When to use the root test. Is this not a good situation to use it? The 2019 Stack Overflow...

Access elements in std::string where positon of string is greater than its size

Why isn't airport relocation done gradually?

Is this food a bread or a loaf?

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

Does it makes sense to buy a new cycle to learn riding?

If the Wish spell is used to duplicate the effect of Simulacrum, are existing duplicates destroyed?

Why is it "Tumoren" and not "Tumore"?

Could JWST stay at L2 "forever"?

How can I fix this gap between bookcases I made?

Geography at the pixel level

A poker game description that does not feel gimmicky

What is this 4-propeller plane?

Inversion Puzzle

Is an up-to-date browser secure on an out-of-date OS?

What is the use of option -o in the useradd command?

Are there any other methods to apply to solving simultaneous equations?

Why is the maximum length of OpenWrt’s root password 8 characters?

Landlord wants to switch my lease to a "Land contract" to "get back at the city"

Is domain driven design an anti-SQL pattern?

Could a US political party gain complete control over the government by removing checks & balances?

How come people say “Would of”?

Does duplicating a spell with Wish count as casting that spell?

What does "rabbited" mean/imply in this sentence?

Is flight data recorder erased after every flight?



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
          2






          active

          oldest

          votes


















          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









            $endgroup$














              Your Answer





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              });
              });
              }, "mathjax-editing");

              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "69"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3181802%2fwhen-to-use-the-root-test-is-this-not-a-good-situation-to-use-it%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              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






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid …



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3181802%2fwhen-to-use-the-root-test-is-this-not-a-good-situation-to-use-it%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              -

                              Popular posts from this blog

                              Gersau Kjelder | Navigasjonsmeny46°59′0″N 8°31′0″E46°59′0″N...

                              Image
                              Tidlegare republikkarSkipingar i 1433Kommunar i SchwyzDistrikt i SchwyzNedleggingar i 1798Stader ved Vierwaldstättersee kommunekantonenSchwyzSveitskm²VierwaldstätterseeLauerz Gersau Kommunevåpenet til Gersau Plassering Styresmakter Land   Sveits Kanton kantonen Schwyz Distrikt Gersau Språk tysk Geografi Flatevidd  - kommune 23,70 km² Innbyggjarar  - kommune (desember 2010)    - folketettleik 2 094   88,4 /km² Koordinatar 46°59′0″N 8°31′0″E  Koordinatar: 46°59′0″N 8°31′0″E  Høgd over havet 435 moh Diverse annan informasjon Postnummer 6442 Nettstad: www.gersau.ch Freistaat und Republik Gersau Fristaten og republikken Gersau By-republikk og alliert protektorat i Det gamle sveitsiske eidsforundet ←   ← 1433–1798 1814–1818 →   → Hovudstad Gersau Styreform Republikk Historisk periode Tidleg moderne tid  - Kjøpte fridom frå
                              Read more

                              Hestehale Innhaldsliste Hestehale på kvinner | Hestehale på menn | Galleri | Sjå òg |...

                              Image
                              Frisyrar frisyrehårhårstrikkhårklemmerøyrethesthovudverkeuropeisk motejobbtreningfletterparykkarungdomsopprørmotkulturmusikararmotorsykkelentusiastarpolitietpolitiskforretningverksemdfiskeflette Hestehale sett bakfrå. Hestehale på ein mannleg indianarleiar. Hestehale er ein frisyre der ein samlar det meste eller alt av langt hår [1] i bakhovudet med ein hårstrikk eller liknande. Avhengig av mote kan det òg vere vanleg å sette opp hestehalen med hårklemmer eller å ha hestehalen hengande på den eine sida av hovudet ved å samle håret ved øyret. Hestehale er ein vanleg frisyre for jenter og for gutar med langt hår. Han har fått namnet sitt etter likskapen med halen på ein hest. For stramme hestehaler kan føre til hårtap, [2] i nokre høve også hovudverk. [3] Innhaldsliste 1 Hestehale på kvinner 2 Hestehale på menn 3 Galleri 4 Sjå òg 5 Bakgrunnsstoff 6 Kjelder Hestehale på kvinner | Ungjente med hestehale som held håret unna ansiktet
                              Read more

                              What is the “three and three hundred thousand syndrome”?Who wrote the book Arena?What five creatures were...

                              Image
                              What happens when the centripetal force is equal and opposite to the centrifugal force? Why is "la Gestapo" feminine? Nested Dynamic SOQL Query How are passwords stolen from companies if they only store hashes? Exposing a company lying about themselves in a tightly knit industry: Is my career at risk on the long run? Can "few" be used as a subject? If so, what is the rule? Why are there no stars visible in cislunar space? categorizing a variable turns it from insignificant to significant Do I need an EFI partition for each 18.04 ubuntu I have on my HD? What are the consequences of changing the number of hours in
                              Read more