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When to use the root test. Is this not a good situation to use it?

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When to use the root test. Is this not a good situation to use it?



The 2019 Stack Overflow Developer Survey Results Are InWhich test would be appropriate to use on this series to show convergence/divergence?Integral test vs root test vs ratio testHow to show convergence or divergence of a series when the ratio test is inconclusive?Root test with nested power function?Confused about using alternating test, ratio test, and root test (please help).Radius and interval of convergence of $sum_{n=1}^{infty}(-1)^nfrac{x^{2n}}{(2n)!}$ by root and ratio test are different?How would I use root/ratio test on $sum_{n=1}^inftyleft(frac{n}{n+1}right)^{n^2}$?How would I know when to use what test for convergence?convergence of a sum fails with root testIntuition for Root Test.












2












$begingroup$


I'm having trouble seeing when to use the root test. nth powers occur, but I think the ratio test is easier:



enter image description here



Here is the problem:



$$sum_{n=1}^{infty} frac{x^n}{n^44^n}$$



So the ratio test seems to work here, but can't the root test be used to? The problem is that the $n^4$ doesnt play well with the root test right?



Here is the beginning of my solution with the ratio test:



$$biggr lbrack frac{a_{n+1}}{a_n} biggr rbrack = biggr lbrack frac{x^{n+1}}{(n+1)^4 * 4^{n+1}} * frac{n^4*4^n}{x^n} biggr rbrack = biggr lbrack frac{x*n^4}{(n+1)^4 * 4} biggr rbrack = frac{x}{4}$$



So I don't think the explanation for when to use the root test is totally right right? I can't really use it here because the $n^4$ causes some problems with the root test right?






sequences-and-series







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    I'm having trouble seeing when to use the root test. nth powers occur, but I think the ratio test is easier:



    enter image description here



    Here is the problem:



    $$sum_{n=1}^{infty} frac{x^n}{n^44^n}$$



    So the ratio test seems to work here, but can't the root test be used to? The problem is that the $n^4$ doesnt play well with the root test right?



    Here is the beginning of my solution with the ratio test:



    $$biggr lbrack frac{a_{n+1}}{a_n} biggr rbrack = biggr lbrack frac{x^{n+1}}{(n+1)^4 * 4^{n+1}} * frac{n^4*4^n}{x^n} biggr rbrack = biggr lbrack frac{x*n^4}{(n+1)^4 * 4} biggr rbrack = frac{x}{4}$$



    So I don't think the explanation for when to use the root test is totally right right? I can't really use it here because the $n^4$ causes some problems with the root test right?






    sequences-and-series







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      I'm having trouble seeing when to use the root test. nth powers occur, but I think the ratio test is easier:



      enter image description here



      Here is the problem:



      $$sum_{n=1}^{infty} frac{x^n}{n^44^n}$$



      So the ratio test seems to work here, but can't the root test be used to? The problem is that the $n^4$ doesnt play well with the root test right?



      Here is the beginning of my solution with the ratio test:



      $$biggr lbrack frac{a_{n+1}}{a_n} biggr rbrack = biggr lbrack frac{x^{n+1}}{(n+1)^4 * 4^{n+1}} * frac{n^4*4^n}{x^n} biggr rbrack = biggr lbrack frac{x*n^4}{(n+1)^4 * 4} biggr rbrack = frac{x}{4}$$



      So I don't think the explanation for when to use the root test is totally right right? I can't really use it here because the $n^4$ causes some problems with the root test right?






      sequences-and-series







      share|cite|improve this question









      $endgroup$




      I'm having trouble seeing when to use the root test. nth powers occur, but I think the ratio test is easier:



      enter image description here



      Here is the problem:



      $$sum_{n=1}^{infty} frac{x^n}{n^44^n}$$



      So the ratio test seems to work here, but can't the root test be used to? The problem is that the $n^4$ doesnt play well with the root test right?



      Here is the beginning of my solution with the ratio test:



      $$biggr lbrack frac{a_{n+1}}{a_n} biggr rbrack = biggr lbrack frac{x^{n+1}}{(n+1)^4 * 4^{n+1}} * frac{n^4*4^n}{x^n} biggr rbrack = biggr lbrack frac{x*n^4}{(n+1)^4 * 4} biggr rbrack = frac{x}{4}$$



      So I don't think the explanation for when to use the root test is totally right right? I can't really use it here because the $n^4$ causes some problems with the root test right?






      sequences-and-series




      sequences-and-series






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 4 hours ago









      Jwan622Jwan622

      2,38011632




      2,38011632






















          2 Answers
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          4












          $begingroup$

          It doesn't cause any problems, because $lim_{ntoinfty}sqrt[n]{n^4}=1.$ Actually, the root test is stronger than the ratio test. Sometimes the root test limit exists, but the ratio test limit does not. However, if they both exist, then they are equal. Which is why if one limit is $1$ you shouldn't try the other, even though the root test is stronger.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            When doing a root test,
            powers of $n$ can be ignored
            because,
            for any fixed $k$,



            $lim_{n to infty} (n^k)^{1/n}
            =1
            $            .



            This is because
            $ (n^k)^{1/n}
            =n^{k/n}
            =e^{k ln(n)/n}
            $
            and
            $lim_{n to infty} frac{ln(n)}{n}
            =0$            .



            An easy,
            but nonelementary proof of this is this:



            $begin{array}\
            ln(n)
            &=int_1^n dfrac{dt}{t}\
            &<int_1^n dfrac{dt}{t^{1/2}}\
            &=2t^{1/2}|_1^n\
            &lt 2sqrt{n}\
            text{so}\
            dfrac{ln(n)}{n}
            &<dfrac{2}{sqrt{n}}\
            end{array}
            $



            Therefore
            $ (n^k)^{1/n}
            =n^{k/n}
            =e^{k ln(n)/n}
            lt e^{2k/sqrt{n}}
            to 1
            $            .






            share|cite|improve this answer









            $endgroup$














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              2 Answers
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              active

              oldest

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              2 Answers
              2






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              active

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              4












              $begingroup$

              It doesn't cause any problems, because $lim_{ntoinfty}sqrt[n]{n^4}=1.$ Actually, the root test is stronger than the ratio test. Sometimes the root test limit exists, but the ratio test limit does not. However, if they both exist, then they are equal. Which is why if one limit is $1$ you shouldn't try the other, even though the root test is stronger.






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$

                It doesn't cause any problems, because $lim_{ntoinfty}sqrt[n]{n^4}=1.$ Actually, the root test is stronger than the ratio test. Sometimes the root test limit exists, but the ratio test limit does not. However, if they both exist, then they are equal. Which is why if one limit is $1$ you shouldn't try the other, even though the root test is stronger.






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  It doesn't cause any problems, because $lim_{ntoinfty}sqrt[n]{n^4}=1.$ Actually, the root test is stronger than the ratio test. Sometimes the root test limit exists, but the ratio test limit does not. However, if they both exist, then they are equal. Which is why if one limit is $1$ you shouldn't try the other, even though the root test is stronger.






                  share|cite|improve this answer









                  $endgroup$



                  It doesn't cause any problems, because $lim_{ntoinfty}sqrt[n]{n^4}=1.$ Actually, the root test is stronger than the ratio test. Sometimes the root test limit exists, but the ratio test limit does not. However, if they both exist, then they are equal. Which is why if one limit is $1$ you shouldn't try the other, even though the root test is stronger.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  MelodyMelody

                  1,07412




                  1,07412























                      2












                      $begingroup$

                      When doing a root test,
                      powers of $n$ can be ignored
                      because,
                      for any fixed $k$,



                      $lim_{n to infty} (n^k)^{1/n}
                      =1
                      $                      .



                      This is because
                      $ (n^k)^{1/n}
                      =n^{k/n}
                      =e^{k ln(n)/n}
                      $
                      and
                      $lim_{n to infty} frac{ln(n)}{n}
                      =0$                      .



                      An easy,
                      but nonelementary proof of this is this:



                      $begin{array}\
                      ln(n)
                      &=int_1^n dfrac{dt}{t}\
                      &<int_1^n dfrac{dt}{t^{1/2}}\
                      &=2t^{1/2}|_1^n\
                      &lt 2sqrt{n}\
                      text{so}\
                      dfrac{ln(n)}{n}
                      &<dfrac{2}{sqrt{n}}\
                      end{array}
                      $



                      Therefore
                      $ (n^k)^{1/n}
                      =n^{k/n}
                      =e^{k ln(n)/n}
                      lt e^{2k/sqrt{n}}
                      to 1
                      $                      .






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        When doing a root test,
                        powers of $n$ can be ignored
                        because,
                        for any fixed $k$,



                        $lim_{n to infty} (n^k)^{1/n}
                        =1
                        $                        .



                        This is because
                        $ (n^k)^{1/n}
                        =n^{k/n}
                        =e^{k ln(n)/n}
                        $
                        and
                        $lim_{n to infty} frac{ln(n)}{n}
                        =0$                        .



                        An easy,
                        but nonelementary proof of this is this:



                        $begin{array}\
                        ln(n)
                        &=int_1^n dfrac{dt}{t}\
                        &<int_1^n dfrac{dt}{t^{1/2}}\
                        &=2t^{1/2}|_1^n\
                        &lt 2sqrt{n}\
                        text{so}\
                        dfrac{ln(n)}{n}
                        &<dfrac{2}{sqrt{n}}\
                        end{array}
                        $



                        Therefore
                        $ (n^k)^{1/n}
                        =n^{k/n}
                        =e^{k ln(n)/n}
                        lt e^{2k/sqrt{n}}
                        to 1
                        $                        .






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          When doing a root test,
                          powers of $n$ can be ignored
                          because,
                          for any fixed $k$,



                          $lim_{n to infty} (n^k)^{1/n}
                          =1
                          $                          .



                          This is because
                          $ (n^k)^{1/n}
                          =n^{k/n}
                          =e^{k ln(n)/n}
                          $
                          and
                          $lim_{n to infty} frac{ln(n)}{n}
                          =0$                          .



                          An easy,
                          but nonelementary proof of this is this:



                          $begin{array}\
                          ln(n)
                          &=int_1^n dfrac{dt}{t}\
                          &<int_1^n dfrac{dt}{t^{1/2}}\
                          &=2t^{1/2}|_1^n\
                          &lt 2sqrt{n}\
                          text{so}\
                          dfrac{ln(n)}{n}
                          &<dfrac{2}{sqrt{n}}\
                          end{array}
                          $



                          Therefore
                          $ (n^k)^{1/n}
                          =n^{k/n}
                          =e^{k ln(n)/n}
                          lt e^{2k/sqrt{n}}
                          to 1
                          $                          .






                          share|cite|improve this answer









                          $endgroup$



                          When doing a root test,
                          powers of $n$ can be ignored
                          because,
                          for any fixed $k$,



                          $lim_{n to infty} (n^k)^{1/n}
                          =1
                          $                          .



                          This is because
                          $ (n^k)^{1/n}
                          =n^{k/n}
                          =e^{k ln(n)/n}
                          $
                          and
                          $lim_{n to infty} frac{ln(n)}{n}
                          =0$                          .



                          An easy,
                          but nonelementary proof of this is this:



                          $begin{array}\
                          ln(n)
                          &=int_1^n dfrac{dt}{t}\
                          &<int_1^n dfrac{dt}{t^{1/2}}\
                          &=2t^{1/2}|_1^n\
                          &lt 2sqrt{n}\
                          text{so}\
                          dfrac{ln(n)}{n}
                          &<dfrac{2}{sqrt{n}}\
                          end{array}
                          $



                          Therefore
                          $ (n^k)^{1/n}
                          =n^{k/n}
                          =e^{k ln(n)/n}
                          lt e^{2k/sqrt{n}}
                          to 1
                          $                          .







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 3 hours ago









                          marty cohenmarty cohen

                          75.2k549130




                          75.2k549130






























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