Finding radius of convergence for $frac{x^n}{nsqrt{n}12^n}$












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$begingroup$


I need help finding the radius of convergence for $frac{x^n}{nsqrt{n}12^n}$



I tried both the root and ratio tests but neither one is getting me anywhere.



Trying the root test I reduced to $lim limits_{n to infty}left(frac{x}{n^{1/2}n^{1/4}12}right)$ which is either incorrect or not helpful.



I need to find radius of convergence, as well as interval of convergence, then determine absolute and conditional convergence.






calculus sequences-and-series







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  • $begingroup$
    Just, btw, the expression in the root test is $(x/12) (n^{- 3/2})^{1/n} = (x/12) n^{-1.5/n},$ not $(x/12) n^{-3/2}.$
    $endgroup$
    – stochasticboy321
    Dec 6 '18 at 1:42


















0












$begingroup$


I need help finding the radius of convergence for $frac{x^n}{nsqrt{n}12^n}$



I tried both the root and ratio tests but neither one is getting me anywhere.



Trying the root test I reduced to $lim limits_{n to infty}left(frac{x}{n^{1/2}n^{1/4}12}right)$ which is either incorrect or not helpful.



I need to find radius of convergence, as well as interval of convergence, then determine absolute and conditional convergence.






calculus sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    Just, btw, the expression in the root test is $(x/12) (n^{- 3/2})^{1/n} = (x/12) n^{-1.5/n},$ not $(x/12) n^{-3/2}.$
    $endgroup$
    – stochasticboy321
    Dec 6 '18 at 1:42
















0












0








0





$begingroup$


I need help finding the radius of convergence for $frac{x^n}{nsqrt{n}12^n}$



I tried both the root and ratio tests but neither one is getting me anywhere.



Trying the root test I reduced to $lim limits_{n to infty}left(frac{x}{n^{1/2}n^{1/4}12}right)$ which is either incorrect or not helpful.



I need to find radius of convergence, as well as interval of convergence, then determine absolute and conditional convergence.






calculus sequences-and-series







share|cite|improve this question











$endgroup$




I need help finding the radius of convergence for $frac{x^n}{nsqrt{n}12^n}$



I tried both the root and ratio tests but neither one is getting me anywhere.



Trying the root test I reduced to $lim limits_{n to infty}left(frac{x}{n^{1/2}n^{1/4}12}right)$ which is either incorrect or not helpful.



I need to find radius of convergence, as well as interval of convergence, then determine absolute and conditional convergence.






calculus sequences-and-series




calculus sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 10:37









user376343

3,3933826




3,3933826










asked Dec 6 '18 at 1:33









SolidSnackDriveSolidSnackDrive

1828




1828












  • $begingroup$
    Just, btw, the expression in the root test is $(x/12) (n^{- 3/2})^{1/n} = (x/12) n^{-1.5/n},$ not $(x/12) n^{-3/2}.$
    $endgroup$
    – stochasticboy321
    Dec 6 '18 at 1:42




















  • $begingroup$
    Just, btw, the expression in the root test is $(x/12) (n^{- 3/2})^{1/n} = (x/12) n^{-1.5/n},$ not $(x/12) n^{-3/2}.$
    $endgroup$
    – stochasticboy321
    Dec 6 '18 at 1:42


















$begingroup$
Just, btw, the expression in the root test is $(x/12) (n^{- 3/2})^{1/n} = (x/12) n^{-1.5/n},$ not $(x/12) n^{-3/2}.$
$endgroup$
– stochasticboy321
Dec 6 '18 at 1:42






$begingroup$
Just, btw, the expression in the root test is $(x/12) (n^{- 3/2})^{1/n} = (x/12) n^{-1.5/n},$ not $(x/12) n^{-3/2}.$
$endgroup$
– stochasticboy321
Dec 6 '18 at 1:42












2 Answers
2






active

oldest

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2












$begingroup$

For ratio test:



$$lim_{n to infty}frac{x^{n+1}}{(n+1)sqrt{n+1}cdot 12^{n+1}}frac{nsqrt{n} cdot 12^n}{x^n}= lim_{n to infty}frac{x}{12}lim_{n to infty}frac{n}{n+1}lim_{n to infty}sqrt{frac{n}{n+1}}$$



Can you evaluate the last two limits?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Second one goes to $1$, third one also $1$?
    $endgroup$
    – SolidSnackDrive
    Dec 6 '18 at 1:50










  • $begingroup$
    yes, $frac{n}{n+1}= frac{1}{1+frac1n}$.
    $endgroup$
    – Siong Thye Goh
    Dec 6 '18 at 1:51



















1












$begingroup$

The ratio test should work. The ratio of the $1/(n sqrt(n))$ part goes to $1$ and the rest gives $|x/12|$.






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

    votes









    2












    $begingroup$

    For ratio test:



    $$lim_{n to infty}frac{x^{n+1}}{(n+1)sqrt{n+1}cdot 12^{n+1}}frac{nsqrt{n} cdot 12^n}{x^n}= lim_{n to infty}frac{x}{12}lim_{n to infty}frac{n}{n+1}lim_{n to infty}sqrt{frac{n}{n+1}}$$



    Can you evaluate the last two limits?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Second one goes to $1$, third one also $1$?
      $endgroup$
      – SolidSnackDrive
      Dec 6 '18 at 1:50










    • $begingroup$
      yes, $frac{n}{n+1}= frac{1}{1+frac1n}$.
      $endgroup$
      – Siong Thye Goh
      Dec 6 '18 at 1:51
















    2












    $begingroup$

    For ratio test:



    $$lim_{n to infty}frac{x^{n+1}}{(n+1)sqrt{n+1}cdot 12^{n+1}}frac{nsqrt{n} cdot 12^n}{x^n}= lim_{n to infty}frac{x}{12}lim_{n to infty}frac{n}{n+1}lim_{n to infty}sqrt{frac{n}{n+1}}$$



    Can you evaluate the last two limits?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Second one goes to $1$, third one also $1$?
      $endgroup$
      – SolidSnackDrive
      Dec 6 '18 at 1:50










    • $begingroup$
      yes, $frac{n}{n+1}= frac{1}{1+frac1n}$.
      $endgroup$
      – Siong Thye Goh
      Dec 6 '18 at 1:51














    2












    2








    2





    $begingroup$

    For ratio test:



    $$lim_{n to infty}frac{x^{n+1}}{(n+1)sqrt{n+1}cdot 12^{n+1}}frac{nsqrt{n} cdot 12^n}{x^n}= lim_{n to infty}frac{x}{12}lim_{n to infty}frac{n}{n+1}lim_{n to infty}sqrt{frac{n}{n+1}}$$



    Can you evaluate the last two limits?






    share|cite|improve this answer









    $endgroup$



    For ratio test:



    $$lim_{n to infty}frac{x^{n+1}}{(n+1)sqrt{n+1}cdot 12^{n+1}}frac{nsqrt{n} cdot 12^n}{x^n}= lim_{n to infty}frac{x}{12}lim_{n to infty}frac{n}{n+1}lim_{n to infty}sqrt{frac{n}{n+1}}$$



    Can you evaluate the last two limits?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 6 '18 at 1:40









    Siong Thye GohSiong Thye Goh

    100k1466117




    100k1466117












    • $begingroup$
      Second one goes to $1$, third one also $1$?
      $endgroup$
      – SolidSnackDrive
      Dec 6 '18 at 1:50










    • $begingroup$
      yes, $frac{n}{n+1}= frac{1}{1+frac1n}$.
      $endgroup$
      – Siong Thye Goh
      Dec 6 '18 at 1:51


















    • $begingroup$
      Second one goes to $1$, third one also $1$?
      $endgroup$
      – SolidSnackDrive
      Dec 6 '18 at 1:50










    • $begingroup$
      yes, $frac{n}{n+1}= frac{1}{1+frac1n}$.
      $endgroup$
      – Siong Thye Goh
      Dec 6 '18 at 1:51
















    $begingroup$
    Second one goes to $1$, third one also $1$?
    $endgroup$
    – SolidSnackDrive
    Dec 6 '18 at 1:50




    $begingroup$
    Second one goes to $1$, third one also $1$?
    $endgroup$
    – SolidSnackDrive
    Dec 6 '18 at 1:50












    $begingroup$
    yes, $frac{n}{n+1}= frac{1}{1+frac1n}$.
    $endgroup$
    – Siong Thye Goh
    Dec 6 '18 at 1:51




    $begingroup$
    yes, $frac{n}{n+1}= frac{1}{1+frac1n}$.
    $endgroup$
    – Siong Thye Goh
    Dec 6 '18 at 1:51











    1












    $begingroup$

    The ratio test should work. The ratio of the $1/(n sqrt(n))$ part goes to $1$ and the rest gives $|x/12|$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The ratio test should work. The ratio of the $1/(n sqrt(n))$ part goes to $1$ and the rest gives $|x/12|$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The ratio test should work. The ratio of the $1/(n sqrt(n))$ part goes to $1$ and the rest gives $|x/12|$.






        share|cite|improve this answer









        $endgroup$



        The ratio test should work. The ratio of the $1/(n sqrt(n))$ part goes to $1$ and the rest gives $|x/12|$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 6 '18 at 1:40









        coffeemathcoffeemath

        2,7361415




        2,7361415






























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