15


















  • $begingroup$
    I realize there is already an accepted answer, but just a a note: the ratio test is not inconclusive. Here we have $$a_n = n! x^{n^2}$$ and so $$lvert a_{n+1}/a_n rvert = (n+1) lvert x rvert ^{2n+1} = (n+1) e^{(2n+1)log(lvert x rvert)}.$$ Now if $lvert x rvert < 1$, so that $log(lvert x vert) < 0$, then this is a decaying exponential and the limit is zero as $n to infty$. If $ lvert x rvert ge 1$, then this limit is $+infty$. Hence the ratio test tells us that we have convergence iff $lvert x rvert < 1$.
    $endgroup$
    – User8128
    Jan 11 at 16:29










  • $begingroup$
    @User8128 thanks
    $endgroup$
    – Cloud JR
    Jan 11 at 17:15
















$begingroup$
I realize there is already an accepted answer, but just a a note: the ratio test is not inconclusive. Here we have $$a_n = n! x^{n^2}$$ and so $$lvert a_{n+1}/a_n rvert = (n+1) lvert x rvert ^{2n+1} = (n+1) e^{(2n+1)log(lvert x rvert)}.$$ Now if $lvert x rvert < 1$, so that $log(lvert x vert) < 0$, then this is a decaying exponential and the limit is zero as $n to infty$. If $ lvert x rvert ge 1$, then this limit is $+infty$. Hence the ratio test tells us that we have convergence iff $lvert x rvert < 1$.
$endgroup$
– User8128
Jan 11 at 16:29




$begingroup$
I realize there is already an accepted answer, but just a a note: the ratio test is not inconclusive. Here we have $$a_n = n! x^{n^2}$$ and so $$lvert a_{n+1}/a_n rvert = (n+1) lvert x rvert ^{2n+1} = (n+1) e^{(2n+1)log(lvert x rvert)}.$$ Now if $lvert x rvert < 1$, so that $log(lvert x vert) < 0$, then this is a decaying exponential and the limit is zero as $n to infty$. If $ lvert x rvert ge 1$, then this limit is $+infty$. Hence the ratio test tells us that we have convergence iff $lvert x rvert < 1$.
$endgroup$
– User8128
Jan 11 at 16:29












$begingroup$
@User8128 thanks
$endgroup$
– Cloud JR
Jan 11 at 17:15




$begingroup$
@User8128 thanks
$endgroup$
– Cloud JR
Jan 11 at 17:15










3 Answers
3






active

oldest

votes


















1












$begingroup$

HINT:



Note that we have



$$ limsup_{ntoinfty}sqrt[n]{left| n!x^{n^2}right|}=limsup_{ntoinfty}(n!)^{1/n}|x|^n$$



Now use Stirling' s Formula



$$n!sim sqrt{2pi n}left( frac ne right)^n$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    does error in stirling approximation tends to 0 as n gets bigger??
    $endgroup$
    – Cloud JR
    Jan 11 at 13:59










  • $begingroup$
    Would the downvoter care to comment?
    $endgroup$
    – Mark Viola
    Jan 11 at 15:56










  • $begingroup$
    No, but $$n!=left(sqrt{2pi n}left( frac ne right)^nright),left(1+O(1/n)right)$$
    $endgroup$
    – Mark Viola
    Jan 11 at 16:00










  • $begingroup$
    idk who downvote but i accept yours !
    $endgroup$
    – Cloud JR
    Jan 11 at 16:22










  • $begingroup$
    Much appreciated!!
    $endgroup$
    – Mark Viola
    Jan 11 at 18:05



















4












$begingroup$

The power series is $sum_0^infty a_m x^m$ where
$$
a_m = begin{cases}
n!, & m = n^2, n in mathbb N^*\
0, & text{otherwise},
end{cases}
$$
so by Hadamard formula,
$$
1/R =varlimsup |a_m|^{1/m} = varlimsup (n!)^{1/n^2} =varlimsup (sqrt{2pi} n^{n+1/2} mathrm e^{-n})^{1/n^2} = lim (2pi)^{1/2n^2} n^{1/n} n^{1/2n^2}mathrm e^{-1/n} = 1,
$$
where we use the Stirling approximation. Hence the radius of convergence is $1$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Nothing wrong with using Stirling's formula, but more simply, for $nge 1 $ we have $1le(n!)^{1/n^2}le (n^n)^{1/n^2}=n^{1/n} to 1$ as $ ntoinfty$.
    $endgroup$
    – DanielWainfleet
    Jan 11 at 6:04








  • 1




    $begingroup$
    @DanielWainfleet Thanks for this trick!
    $endgroup$
    – xbh
    Jan 11 at 6:06










  • $begingroup$
    @xbh does error in stirling approximation tends to 0 as n gets bigger?? also is it valid to take$1/n^2$ th root of x... please explain
    $endgroup$
    – Cloud JR
    Jan 11 at 13:58












  • $begingroup$
    en.wikipedia.org/wiki/Cauchy%E2%80%93Hadamard_theorem. its not mentioned in hereto take 1/n^2 root , im confused.
    $endgroup$
    – Cloud JR
    Jan 11 at 14:04








  • 1




    $begingroup$
    @CloudJR I think my answer has said everything you need. The given series is a power series that a lot of terms lost, and $(n!)^{1/n}$ is not correct.
    $endgroup$
    – xbh
    Jan 11 at 14:41



















1












$begingroup$

Your formula is wrong because the exponent of $x$ is $n^2,$ but you can use the Weak form of Stirling’s Formula
begin{align} dfrac{n^n}{e^{n-1}}<n!<dfrac{n^{n+1}}{e^{n-1}} . end{align}
This implies that
begin{align} sum^{infty}_{n=0} n!{x^{n^2}}leq sum^{infty}_{n=0} dfrac{n^{n+1}}{e^{n-1}} {x^{n^2}}. end{align}
So, fix $xin Bbb{R},$ then by D'Alembert's ratio test
begin{align} limlimits_{nto infty}left|dfrac{n^{n+2}{x^{(n+1)^2}}}{e^{n}} dfrac{e^{n-1}} {n^{n+1} x^{n^2}}right|&=limlimits_{nto infty}left|dfrac{n }{e} {x^{(n+1)^2-n^2}}right|\&=limlimits_{nto infty}left|dfrac{n }{e} {x^{2n+1}}right|\&=dfrac{left|x right|}{e} limlimits_{nto infty}n, {left|xright|}^{2n}. end{align}



CASE I: $|x|<1$



Since $|x|<1$, then ${left|xright|}^{2n}to 0$, and so, there exists $Nin Bbb{N}$ such that
begin{align} {left|xright|}^{2n}<dfrac{1}{n^2},forall;ngeq N .end{align}
Thus,
begin{align} limlimits_{nto infty}left|dfrac{n^{n+2}{x^{(n+1)^2}}}{e^{n}} dfrac{e^{n-1}} {n^{n+1} x^{n^2}}right|&=dfrac{left|x right|}{e} limlimits_{nto infty}n, {left|xright|}^{2n}leq dfrac{left|x right|}{e} limlimits_{nto infty}n, left(dfrac{1}{n^2}right)=0<1. end{align}
Hence, the series converges when $|x|<1.$



CASE II: $|x|geq 1$
begin{align} limlimits_{nto infty}n!{x^{n^2}}=infty end{align}
which implies that the series cannot not converge. Consequently, the radius of convergence is $1$.






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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    HINT:



    Note that we have



    $$ limsup_{ntoinfty}sqrt[n]{left| n!x^{n^2}right|}=limsup_{ntoinfty}(n!)^{1/n}|x|^n$$



    Now use Stirling' s Formula



    $$n!sim sqrt{2pi n}left( frac ne right)^n$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      does error in stirling approximation tends to 0 as n gets bigger??
      $endgroup$
      – Cloud JR
      Jan 11 at 13:59










    • $begingroup$
      Would the downvoter care to comment?
      $endgroup$
      – Mark Viola
      Jan 11 at 15:56










    • $begingroup$
      No, but $$n!=left(sqrt{2pi n}left( frac ne right)^nright),left(1+O(1/n)right)$$
      $endgroup$
      – Mark Viola
      Jan 11 at 16:00










    • $begingroup$
      idk who downvote but i accept yours !
      $endgroup$
      – Cloud JR
      Jan 11 at 16:22










    • $begingroup$
      Much appreciated!!
      $endgroup$
      – Mark Viola
      Jan 11 at 18:05
















    1












    $begingroup$

    HINT:



    Note that we have



    $$ limsup_{ntoinfty}sqrt[n]{left| n!x^{n^2}right|}=limsup_{ntoinfty}(n!)^{1/n}|x|^n$$



    Now use Stirling' s Formula



    $$n!sim sqrt{2pi n}left( frac ne right)^n$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      does error in stirling approximation tends to 0 as n gets bigger??
      $endgroup$
      – Cloud JR
      Jan 11 at 13:59










    • $begingroup$
      Would the downvoter care to comment?
      $endgroup$
      – Mark Viola
      Jan 11 at 15:56










    • $begingroup$
      No, but $$n!=left(sqrt{2pi n}left( frac ne right)^nright),left(1+O(1/n)right)$$
      $endgroup$
      – Mark Viola
      Jan 11 at 16:00










    • $begingroup$
      idk who downvote but i accept yours !
      $endgroup$
      – Cloud JR
      Jan 11 at 16:22










    • $begingroup$
      Much appreciated!!
      $endgroup$
      – Mark Viola
      Jan 11 at 18:05














    1












    1








    1





    $begingroup$

    HINT:



    Note that we have



    $$ limsup_{ntoinfty}sqrt[n]{left| n!x^{n^2}right|}=limsup_{ntoinfty}(n!)^{1/n}|x|^n$$



    Now use Stirling' s Formula



    $$n!sim sqrt{2pi n}left( frac ne right)^n$$






    share|cite|improve this answer









    $endgroup$



    HINT:



    Note that we have



    $$ limsup_{ntoinfty}sqrt[n]{left| n!x^{n^2}right|}=limsup_{ntoinfty}(n!)^{1/n}|x|^n$$



    Now use Stirling' s Formula



    $$n!sim sqrt{2pi n}left( frac ne right)^n$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 11 at 4:24









    Mark ViolaMark Viola

    134k1278177




    134k1278177












    • $begingroup$
      does error in stirling approximation tends to 0 as n gets bigger??
      $endgroup$
      – Cloud JR
      Jan 11 at 13:59










    • $begingroup$
      Would the downvoter care to comment?
      $endgroup$
      – Mark Viola
      Jan 11 at 15:56










    • $begingroup$
      No, but $$n!=left(sqrt{2pi n}left( frac ne right)^nright),left(1+O(1/n)right)$$
      $endgroup$
      – Mark Viola
      Jan 11 at 16:00










    • $begingroup$
      idk who downvote but i accept yours !
      $endgroup$
      – Cloud JR
      Jan 11 at 16:22










    • $begingroup$
      Much appreciated!!
      $endgroup$
      – Mark Viola
      Jan 11 at 18:05


















    • $begingroup$
      does error in stirling approximation tends to 0 as n gets bigger??
      $endgroup$
      – Cloud JR
      Jan 11 at 13:59










    • $begingroup$
      Would the downvoter care to comment?
      $endgroup$
      – Mark Viola
      Jan 11 at 15:56










    • $begingroup$
      No, but $$n!=left(sqrt{2pi n}left( frac ne right)^nright),left(1+O(1/n)right)$$
      $endgroup$
      – Mark Viola
      Jan 11 at 16:00










    • $begingroup$
      idk who downvote but i accept yours !
      $endgroup$
      – Cloud JR
      Jan 11 at 16:22










    • $begingroup$
      Much appreciated!!
      $endgroup$
      – Mark Viola
      Jan 11 at 18:05
















    $begingroup$
    does error in stirling approximation tends to 0 as n gets bigger??
    $endgroup$
    – Cloud JR
    Jan 11 at 13:59




    $begingroup$
    does error in stirling approximation tends to 0 as n gets bigger??
    $endgroup$
    – Cloud JR
    Jan 11 at 13:59












    $begingroup$
    Would the downvoter care to comment?
    $endgroup$
    – Mark Viola
    Jan 11 at 15:56




    $begingroup$
    Would the downvoter care to comment?
    $endgroup$
    – Mark Viola
    Jan 11 at 15:56












    $begingroup$
    No, but $$n!=left(sqrt{2pi n}left( frac ne right)^nright),left(1+O(1/n)right)$$
    $endgroup$
    – Mark Viola
    Jan 11 at 16:00




    $begingroup$
    No, but $$n!=left(sqrt{2pi n}left( frac ne right)^nright),left(1+O(1/n)right)$$
    $endgroup$
    – Mark Viola
    Jan 11 at 16:00












    $begingroup$
    idk who downvote but i accept yours !
    $endgroup$
    – Cloud JR
    Jan 11 at 16:22




    $begingroup$
    idk who downvote but i accept yours !
    $endgroup$
    – Cloud JR
    Jan 11 at 16:22












    $begingroup$
    Much appreciated!!
    $endgroup$
    – Mark Viola
    Jan 11 at 18:05




    $begingroup$
    Much appreciated!!
    $endgroup$
    – Mark Viola
    Jan 11 at 18:05











    4












    $begingroup$

    The power series is $sum_0^infty a_m x^m$ where
    $$
    a_m = begin{cases}
    n!, & m = n^2, n in mathbb N^*\
    0, & text{otherwise},
    end{cases}
    $$
    so by Hadamard formula,
    $$
    1/R =varlimsup |a_m|^{1/m} = varlimsup (n!)^{1/n^2} =varlimsup (sqrt{2pi} n^{n+1/2} mathrm e^{-n})^{1/n^2} = lim (2pi)^{1/2n^2} n^{1/n} n^{1/2n^2}mathrm e^{-1/n} = 1,
    $$
    where we use the Stirling approximation. Hence the radius of convergence is $1$.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Nothing wrong with using Stirling's formula, but more simply, for $nge 1 $ we have $1le(n!)^{1/n^2}le (n^n)^{1/n^2}=n^{1/n} to 1$ as $ ntoinfty$.
      $endgroup$
      – DanielWainfleet
      Jan 11 at 6:04








    • 1




      $begingroup$
      @DanielWainfleet Thanks for this trick!
      $endgroup$
      – xbh
      Jan 11 at 6:06










    • $begingroup$
      @xbh does error in stirling approximation tends to 0 as n gets bigger?? also is it valid to take$1/n^2$ th root of x... please explain
      $endgroup$
      – Cloud JR
      Jan 11 at 13:58












    • $begingroup$
      en.wikipedia.org/wiki/Cauchy%E2%80%93Hadamard_theorem. its not mentioned in hereto take 1/n^2 root , im confused.
      $endgroup$
      – Cloud JR
      Jan 11 at 14:04








    • 1




      $begingroup$
      @CloudJR I think my answer has said everything you need. The given series is a power series that a lot of terms lost, and $(n!)^{1/n}$ is not correct.
      $endgroup$
      – xbh
      Jan 11 at 14:41
















    4












    $begingroup$

    The power series is $sum_0^infty a_m x^m$ where
    $$
    a_m = begin{cases}
    n!, & m = n^2, n in mathbb N^*\
    0, & text{otherwise},
    end{cases}
    $$
    so by Hadamard formula,
    $$
    1/R =varlimsup |a_m|^{1/m} = varlimsup (n!)^{1/n^2} =varlimsup (sqrt{2pi} n^{n+1/2} mathrm e^{-n})^{1/n^2} = lim (2pi)^{1/2n^2} n^{1/n} n^{1/2n^2}mathrm e^{-1/n} = 1,
    $$
    where we use the Stirling approximation. Hence the radius of convergence is $1$.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Nothing wrong with using Stirling's formula, but more simply, for $nge 1 $ we have $1le(n!)^{1/n^2}le (n^n)^{1/n^2}=n^{1/n} to 1$ as $ ntoinfty$.
      $endgroup$
      – DanielWainfleet
      Jan 11 at 6:04








    • 1




      $begingroup$
      @DanielWainfleet Thanks for this trick!
      $endgroup$
      – xbh
      Jan 11 at 6:06










    • $begingroup$
      @xbh does error in stirling approximation tends to 0 as n gets bigger?? also is it valid to take$1/n^2$ th root of x... please explain
      $endgroup$
      – Cloud JR
      Jan 11 at 13:58












    • $begingroup$
      en.wikipedia.org/wiki/Cauchy%E2%80%93Hadamard_theorem. its not mentioned in hereto take 1/n^2 root , im confused.
      $endgroup$
      – Cloud JR
      Jan 11 at 14:04








    • 1




      $begingroup$
      @CloudJR I think my answer has said everything you need. The given series is a power series that a lot of terms lost, and $(n!)^{1/n}$ is not correct.
      $endgroup$
      – xbh
      Jan 11 at 14:41














    4












    4








    4





    $begingroup$

    The power series is $sum_0^infty a_m x^m$ where
    $$
    a_m = begin{cases}
    n!, & m = n^2, n in mathbb N^*\
    0, & text{otherwise},
    end{cases}
    $$
    so by Hadamard formula,
    $$
    1/R =varlimsup |a_m|^{1/m} = varlimsup (n!)^{1/n^2} =varlimsup (sqrt{2pi} n^{n+1/2} mathrm e^{-n})^{1/n^2} = lim (2pi)^{1/2n^2} n^{1/n} n^{1/2n^2}mathrm e^{-1/n} = 1,
    $$
    where we use the Stirling approximation. Hence the radius of convergence is $1$.






    share|cite|improve this answer









    $endgroup$



    The power series is $sum_0^infty a_m x^m$ where
    $$
    a_m = begin{cases}
    n!, & m = n^2, n in mathbb N^*\
    0, & text{otherwise},
    end{cases}
    $$
    so by Hadamard formula,
    $$
    1/R =varlimsup |a_m|^{1/m} = varlimsup (n!)^{1/n^2} =varlimsup (sqrt{2pi} n^{n+1/2} mathrm e^{-n})^{1/n^2} = lim (2pi)^{1/2n^2} n^{1/n} n^{1/2n^2}mathrm e^{-1/n} = 1,
    $$
    where we use the Stirling approximation. 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 5:49









    xbhxbh

    6,3201522




    6,3201522








    • 1




      $begingroup$
      Nothing wrong with using Stirling's formula, but more simply, for $nge 1 $ we have $1le(n!)^{1/n^2}le (n^n)^{1/n^2}=n^{1/n} to 1$ as $ ntoinfty$.
      $endgroup$
      – DanielWainfleet
      Jan 11 at 6:04








    • 1




      $begingroup$
      @DanielWainfleet Thanks for this trick!
      $endgroup$
      – xbh
      Jan 11 at 6:06










    • $begingroup$
      @xbh does error in stirling approximation tends to 0 as n gets bigger?? also is it valid to take$1/n^2$ th root of x... please explain
      $endgroup$
      – Cloud JR
      Jan 11 at 13:58












    • $begingroup$
      en.wikipedia.org/wiki/Cauchy%E2%80%93Hadamard_theorem. its not mentioned in hereto take 1/n^2 root , im confused.
      $endgroup$
      – Cloud JR
      Jan 11 at 14:04








    • 1




      $begingroup$
      @CloudJR I think my answer has said everything you need. The given series is a power series that a lot of terms lost, and $(n!)^{1/n}$ is not correct.
      $endgroup$
      – xbh
      Jan 11 at 14:41














    • 1




      $begingroup$
      Nothing wrong with using Stirling's formula, but more simply, for $nge 1 $ we have $1le(n!)^{1/n^2}le (n^n)^{1/n^2}=n^{1/n} to 1$ as $ ntoinfty$.
      $endgroup$
      – DanielWainfleet
      Jan 11 at 6:04








    • 1




      $begingroup$
      @DanielWainfleet Thanks for this trick!
      $endgroup$
      – xbh
      Jan 11 at 6:06










    • $begingroup$
      @xbh does error in stirling approximation tends to 0 as n gets bigger?? also is it valid to take$1/n^2$ th root of x... please explain
      $endgroup$
      – Cloud JR
      Jan 11 at 13:58












    • $begingroup$
      en.wikipedia.org/wiki/Cauchy%E2%80%93Hadamard_theorem. its not mentioned in hereto take 1/n^2 root , im confused.
      $endgroup$
      – Cloud JR
      Jan 11 at 14:04








    • 1




      $begingroup$
      @CloudJR I think my answer has said everything you need. The given series is a power series that a lot of terms lost, and $(n!)^{1/n}$ is not correct.
      $endgroup$
      – xbh
      Jan 11 at 14:41








    1




    1




    $begingroup$
    Nothing wrong with using Stirling's formula, but more simply, for $nge 1 $ we have $1le(n!)^{1/n^2}le (n^n)^{1/n^2}=n^{1/n} to 1$ as $ ntoinfty$.
    $endgroup$
    – DanielWainfleet
    Jan 11 at 6:04






    $begingroup$
    Nothing wrong with using Stirling's formula, but more simply, for $nge 1 $ we have $1le(n!)^{1/n^2}le (n^n)^{1/n^2}=n^{1/n} to 1$ as $ ntoinfty$.
    $endgroup$
    – DanielWainfleet
    Jan 11 at 6:04






    1




    1




    $begingroup$
    @DanielWainfleet Thanks for this trick!
    $endgroup$
    – xbh
    Jan 11 at 6:06




    $begingroup$
    @DanielWainfleet Thanks for this trick!
    $endgroup$
    – xbh
    Jan 11 at 6:06












    $begingroup$
    @xbh does error in stirling approximation tends to 0 as n gets bigger?? also is it valid to take$1/n^2$ th root of x... please explain
    $endgroup$
    – Cloud JR
    Jan 11 at 13:58






    $begingroup$
    @xbh does error in stirling approximation tends to 0 as n gets bigger?? also is it valid to take$1/n^2$ th root of x... please explain
    $endgroup$
    – Cloud JR
    Jan 11 at 13:58














    $begingroup$
    en.wikipedia.org/wiki/Cauchy%E2%80%93Hadamard_theorem. its not mentioned in hereto take 1/n^2 root , im confused.
    $endgroup$
    – Cloud JR
    Jan 11 at 14:04






    $begingroup$
    en.wikipedia.org/wiki/Cauchy%E2%80%93Hadamard_theorem. its not mentioned in hereto take 1/n^2 root , im confused.
    $endgroup$
    – Cloud JR
    Jan 11 at 14:04






    1




    1




    $begingroup$
    @CloudJR I think my answer has said everything you need. The given series is a power series that a lot of terms lost, and $(n!)^{1/n}$ is not correct.
    $endgroup$
    – xbh
    Jan 11 at 14:41




    $begingroup$
    @CloudJR I think my answer has said everything you need. The given series is a power series that a lot of terms lost, and $(n!)^{1/n}$ is not correct.
    $endgroup$
    – xbh
    Jan 11 at 14:41











    1












    $begingroup$

    Your formula is wrong because the exponent of $x$ is $n^2,$ but you can use the Weak form of Stirling’s Formula
    begin{align} dfrac{n^n}{e^{n-1}}<n!<dfrac{n^{n+1}}{e^{n-1}} . end{align}
    This implies that
    begin{align} sum^{infty}_{n=0} n!{x^{n^2}}leq sum^{infty}_{n=0} dfrac{n^{n+1}}{e^{n-1}} {x^{n^2}}. end{align}
    So, fix $xin Bbb{R},$ then by D'Alembert's ratio test
    begin{align} limlimits_{nto infty}left|dfrac{n^{n+2}{x^{(n+1)^2}}}{e^{n}} dfrac{e^{n-1}} {n^{n+1} x^{n^2}}right|&=limlimits_{nto infty}left|dfrac{n }{e} {x^{(n+1)^2-n^2}}right|\&=limlimits_{nto infty}left|dfrac{n }{e} {x^{2n+1}}right|\&=dfrac{left|x right|}{e} limlimits_{nto infty}n, {left|xright|}^{2n}. end{align}



    CASE I: $|x|<1$



    Since $|x|<1$, then ${left|xright|}^{2n}to 0$, and so, there exists $Nin Bbb{N}$ such that
    begin{align} {left|xright|}^{2n}<dfrac{1}{n^2},forall;ngeq N .end{align}
    Thus,
    begin{align} limlimits_{nto infty}left|dfrac{n^{n+2}{x^{(n+1)^2}}}{e^{n}} dfrac{e^{n-1}} {n^{n+1} x^{n^2}}right|&=dfrac{left|x right|}{e} limlimits_{nto infty}n, {left|xright|}^{2n}leq dfrac{left|x right|}{e} limlimits_{nto infty}n, left(dfrac{1}{n^2}right)=0<1. end{align}
    Hence, the series converges when $|x|<1.$



    CASE II: $|x|geq 1$
    begin{align} limlimits_{nto infty}n!{x^{n^2}}=infty end{align}
    which implies that the series cannot not converge. Consequently, the radius of convergence is $1$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Your formula is wrong because the exponent of $x$ is $n^2,$ but you can use the Weak form of Stirling’s Formula
      begin{align} dfrac{n^n}{e^{n-1}}<n!<dfrac{n^{n+1}}{e^{n-1}} . end{align}
      This implies that
      begin{align} sum^{infty}_{n=0} n!{x^{n^2}}leq sum^{infty}_{n=0} dfrac{n^{n+1}}{e^{n-1}} {x^{n^2}}. end{align}
      So, fix $xin Bbb{R},$ then by D'Alembert's ratio test
      begin{align} limlimits_{nto infty}left|dfrac{n^{n+2}{x^{(n+1)^2}}}{e^{n}} dfrac{e^{n-1}} {n^{n+1} x^{n^2}}right|&=limlimits_{nto infty}left|dfrac{n }{e} {x^{(n+1)^2-n^2}}right|\&=limlimits_{nto infty}left|dfrac{n }{e} {x^{2n+1}}right|\&=dfrac{left|x right|}{e} limlimits_{nto infty}n, {left|xright|}^{2n}. end{align}



      CASE I: $|x|<1$



      Since $|x|<1$, then ${left|xright|}^{2n}to 0$, and so, there exists $Nin Bbb{N}$ such that
      begin{align} {left|xright|}^{2n}<dfrac{1}{n^2},forall;ngeq N .end{align}
      Thus,
      begin{align} limlimits_{nto infty}left|dfrac{n^{n+2}{x^{(n+1)^2}}}{e^{n}} dfrac{e^{n-1}} {n^{n+1} x^{n^2}}right|&=dfrac{left|x right|}{e} limlimits_{nto infty}n, {left|xright|}^{2n}leq dfrac{left|x right|}{e} limlimits_{nto infty}n, left(dfrac{1}{n^2}right)=0<1. end{align}
      Hence, the series converges when $|x|<1.$



      CASE II: $|x|geq 1$
      begin{align} limlimits_{nto infty}n!{x^{n^2}}=infty end{align}
      which implies that the series cannot not converge. Consequently, the radius of convergence is $1$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Your formula is wrong because the exponent of $x$ is $n^2,$ but you can use the Weak form of Stirling’s Formula
        begin{align} dfrac{n^n}{e^{n-1}}<n!<dfrac{n^{n+1}}{e^{n-1}} . end{align}
        This implies that
        begin{align} sum^{infty}_{n=0} n!{x^{n^2}}leq sum^{infty}_{n=0} dfrac{n^{n+1}}{e^{n-1}} {x^{n^2}}. end{align}
        So, fix $xin Bbb{R},$ then by D'Alembert's ratio test
        begin{align} limlimits_{nto infty}left|dfrac{n^{n+2}{x^{(n+1)^2}}}{e^{n}} dfrac{e^{n-1}} {n^{n+1} x^{n^2}}right|&=limlimits_{nto infty}left|dfrac{n }{e} {x^{(n+1)^2-n^2}}right|\&=limlimits_{nto infty}left|dfrac{n }{e} {x^{2n+1}}right|\&=dfrac{left|x right|}{e} limlimits_{nto infty}n, {left|xright|}^{2n}. end{align}



        CASE I: $|x|<1$



        Since $|x|<1$, then ${left|xright|}^{2n}to 0$, and so, there exists $Nin Bbb{N}$ such that
        begin{align} {left|xright|}^{2n}<dfrac{1}{n^2},forall;ngeq N .end{align}
        Thus,
        begin{align} limlimits_{nto infty}left|dfrac{n^{n+2}{x^{(n+1)^2}}}{e^{n}} dfrac{e^{n-1}} {n^{n+1} x^{n^2}}right|&=dfrac{left|x right|}{e} limlimits_{nto infty}n, {left|xright|}^{2n}leq dfrac{left|x right|}{e} limlimits_{nto infty}n, left(dfrac{1}{n^2}right)=0<1. end{align}
        Hence, the series converges when $|x|<1.$



        CASE II: $|x|geq 1$
        begin{align} limlimits_{nto infty}n!{x^{n^2}}=infty end{align}
        which implies that the series cannot not converge. Consequently, the radius of convergence is $1$.






        share|cite|improve this answer











        $endgroup$



        Your formula is wrong because the exponent of $x$ is $n^2,$ but you can use the Weak form of Stirling’s Formula
        begin{align} dfrac{n^n}{e^{n-1}}<n!<dfrac{n^{n+1}}{e^{n-1}} . end{align}
        This implies that
        begin{align} sum^{infty}_{n=0} n!{x^{n^2}}leq sum^{infty}_{n=0} dfrac{n^{n+1}}{e^{n-1}} {x^{n^2}}. end{align}
        So, fix $xin Bbb{R},$ then by D'Alembert's ratio test
        begin{align} limlimits_{nto infty}left|dfrac{n^{n+2}{x^{(n+1)^2}}}{e^{n}} dfrac{e^{n-1}} {n^{n+1} x^{n^2}}right|&=limlimits_{nto infty}left|dfrac{n }{e} {x^{(n+1)^2-n^2}}right|\&=limlimits_{nto infty}left|dfrac{n }{e} {x^{2n+1}}right|\&=dfrac{left|x right|}{e} limlimits_{nto infty}n, {left|xright|}^{2n}. end{align}



        CASE I: $|x|<1$



        Since $|x|<1$, then ${left|xright|}^{2n}to 0$, and so, there exists $Nin Bbb{N}$ such that
        begin{align} {left|xright|}^{2n}<dfrac{1}{n^2},forall;ngeq N .end{align}
        Thus,
        begin{align} limlimits_{nto infty}left|dfrac{n^{n+2}{x^{(n+1)^2}}}{e^{n}} dfrac{e^{n-1}} {n^{n+1} x^{n^2}}right|&=dfrac{left|x right|}{e} limlimits_{nto infty}n, {left|xright|}^{2n}leq dfrac{left|x right|}{e} limlimits_{nto infty}n, left(dfrac{1}{n^2}right)=0<1. end{align}
        Hence, the series converges when $|x|<1.$



        CASE II: $|x|geq 1$
        begin{align} limlimits_{nto infty}n!{x^{n^2}}=infty end{align}
        which implies that the series cannot not converge. Consequently, the radius of convergence is $1$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 11 at 13:58

























        answered Jan 11 at 13:51









        Omojola MichealOmojola Micheal

        2,064424




        2,064424






























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