Radius of convergence - ratio test for power series/real numbers












0












$begingroup$


Having trouble with when to apply the ratio test for power series and when to apply the ratio test for real numbers.



For example, find radius of convergence of these....



$sum_{n=0}^{infty}(-1)^n (x+3)^n$



For this I would use ratio test for power series and get radius of convergence = 1, fine.



but for



$sum_{n=0}^{infty}n!(n+1)!(x+1)^{2n}$



Why do we need to use the ratio test for real numbers, and define for a fixed $x$? and why can't we use the ratio test for power series?



Is it because the $(x+1)^{2n}$ part is $2n$ and not $n$?






sequences-and-series analysis convergence power-series







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$endgroup$








  • 2




    $begingroup$
    $sum_{n=0}^{infty} n!(n+1)!(x+1)^{2n}$ is a power series in $(x+1)^2$. The ratio test for power series will work fine and give a radius of $0$. Latex is explained at [maa.org/latex/ltx-2.html].
    $endgroup$
    – Angela Richardson
    May 15 '13 at 12:42










  • $begingroup$
    but its 2n...so all the odd powers of x are missing. I think that's maybe why. radius is 0 though...
    $endgroup$
    – Tom
    May 15 '13 at 13:02










  • $begingroup$
    when coefficients are zero, you cannot put them in the denominator as required in the ratio test. In some cases (as here) you can recognize it as a power series in another variable and use the ratio test on that one.
    $endgroup$
    – GEdgar
    May 15 '13 at 14:07
















0












$begingroup$


Having trouble with when to apply the ratio test for power series and when to apply the ratio test for real numbers.



For example, find radius of convergence of these....



$sum_{n=0}^{infty}(-1)^n (x+3)^n$



For this I would use ratio test for power series and get radius of convergence = 1, fine.



but for



$sum_{n=0}^{infty}n!(n+1)!(x+1)^{2n}$



Why do we need to use the ratio test for real numbers, and define for a fixed $x$? and why can't we use the ratio test for power series?



Is it because the $(x+1)^{2n}$ part is $2n$ and not $n$?






sequences-and-series analysis convergence power-series







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    $sum_{n=0}^{infty} n!(n+1)!(x+1)^{2n}$ is a power series in $(x+1)^2$. The ratio test for power series will work fine and give a radius of $0$. Latex is explained at [maa.org/latex/ltx-2.html].
    $endgroup$
    – Angela Richardson
    May 15 '13 at 12:42










  • $begingroup$
    but its 2n...so all the odd powers of x are missing. I think that's maybe why. radius is 0 though...
    $endgroup$
    – Tom
    May 15 '13 at 13:02










  • $begingroup$
    when coefficients are zero, you cannot put them in the denominator as required in the ratio test. In some cases (as here) you can recognize it as a power series in another variable and use the ratio test on that one.
    $endgroup$
    – GEdgar
    May 15 '13 at 14:07














0












0








0





$begingroup$


Having trouble with when to apply the ratio test for power series and when to apply the ratio test for real numbers.



For example, find radius of convergence of these....



$sum_{n=0}^{infty}(-1)^n (x+3)^n$



For this I would use ratio test for power series and get radius of convergence = 1, fine.



but for



$sum_{n=0}^{infty}n!(n+1)!(x+1)^{2n}$



Why do we need to use the ratio test for real numbers, and define for a fixed $x$? and why can't we use the ratio test for power series?



Is it because the $(x+1)^{2n}$ part is $2n$ and not $n$?






sequences-and-series analysis convergence power-series







share|cite|improve this question











$endgroup$




Having trouble with when to apply the ratio test for power series and when to apply the ratio test for real numbers.



For example, find radius of convergence of these....



$sum_{n=0}^{infty}(-1)^n (x+3)^n$



For this I would use ratio test for power series and get radius of convergence = 1, fine.



but for



$sum_{n=0}^{infty}n!(n+1)!(x+1)^{2n}$



Why do we need to use the ratio test for real numbers, and define for a fixed $x$? and why can't we use the ratio test for power series?



Is it because the $(x+1)^{2n}$ part is $2n$ and not $n$?






sequences-and-series analysis convergence power-series




sequences-and-series analysis convergence power-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 '18 at 8:34









Martin Sleziak

44.7k8117272




44.7k8117272










asked May 15 '13 at 12:30









TomTom

1




1








  • 2




    $begingroup$
    $sum_{n=0}^{infty} n!(n+1)!(x+1)^{2n}$ is a power series in $(x+1)^2$. The ratio test for power series will work fine and give a radius of $0$. Latex is explained at [maa.org/latex/ltx-2.html].
    $endgroup$
    – Angela Richardson
    May 15 '13 at 12:42










  • $begingroup$
    but its 2n...so all the odd powers of x are missing. I think that's maybe why. radius is 0 though...
    $endgroup$
    – Tom
    May 15 '13 at 13:02










  • $begingroup$
    when coefficients are zero, you cannot put them in the denominator as required in the ratio test. In some cases (as here) you can recognize it as a power series in another variable and use the ratio test on that one.
    $endgroup$
    – GEdgar
    May 15 '13 at 14:07














  • 2




    $begingroup$
    $sum_{n=0}^{infty} n!(n+1)!(x+1)^{2n}$ is a power series in $(x+1)^2$. The ratio test for power series will work fine and give a radius of $0$. Latex is explained at [maa.org/latex/ltx-2.html].
    $endgroup$
    – Angela Richardson
    May 15 '13 at 12:42










  • $begingroup$
    but its 2n...so all the odd powers of x are missing. I think that's maybe why. radius is 0 though...
    $endgroup$
    – Tom
    May 15 '13 at 13:02










  • $begingroup$
    when coefficients are zero, you cannot put them in the denominator as required in the ratio test. In some cases (as here) you can recognize it as a power series in another variable and use the ratio test on that one.
    $endgroup$
    – GEdgar
    May 15 '13 at 14:07








2




2




$begingroup$
$sum_{n=0}^{infty} n!(n+1)!(x+1)^{2n}$ is a power series in $(x+1)^2$. The ratio test for power series will work fine and give a radius of $0$. Latex is explained at [maa.org/latex/ltx-2.html].
$endgroup$
– Angela Richardson
May 15 '13 at 12:42




$begingroup$
$sum_{n=0}^{infty} n!(n+1)!(x+1)^{2n}$ is a power series in $(x+1)^2$. The ratio test for power series will work fine and give a radius of $0$. Latex is explained at [maa.org/latex/ltx-2.html].
$endgroup$
– Angela Richardson
May 15 '13 at 12:42












$begingroup$
but its 2n...so all the odd powers of x are missing. I think that's maybe why. radius is 0 though...
$endgroup$
– Tom
May 15 '13 at 13:02




$begingroup$
but its 2n...so all the odd powers of x are missing. I think that's maybe why. radius is 0 though...
$endgroup$
– Tom
May 15 '13 at 13:02












$begingroup$
when coefficients are zero, you cannot put them in the denominator as required in the ratio test. In some cases (as here) you can recognize it as a power series in another variable and use the ratio test on that one.
$endgroup$
– GEdgar
May 15 '13 at 14:07




$begingroup$
when coefficients are zero, you cannot put them in the denominator as required in the ratio test. In some cases (as here) you can recognize it as a power series in another variable and use the ratio test on that one.
$endgroup$
– GEdgar
May 15 '13 at 14:07










1 Answer
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$begingroup$

Let $$sum _{n=0} ^{infty} n!(n+1)!(x+1)^{2n}$$ wants be convegent, so according to tests, we should have:



$$lim_{nto infty}{Big|frac{(n + 1)! (n+2)! (x+1)^{(2n + 2)}}{n!(n+1)!(x+1)^{2n}}Big|}<1$$ So,



$$lim_{nto infty}{|(n + 1)(n + 2)x^2|}<1$$



Since for each $xne0$ we see $lim_{nto infty}{|(n+1)(n+2)x^2|}=infty$ so only for $x=0$ our series will be convergent and hence radius of convergence is equal to zero.






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    $begingroup$

    Let $$sum _{n=0} ^{infty} n!(n+1)!(x+1)^{2n}$$ wants be convegent, so according to tests, we should have:



    $$lim_{nto infty}{Big|frac{(n + 1)! (n+2)! (x+1)^{(2n + 2)}}{n!(n+1)!(x+1)^{2n}}Big|}<1$$ So,



    $$lim_{nto infty}{|(n + 1)(n + 2)x^2|}<1$$



    Since for each $xne0$ we see $lim_{nto infty}{|(n+1)(n+2)x^2|}=infty$ so only for $x=0$ our series will be convergent and hence radius of convergence is equal to zero.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      Let $$sum _{n=0} ^{infty} n!(n+1)!(x+1)^{2n}$$ wants be convegent, so according to tests, we should have:



      $$lim_{nto infty}{Big|frac{(n + 1)! (n+2)! (x+1)^{(2n + 2)}}{n!(n+1)!(x+1)^{2n}}Big|}<1$$ So,



      $$lim_{nto infty}{|(n + 1)(n + 2)x^2|}<1$$



      Since for each $xne0$ we see $lim_{nto infty}{|(n+1)(n+2)x^2|}=infty$ so only for $x=0$ our series will be convergent and hence radius of convergence is equal to zero.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Let $$sum _{n=0} ^{infty} n!(n+1)!(x+1)^{2n}$$ wants be convegent, so according to tests, we should have:



        $$lim_{nto infty}{Big|frac{(n + 1)! (n+2)! (x+1)^{(2n + 2)}}{n!(n+1)!(x+1)^{2n}}Big|}<1$$ So,



        $$lim_{nto infty}{|(n + 1)(n + 2)x^2|}<1$$



        Since for each $xne0$ we see $lim_{nto infty}{|(n+1)(n+2)x^2|}=infty$ so only for $x=0$ our series will be convergent and hence radius of convergence is equal to zero.






        share|cite|improve this answer











        $endgroup$



        Let $$sum _{n=0} ^{infty} n!(n+1)!(x+1)^{2n}$$ wants be convegent, so according to tests, we should have:



        $$lim_{nto infty}{Big|frac{(n + 1)! (n+2)! (x+1)^{(2n + 2)}}{n!(n+1)!(x+1)^{2n}}Big|}<1$$ So,



        $$lim_{nto infty}{|(n + 1)(n + 2)x^2|}<1$$



        Since for each $xne0$ we see $lim_{nto infty}{|(n+1)(n+2)x^2|}=infty$ so only for $x=0$ our series will be convergent and hence radius of convergence is equal to zero.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 1 '18 at 8:34









        Artem Lenskiy

        32




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        answered May 15 '13 at 12:49









        SomayeSomaye

        2,578712




        2,578712






























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