Find radius of convergence of the power series.












-1












$begingroup$


Find the radius of convergence of power series $$ sum_{n=0}^{infty} 2^{2n} x^{n^2}$$
A)1



B)2



C) 4



D)1/4



I try to apply ratio and root test ( Cauchy–Hadamard theorem ) .but they don't seem promising as I was left with $|x^n|$ .



But I find solution by trial and error . I substitute 1 for x , series diverges , so 2,4 cannot be radius of convergence, and I substitute 1/2 the series converges .so the answer is 1.



Please help me how to proceed in problem like this.






sequences-and-series convergence power-series divergent-series absolute-convergence







share|cite|improve this question









$endgroup$

















    -1












    $begingroup$


    Find the radius of convergence of power series $$ sum_{n=0}^{infty} 2^{2n} x^{n^2}$$
    A)1



    B)2



    C) 4



    D)1/4



    I try to apply ratio and root test ( Cauchy–Hadamard theorem ) .but they don't seem promising as I was left with $|x^n|$ .



    But I find solution by trial and error . I substitute 1 for x , series diverges , so 2,4 cannot be radius of convergence, and I substitute 1/2 the series converges .so the answer is 1.



    Please help me how to proceed in problem like this.






    sequences-and-series convergence power-series divergent-series absolute-convergence







    share|cite|improve this question









    $endgroup$















      -1












      -1








      -1





      $begingroup$


      Find the radius of convergence of power series $$ sum_{n=0}^{infty} 2^{2n} x^{n^2}$$
      A)1



      B)2



      C) 4



      D)1/4



      I try to apply ratio and root test ( Cauchy–Hadamard theorem ) .but they don't seem promising as I was left with $|x^n|$ .



      But I find solution by trial and error . I substitute 1 for x , series diverges , so 2,4 cannot be radius of convergence, and I substitute 1/2 the series converges .so the answer is 1.



      Please help me how to proceed in problem like this.






      sequences-and-series convergence power-series divergent-series absolute-convergence







      share|cite|improve this question









      $endgroup$




      Find the radius of convergence of power series $$ sum_{n=0}^{infty} 2^{2n} x^{n^2}$$
      A)1



      B)2



      C) 4



      D)1/4



      I try to apply ratio and root test ( Cauchy–Hadamard theorem ) .but they don't seem promising as I was left with $|x^n|$ .



      But I find solution by trial and error . I substitute 1 for x , series diverges , so 2,4 cannot be radius of convergence, and I substitute 1/2 the series converges .so the answer is 1.



      Please help me how to proceed in problem like this.






      sequences-and-series convergence power-series divergent-series absolute-convergence




      sequences-and-series convergence power-series divergent-series absolute-convergence






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 1 at 15:08









      Cloud JRCloud JR

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          $begingroup$

          Since$$leftlvertfrac{2^{2(n+1)}x^{(n+1)^2}}{2^{2n}x^{n^2}}rightrvert=2^2lvert xrvert^{2n+1},$$the series converges absolutely if $lvert xrvert<1$ and diverges if $lvert xrvert>1$. So, the radius of convergence is $1$.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Hint: what does $x^n$ do as $n to infty$ if $|x| < 1$? If $|x| > 1$?






            share|cite|improve this answer









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              2 Answers
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              $begingroup$

              Since$$leftlvertfrac{2^{2(n+1)}x^{(n+1)^2}}{2^{2n}x^{n^2}}rightrvert=2^2lvert xrvert^{2n+1},$$the series converges absolutely if $lvert xrvert<1$ and diverges if $lvert xrvert>1$. So, the radius of convergence is $1$.






              share|cite|improve this answer









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                2












                $begingroup$

                Since$$leftlvertfrac{2^{2(n+1)}x^{(n+1)^2}}{2^{2n}x^{n^2}}rightrvert=2^2lvert xrvert^{2n+1},$$the series converges absolutely if $lvert xrvert<1$ and diverges if $lvert xrvert>1$. So, the radius of convergence is $1$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Since$$leftlvertfrac{2^{2(n+1)}x^{(n+1)^2}}{2^{2n}x^{n^2}}rightrvert=2^2lvert xrvert^{2n+1},$$the series converges absolutely if $lvert xrvert<1$ and diverges if $lvert xrvert>1$. So, the radius of convergence is $1$.






                  share|cite|improve this answer









                  $endgroup$



                  Since$$leftlvertfrac{2^{2(n+1)}x^{(n+1)^2}}{2^{2n}x^{n^2}}rightrvert=2^2lvert xrvert^{2n+1},$$the series converges absolutely if $lvert xrvert<1$ and diverges if $lvert xrvert>1$. So, the radius of convergence is $1$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 1 at 15:15









                  José Carlos SantosJosé Carlos Santos

                  153k22123225




                  153k22123225























                      1












                      $begingroup$

                      Hint: what does $x^n$ do as $n to infty$ if $|x| < 1$? If $|x| > 1$?






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        Hint: what does $x^n$ do as $n to infty$ if $|x| < 1$? If $|x| > 1$?






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Hint: what does $x^n$ do as $n to infty$ if $|x| < 1$? If $|x| > 1$?






                          share|cite|improve this answer









                          $endgroup$



                          Hint: what does $x^n$ do as $n to infty$ if $|x| < 1$? If $|x| > 1$?







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 1 at 15:13









                          Robert IsraelRobert Israel

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                          319k23209459






























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