Radius of convergence for three series












0












$begingroup$


I need to find the radius of convergence of the 3 following series, but there are no solutions, so I don't know if my steps are correct.



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

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

3) $sum_{n=0}^{infty} frac{1}{1+n3^n}x^n$





1) For the first, $a_n=1$ iff $exists k : k!=n, k in mathbb{N}$, zero elsewhere. So the radius is given by $frac{1}{lim : sup_{n to infty}sqrt[n]{|a_n|}}$ and the limit superior is max{1,1,0,0,0,1,0,0,....}=1



Thus the radius of convergence is 1.

On the border, we have 1 and -1 to an even power (factorial), which always yields 1, and not 0. So it diverges for both endpoints.



2) Here, we use $rho = frac{a_n}{a_{n+1}}$ So we get : $lim_{n to infty}(frac{1+2^{n+1}}{1+2^{n}})=2$ On the border, we have $frac{2^n}{1+2^n}=frac{2^n}{2^n}frac{1}{frac{1}{2^n}+1}=1$ for $n to infty$, so not toward 0, and thus it diverges.

We also have $frac{(-1)^n(2)^n}{1+2^n}$, which is an alternating serie. According to the alternating serie test, it should be converging toward 0 for $n to infty$ which isn't the case here, so it also diverges for $x=-2$



3) Here, we get $frac{1+3^{n+1}(n+1)}{1+3^nn}= frac{3^nn}{3^nn}frac{frac{1}{3^nn}+frac{3(n+1)}{n}}{1+ frac{1}{3^nn}}=frac{3}{1}=3$ for $n to infty$


On the border, we have $frac{3^n}{1+n3^n}=frac{1}{frac{1}{3^n}+n}$ and $frac{(-1)^n3^n}{1+n3^n}=frac{(-1)^n}{frac{1}{3^n}+n}$. Both cases give the harmonic serie, which diverges. So no convergence on the border.





Are my results correct ? If not, what did I do wrong or forgot ?

Thanks for your help !










share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I need to find the radius of convergence of the 3 following series, but there are no solutions, so I don't know if my steps are correct.



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

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

    3) $sum_{n=0}^{infty} frac{1}{1+n3^n}x^n$





    1) For the first, $a_n=1$ iff $exists k : k!=n, k in mathbb{N}$, zero elsewhere. So the radius is given by $frac{1}{lim : sup_{n to infty}sqrt[n]{|a_n|}}$ and the limit superior is max{1,1,0,0,0,1,0,0,....}=1



    Thus the radius of convergence is 1.

    On the border, we have 1 and -1 to an even power (factorial), which always yields 1, and not 0. So it diverges for both endpoints.



    2) Here, we use $rho = frac{a_n}{a_{n+1}}$ So we get : $lim_{n to infty}(frac{1+2^{n+1}}{1+2^{n}})=2$ On the border, we have $frac{2^n}{1+2^n}=frac{2^n}{2^n}frac{1}{frac{1}{2^n}+1}=1$ for $n to infty$, so not toward 0, and thus it diverges.

    We also have $frac{(-1)^n(2)^n}{1+2^n}$, which is an alternating serie. According to the alternating serie test, it should be converging toward 0 for $n to infty$ which isn't the case here, so it also diverges for $x=-2$



    3) Here, we get $frac{1+3^{n+1}(n+1)}{1+3^nn}= frac{3^nn}{3^nn}frac{frac{1}{3^nn}+frac{3(n+1)}{n}}{1+ frac{1}{3^nn}}=frac{3}{1}=3$ for $n to infty$


    On the border, we have $frac{3^n}{1+n3^n}=frac{1}{frac{1}{3^n}+n}$ and $frac{(-1)^n3^n}{1+n3^n}=frac{(-1)^n}{frac{1}{3^n}+n}$. Both cases give the harmonic serie, which diverges. So no convergence on the border.





    Are my results correct ? If not, what did I do wrong or forgot ?

    Thanks for your help !










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I need to find the radius of convergence of the 3 following series, but there are no solutions, so I don't know if my steps are correct.



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

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

      3) $sum_{n=0}^{infty} frac{1}{1+n3^n}x^n$





      1) For the first, $a_n=1$ iff $exists k : k!=n, k in mathbb{N}$, zero elsewhere. So the radius is given by $frac{1}{lim : sup_{n to infty}sqrt[n]{|a_n|}}$ and the limit superior is max{1,1,0,0,0,1,0,0,....}=1



      Thus the radius of convergence is 1.

      On the border, we have 1 and -1 to an even power (factorial), which always yields 1, and not 0. So it diverges for both endpoints.



      2) Here, we use $rho = frac{a_n}{a_{n+1}}$ So we get : $lim_{n to infty}(frac{1+2^{n+1}}{1+2^{n}})=2$ On the border, we have $frac{2^n}{1+2^n}=frac{2^n}{2^n}frac{1}{frac{1}{2^n}+1}=1$ for $n to infty$, so not toward 0, and thus it diverges.

      We also have $frac{(-1)^n(2)^n}{1+2^n}$, which is an alternating serie. According to the alternating serie test, it should be converging toward 0 for $n to infty$ which isn't the case here, so it also diverges for $x=-2$



      3) Here, we get $frac{1+3^{n+1}(n+1)}{1+3^nn}= frac{3^nn}{3^nn}frac{frac{1}{3^nn}+frac{3(n+1)}{n}}{1+ frac{1}{3^nn}}=frac{3}{1}=3$ for $n to infty$


      On the border, we have $frac{3^n}{1+n3^n}=frac{1}{frac{1}{3^n}+n}$ and $frac{(-1)^n3^n}{1+n3^n}=frac{(-1)^n}{frac{1}{3^n}+n}$. Both cases give the harmonic serie, which diverges. So no convergence on the border.





      Are my results correct ? If not, what did I do wrong or forgot ?

      Thanks for your help !










      share|cite|improve this question











      $endgroup$




      I need to find the radius of convergence of the 3 following series, but there are no solutions, so I don't know if my steps are correct.



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

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

      3) $sum_{n=0}^{infty} frac{1}{1+n3^n}x^n$





      1) For the first, $a_n=1$ iff $exists k : k!=n, k in mathbb{N}$, zero elsewhere. So the radius is given by $frac{1}{lim : sup_{n to infty}sqrt[n]{|a_n|}}$ and the limit superior is max{1,1,0,0,0,1,0,0,....}=1



      Thus the radius of convergence is 1.

      On the border, we have 1 and -1 to an even power (factorial), which always yields 1, and not 0. So it diverges for both endpoints.



      2) Here, we use $rho = frac{a_n}{a_{n+1}}$ So we get : $lim_{n to infty}(frac{1+2^{n+1}}{1+2^{n}})=2$ On the border, we have $frac{2^n}{1+2^n}=frac{2^n}{2^n}frac{1}{frac{1}{2^n}+1}=1$ for $n to infty$, so not toward 0, and thus it diverges.

      We also have $frac{(-1)^n(2)^n}{1+2^n}$, which is an alternating serie. According to the alternating serie test, it should be converging toward 0 for $n to infty$ which isn't the case here, so it also diverges for $x=-2$



      3) Here, we get $frac{1+3^{n+1}(n+1)}{1+3^nn}= frac{3^nn}{3^nn}frac{frac{1}{3^nn}+frac{3(n+1)}{n}}{1+ frac{1}{3^nn}}=frac{3}{1}=3$ for $n to infty$


      On the border, we have $frac{3^n}{1+n3^n}=frac{1}{frac{1}{3^n}+n}$ and $frac{(-1)^n3^n}{1+n3^n}=frac{(-1)^n}{frac{1}{3^n}+n}$. Both cases give the harmonic serie, which diverges. So no convergence on the border.





      Are my results correct ? If not, what did I do wrong or forgot ?

      Thanks for your help !







      real-analysis sequences-and-series convergence power-series divergent-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 20 '18 at 13:04









      David G. Stork

      12.1k41735




      12.1k41735










      asked Dec 20 '18 at 12:58









      PoujhPoujh

      6111516




      6111516






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          For the first



          the ratio test gives the limit



          $$lim_{nto+infty}leftlvert frac{x^{(n+1)!}}{x^{n!}}rightrvert=lim_{nto+infty} lvert xrvert^{ncdot n!}$$



          if $lvert xrvert<1$, the limit is zero.
          if $lvert xrvert>1$ the limit is infinity.



          the radius is $R=1$.



          When you look for the radius only, you do not need the convergence on the border.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Something is wrong with the total rep.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:13










          • $begingroup$
            What do you mean exactly with "wrong with the total rep." ? Ans just for 1), your method seems in fact a lot easier but would my method also be correct for 1) ?
            $endgroup$
            – Poujh
            Dec 20 '18 at 13:15










          • $begingroup$
            @Poujh This post if for managers.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:16






          • 1




            $begingroup$
            @Poujh Your answerS are correct.but no need for the border.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:19






          • 1




            $begingroup$
            I’m sorry, I don’t follow. How is $frac{x^{(n+1)!}}{x^{n!}} not equal to $frac{x^{n*n!}$
            $endgroup$
            – R.Jackson
            Dec 20 '18 at 13:42



















          1












          $begingroup$

          $(1)$ $$displaystylesum_{n=0}^{infty}x^{n!}<sum_{n=0}^{infty}x^n$$ and $displaystylesum_{n=0}^{infty}x^n$ converges for $|x|<1$.
          So Radius of convergence is $1$



          $(2)$ Use D'Alembert's ratio test: $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+2^n}{1+2^{n+1}}right||x|<1 implies |x|<2$$
          Radius of convergence is $2$.



          $(3)$ $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+n3^n}{1+(n+1)3^{n+1}}right||x|<1 implies |x|<3$$ Radius of convergence is $3$



          (When you require radius of convergence, convergence of boundary is not necessary)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Just for 1), do you think my method would also be correct (even if yours is easier) ?
            $endgroup$
            – Poujh
            Dec 20 '18 at 13:19








          • 1




            $begingroup$
            @Poujh: Yes it's correct. Your argument for $(1)$ is better than mine.
            $endgroup$
            – Yadati Kiran
            Dec 20 '18 at 13:22














          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%2f3047504%2fradius-of-convergence-for-three-series%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









          2












          $begingroup$

          For the first



          the ratio test gives the limit



          $$lim_{nto+infty}leftlvert frac{x^{(n+1)!}}{x^{n!}}rightrvert=lim_{nto+infty} lvert xrvert^{ncdot n!}$$



          if $lvert xrvert<1$, the limit is zero.
          if $lvert xrvert>1$ the limit is infinity.



          the radius is $R=1$.



          When you look for the radius only, you do not need the convergence on the border.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Something is wrong with the total rep.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:13










          • $begingroup$
            What do you mean exactly with "wrong with the total rep." ? Ans just for 1), your method seems in fact a lot easier but would my method also be correct for 1) ?
            $endgroup$
            – Poujh
            Dec 20 '18 at 13:15










          • $begingroup$
            @Poujh This post if for managers.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:16






          • 1




            $begingroup$
            @Poujh Your answerS are correct.but no need for the border.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:19






          • 1




            $begingroup$
            I’m sorry, I don’t follow. How is $frac{x^{(n+1)!}}{x^{n!}} not equal to $frac{x^{n*n!}$
            $endgroup$
            – R.Jackson
            Dec 20 '18 at 13:42
















          2












          $begingroup$

          For the first



          the ratio test gives the limit



          $$lim_{nto+infty}leftlvert frac{x^{(n+1)!}}{x^{n!}}rightrvert=lim_{nto+infty} lvert xrvert^{ncdot n!}$$



          if $lvert xrvert<1$, the limit is zero.
          if $lvert xrvert>1$ the limit is infinity.



          the radius is $R=1$.



          When you look for the radius only, you do not need the convergence on the border.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Something is wrong with the total rep.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:13










          • $begingroup$
            What do you mean exactly with "wrong with the total rep." ? Ans just for 1), your method seems in fact a lot easier but would my method also be correct for 1) ?
            $endgroup$
            – Poujh
            Dec 20 '18 at 13:15










          • $begingroup$
            @Poujh This post if for managers.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:16






          • 1




            $begingroup$
            @Poujh Your answerS are correct.but no need for the border.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:19






          • 1




            $begingroup$
            I’m sorry, I don’t follow. How is $frac{x^{(n+1)!}}{x^{n!}} not equal to $frac{x^{n*n!}$
            $endgroup$
            – R.Jackson
            Dec 20 '18 at 13:42














          2












          2








          2





          $begingroup$

          For the first



          the ratio test gives the limit



          $$lim_{nto+infty}leftlvert frac{x^{(n+1)!}}{x^{n!}}rightrvert=lim_{nto+infty} lvert xrvert^{ncdot n!}$$



          if $lvert xrvert<1$, the limit is zero.
          if $lvert xrvert>1$ the limit is infinity.



          the radius is $R=1$.



          When you look for the radius only, you do not need the convergence on the border.






          share|cite|improve this answer











          $endgroup$



          For the first



          the ratio test gives the limit



          $$lim_{nto+infty}leftlvert frac{x^{(n+1)!}}{x^{n!}}rightrvert=lim_{nto+infty} lvert xrvert^{ncdot n!}$$



          if $lvert xrvert<1$, the limit is zero.
          if $lvert xrvert>1$ the limit is infinity.



          the radius is $R=1$.



          When you look for the radius only, you do not need the convergence on the border.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 20 '18 at 17:30









          user10354138

          7,5472925




          7,5472925










          answered Dec 20 '18 at 13:09









          hamam_Abdallahhamam_Abdallah

          38.1k21634




          38.1k21634












          • $begingroup$
            Something is wrong with the total rep.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:13










          • $begingroup$
            What do you mean exactly with "wrong with the total rep." ? Ans just for 1), your method seems in fact a lot easier but would my method also be correct for 1) ?
            $endgroup$
            – Poujh
            Dec 20 '18 at 13:15










          • $begingroup$
            @Poujh This post if for managers.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:16






          • 1




            $begingroup$
            @Poujh Your answerS are correct.but no need for the border.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:19






          • 1




            $begingroup$
            I’m sorry, I don’t follow. How is $frac{x^{(n+1)!}}{x^{n!}} not equal to $frac{x^{n*n!}$
            $endgroup$
            – R.Jackson
            Dec 20 '18 at 13:42


















          • $begingroup$
            Something is wrong with the total rep.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:13










          • $begingroup$
            What do you mean exactly with "wrong with the total rep." ? Ans just for 1), your method seems in fact a lot easier but would my method also be correct for 1) ?
            $endgroup$
            – Poujh
            Dec 20 '18 at 13:15










          • $begingroup$
            @Poujh This post if for managers.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:16






          • 1




            $begingroup$
            @Poujh Your answerS are correct.but no need for the border.
            $endgroup$
            – hamam_Abdallah
            Dec 20 '18 at 13:19






          • 1




            $begingroup$
            I’m sorry, I don’t follow. How is $frac{x^{(n+1)!}}{x^{n!}} not equal to $frac{x^{n*n!}$
            $endgroup$
            – R.Jackson
            Dec 20 '18 at 13:42
















          $begingroup$
          Something is wrong with the total rep.
          $endgroup$
          – hamam_Abdallah
          Dec 20 '18 at 13:13




          $begingroup$
          Something is wrong with the total rep.
          $endgroup$
          – hamam_Abdallah
          Dec 20 '18 at 13:13












          $begingroup$
          What do you mean exactly with "wrong with the total rep." ? Ans just for 1), your method seems in fact a lot easier but would my method also be correct for 1) ?
          $endgroup$
          – Poujh
          Dec 20 '18 at 13:15




          $begingroup$
          What do you mean exactly with "wrong with the total rep." ? Ans just for 1), your method seems in fact a lot easier but would my method also be correct for 1) ?
          $endgroup$
          – Poujh
          Dec 20 '18 at 13:15












          $begingroup$
          @Poujh This post if for managers.
          $endgroup$
          – hamam_Abdallah
          Dec 20 '18 at 13:16




          $begingroup$
          @Poujh This post if for managers.
          $endgroup$
          – hamam_Abdallah
          Dec 20 '18 at 13:16




          1




          1




          $begingroup$
          @Poujh Your answerS are correct.but no need for the border.
          $endgroup$
          – hamam_Abdallah
          Dec 20 '18 at 13:19




          $begingroup$
          @Poujh Your answerS are correct.but no need for the border.
          $endgroup$
          – hamam_Abdallah
          Dec 20 '18 at 13:19




          1




          1




          $begingroup$
          I’m sorry, I don’t follow. How is $frac{x^{(n+1)!}}{x^{n!}} not equal to $frac{x^{n*n!}$
          $endgroup$
          – R.Jackson
          Dec 20 '18 at 13:42




          $begingroup$
          I’m sorry, I don’t follow. How is $frac{x^{(n+1)!}}{x^{n!}} not equal to $frac{x^{n*n!}$
          $endgroup$
          – R.Jackson
          Dec 20 '18 at 13:42











          1












          $begingroup$

          $(1)$ $$displaystylesum_{n=0}^{infty}x^{n!}<sum_{n=0}^{infty}x^n$$ and $displaystylesum_{n=0}^{infty}x^n$ converges for $|x|<1$.
          So Radius of convergence is $1$



          $(2)$ Use D'Alembert's ratio test: $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+2^n}{1+2^{n+1}}right||x|<1 implies |x|<2$$
          Radius of convergence is $2$.



          $(3)$ $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+n3^n}{1+(n+1)3^{n+1}}right||x|<1 implies |x|<3$$ Radius of convergence is $3$



          (When you require radius of convergence, convergence of boundary is not necessary)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Just for 1), do you think my method would also be correct (even if yours is easier) ?
            $endgroup$
            – Poujh
            Dec 20 '18 at 13:19








          • 1




            $begingroup$
            @Poujh: Yes it's correct. Your argument for $(1)$ is better than mine.
            $endgroup$
            – Yadati Kiran
            Dec 20 '18 at 13:22


















          1












          $begingroup$

          $(1)$ $$displaystylesum_{n=0}^{infty}x^{n!}<sum_{n=0}^{infty}x^n$$ and $displaystylesum_{n=0}^{infty}x^n$ converges for $|x|<1$.
          So Radius of convergence is $1$



          $(2)$ Use D'Alembert's ratio test: $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+2^n}{1+2^{n+1}}right||x|<1 implies |x|<2$$
          Radius of convergence is $2$.



          $(3)$ $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+n3^n}{1+(n+1)3^{n+1}}right||x|<1 implies |x|<3$$ Radius of convergence is $3$



          (When you require radius of convergence, convergence of boundary is not necessary)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Just for 1), do you think my method would also be correct (even if yours is easier) ?
            $endgroup$
            – Poujh
            Dec 20 '18 at 13:19








          • 1




            $begingroup$
            @Poujh: Yes it's correct. Your argument for $(1)$ is better than mine.
            $endgroup$
            – Yadati Kiran
            Dec 20 '18 at 13:22
















          1












          1








          1





          $begingroup$

          $(1)$ $$displaystylesum_{n=0}^{infty}x^{n!}<sum_{n=0}^{infty}x^n$$ and $displaystylesum_{n=0}^{infty}x^n$ converges for $|x|<1$.
          So Radius of convergence is $1$



          $(2)$ Use D'Alembert's ratio test: $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+2^n}{1+2^{n+1}}right||x|<1 implies |x|<2$$
          Radius of convergence is $2$.



          $(3)$ $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+n3^n}{1+(n+1)3^{n+1}}right||x|<1 implies |x|<3$$ Radius of convergence is $3$



          (When you require radius of convergence, convergence of boundary is not necessary)






          share|cite|improve this answer











          $endgroup$



          $(1)$ $$displaystylesum_{n=0}^{infty}x^{n!}<sum_{n=0}^{infty}x^n$$ and $displaystylesum_{n=0}^{infty}x^n$ converges for $|x|<1$.
          So Radius of convergence is $1$



          $(2)$ Use D'Alembert's ratio test: $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+2^n}{1+2^{n+1}}right||x|<1 implies |x|<2$$
          Radius of convergence is $2$.



          $(3)$ $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+n3^n}{1+(n+1)3^{n+1}}right||x|<1 implies |x|<3$$ Radius of convergence is $3$



          (When you require radius of convergence, convergence of boundary is not necessary)







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 20 '18 at 13:30

























          answered Dec 20 '18 at 13:10









          Yadati KiranYadati Kiran

          2,1161622




          2,1161622












          • $begingroup$
            Just for 1), do you think my method would also be correct (even if yours is easier) ?
            $endgroup$
            – Poujh
            Dec 20 '18 at 13:19








          • 1




            $begingroup$
            @Poujh: Yes it's correct. Your argument for $(1)$ is better than mine.
            $endgroup$
            – Yadati Kiran
            Dec 20 '18 at 13:22




















          • $begingroup$
            Just for 1), do you think my method would also be correct (even if yours is easier) ?
            $endgroup$
            – Poujh
            Dec 20 '18 at 13:19








          • 1




            $begingroup$
            @Poujh: Yes it's correct. Your argument for $(1)$ is better than mine.
            $endgroup$
            – Yadati Kiran
            Dec 20 '18 at 13:22


















          $begingroup$
          Just for 1), do you think my method would also be correct (even if yours is easier) ?
          $endgroup$
          – Poujh
          Dec 20 '18 at 13:19






          $begingroup$
          Just for 1), do you think my method would also be correct (even if yours is easier) ?
          $endgroup$
          – Poujh
          Dec 20 '18 at 13:19






          1




          1




          $begingroup$
          @Poujh: Yes it's correct. Your argument for $(1)$ is better than mine.
          $endgroup$
          – Yadati Kiran
          Dec 20 '18 at 13:22






          $begingroup$
          @Poujh: Yes it's correct. Your argument for $(1)$ is better than mine.
          $endgroup$
          – Yadati Kiran
          Dec 20 '18 at 13:22




















          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%2f3047504%2fradius-of-convergence-for-three-series%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

          Plaza Victoria

          Puebla de Zaragoza

          Musa