Radius of convergence of $sumlimits_{n=1}^infty((frac{1}{4})^n+(frac{1}{3})^n)x^n$












1












$begingroup$


according to wolfram alpha the radius is 3. I'm struggeling with the proof and would be very glad if someone could take a short look. Here my approach using the root test:



$a_n:=sqrt[n]{(frac{1}{4})^n+(frac{1}{3})^n}<sqrt[n]{(frac{1}{3})^n+(frac{1}{3})^n}=sqrt[n]{2(frac{1}{3})^n}=sqrt[n]2(frac{1}{3})=:b_n$



$implies limsuplimits_{nrightarrowinfty}a_n<limsuplimits_{nrightarrowinfty}b_n=frac{1}{3}$



$left(implies forall x in mathbb{R}: |x|<3 implies exists alpha in mathbb{R}: limlimits_{nrightarrowinfty}sum_{n=1}^infty b_nx^n=alpha right)$



But that does not necessarily mean that $limlimits_{nrightarrowinfty}sum_{n=1}^infty a_nx^n$ has the same radius of convergence, does it? The radius could still be smaller, right?



Hints to a better proof are greatly appreciated! :)






real-analysis sequences-and-series power-series







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    according to wolfram alpha the radius is 3. I'm struggeling with the proof and would be very glad if someone could take a short look. Here my approach using the root test:



    $a_n:=sqrt[n]{(frac{1}{4})^n+(frac{1}{3})^n}<sqrt[n]{(frac{1}{3})^n+(frac{1}{3})^n}=sqrt[n]{2(frac{1}{3})^n}=sqrt[n]2(frac{1}{3})=:b_n$



    $implies limsuplimits_{nrightarrowinfty}a_n<limsuplimits_{nrightarrowinfty}b_n=frac{1}{3}$



    $left(implies forall x in mathbb{R}: |x|<3 implies exists alpha in mathbb{R}: limlimits_{nrightarrowinfty}sum_{n=1}^infty b_nx^n=alpha right)$



    But that does not necessarily mean that $limlimits_{nrightarrowinfty}sum_{n=1}^infty a_nx^n$ has the same radius of convergence, does it? The radius could still be smaller, right?



    Hints to a better proof are greatly appreciated! :)






    real-analysis sequences-and-series power-series







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      according to wolfram alpha the radius is 3. I'm struggeling with the proof and would be very glad if someone could take a short look. Here my approach using the root test:



      $a_n:=sqrt[n]{(frac{1}{4})^n+(frac{1}{3})^n}<sqrt[n]{(frac{1}{3})^n+(frac{1}{3})^n}=sqrt[n]{2(frac{1}{3})^n}=sqrt[n]2(frac{1}{3})=:b_n$



      $implies limsuplimits_{nrightarrowinfty}a_n<limsuplimits_{nrightarrowinfty}b_n=frac{1}{3}$



      $left(implies forall x in mathbb{R}: |x|<3 implies exists alpha in mathbb{R}: limlimits_{nrightarrowinfty}sum_{n=1}^infty b_nx^n=alpha right)$



      But that does not necessarily mean that $limlimits_{nrightarrowinfty}sum_{n=1}^infty a_nx^n$ has the same radius of convergence, does it? The radius could still be smaller, right?



      Hints to a better proof are greatly appreciated! :)






      real-analysis sequences-and-series power-series







      share|cite|improve this question











      $endgroup$




      according to wolfram alpha the radius is 3. I'm struggeling with the proof and would be very glad if someone could take a short look. Here my approach using the root test:



      $a_n:=sqrt[n]{(frac{1}{4})^n+(frac{1}{3})^n}<sqrt[n]{(frac{1}{3})^n+(frac{1}{3})^n}=sqrt[n]{2(frac{1}{3})^n}=sqrt[n]2(frac{1}{3})=:b_n$



      $implies limsuplimits_{nrightarrowinfty}a_n<limsuplimits_{nrightarrowinfty}b_n=frac{1}{3}$



      $left(implies forall x in mathbb{R}: |x|<3 implies exists alpha in mathbb{R}: limlimits_{nrightarrowinfty}sum_{n=1}^infty b_nx^n=alpha right)$



      But that does not necessarily mean that $limlimits_{nrightarrowinfty}sum_{n=1}^infty a_nx^n$ has the same radius of convergence, does it? The radius could still be smaller, right?



      Hints to a better proof are greatly appreciated! :)






      real-analysis sequences-and-series power-series




      real-analysis sequences-and-series power-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 4 '18 at 22:18









      Did

      246k23221457




      246k23221457










      asked Dec 4 '18 at 22:00









      DDevelopsDDevelops

      533




      533






















          3 Answers
          3






          active

          oldest

          votes


















          2












          $begingroup$

          Better if you compute $lim_{ntoinfty}frac{a_{n+1}}{a_n}$. In your case,
          $$frac{1/4^{n+1}+1/3^{n+1}}{1/4^n+1/3^n}cdot frac{3^{n+1}}{3^{n+1}} =frac{(3/4)^{n+1}+1}{3cdot (3/4)^n+3}overset{ntoinfty}{longrightarrow} frac{1}{3}. $$



          Hence, $R=3$.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            You can use also the ratio test to get the radius of convergence, I mean, if $a_nneq 0$ for all $nge N$ for enough large $NinBbb N$ and the limit



            $$rho=lim_{ntoinfty}frac{|a_n|}{|a_{n+1}|}$$



            exists then $rho$ is the radius of convergence of the power series. In this case we have that



            $$frac{|a_n|}{|a_{n+1}|}=frac{frac{4^n+3^n}{4^n 3^n}}{frac{4^{n+1}+3^{n+1}}{4^{n+1}3^{n+1}}}=12frac{4^n+3^n}{4^{n+1}+3^{n+1}}=12frac{1+(3/4)^n}{4+3(3/4)^n}tofrac{12}{4}=3$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              This is the ratio test, not the root test (and yes, it works smoothly in the present case).
              $endgroup$
              – Did
              Dec 4 '18 at 22:16





















            1












            $begingroup$

            We have that



            $$|x|cdotsqrt[n]{left(frac{1}{4}right)^n+left(frac{1}{3}right)^n}=|x|cdotfrac13sqrt[n]{left(frac{3}{4}right)^n+1} to|x|cdotfrac13cdot 1 = |x|cdotfrac13<1$$



            and therefore the radius of convergence is $3$.



            Moreover note that




            • for $x=3 implies left(left(frac{1}{4}right)^n+left(frac{1}{3}right)^nright)x^n=left(frac{3}{4}right)^n+1$


            • for $x=-3 implies left(left(frac{1}{4}right)^n+left(frac{1}{3}right)^nright)x^n=left(-frac{3}{4}right)^n+(-1)^n$







            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Why the downvote, something wrong?
              $endgroup$
              – gimusi
              Dec 4 '18 at 22:21











            Your Answer





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            3 Answers
            3






            active

            oldest

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            3 Answers
            3






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Better if you compute $lim_{ntoinfty}frac{a_{n+1}}{a_n}$. In your case,
            $$frac{1/4^{n+1}+1/3^{n+1}}{1/4^n+1/3^n}cdot frac{3^{n+1}}{3^{n+1}} =frac{(3/4)^{n+1}+1}{3cdot (3/4)^n+3}overset{ntoinfty}{longrightarrow} frac{1}{3}. $$



            Hence, $R=3$.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Better if you compute $lim_{ntoinfty}frac{a_{n+1}}{a_n}$. In your case,
              $$frac{1/4^{n+1}+1/3^{n+1}}{1/4^n+1/3^n}cdot frac{3^{n+1}}{3^{n+1}} =frac{(3/4)^{n+1}+1}{3cdot (3/4)^n+3}overset{ntoinfty}{longrightarrow} frac{1}{3}. $$



              Hence, $R=3$.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Better if you compute $lim_{ntoinfty}frac{a_{n+1}}{a_n}$. In your case,
                $$frac{1/4^{n+1}+1/3^{n+1}}{1/4^n+1/3^n}cdot frac{3^{n+1}}{3^{n+1}} =frac{(3/4)^{n+1}+1}{3cdot (3/4)^n+3}overset{ntoinfty}{longrightarrow} frac{1}{3}. $$



                Hence, $R=3$.






                share|cite|improve this answer









                $endgroup$



                Better if you compute $lim_{ntoinfty}frac{a_{n+1}}{a_n}$. In your case,
                $$frac{1/4^{n+1}+1/3^{n+1}}{1/4^n+1/3^n}cdot frac{3^{n+1}}{3^{n+1}} =frac{(3/4)^{n+1}+1}{3cdot (3/4)^n+3}overset{ntoinfty}{longrightarrow} frac{1}{3}. $$



                Hence, $R=3$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 4 '18 at 22:11









                Tito EliatronTito Eliatron

                1,448622




                1,448622























                    1












                    $begingroup$

                    You can use also the ratio test to get the radius of convergence, I mean, if $a_nneq 0$ for all $nge N$ for enough large $NinBbb N$ and the limit



                    $$rho=lim_{ntoinfty}frac{|a_n|}{|a_{n+1}|}$$



                    exists then $rho$ is the radius of convergence of the power series. In this case we have that



                    $$frac{|a_n|}{|a_{n+1}|}=frac{frac{4^n+3^n}{4^n 3^n}}{frac{4^{n+1}+3^{n+1}}{4^{n+1}3^{n+1}}}=12frac{4^n+3^n}{4^{n+1}+3^{n+1}}=12frac{1+(3/4)^n}{4+3(3/4)^n}tofrac{12}{4}=3$$






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      This is the ratio test, not the root test (and yes, it works smoothly in the present case).
                      $endgroup$
                      – Did
                      Dec 4 '18 at 22:16


















                    1












                    $begingroup$

                    You can use also the ratio test to get the radius of convergence, I mean, if $a_nneq 0$ for all $nge N$ for enough large $NinBbb N$ and the limit



                    $$rho=lim_{ntoinfty}frac{|a_n|}{|a_{n+1}|}$$



                    exists then $rho$ is the radius of convergence of the power series. In this case we have that



                    $$frac{|a_n|}{|a_{n+1}|}=frac{frac{4^n+3^n}{4^n 3^n}}{frac{4^{n+1}+3^{n+1}}{4^{n+1}3^{n+1}}}=12frac{4^n+3^n}{4^{n+1}+3^{n+1}}=12frac{1+(3/4)^n}{4+3(3/4)^n}tofrac{12}{4}=3$$






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      This is the ratio test, not the root test (and yes, it works smoothly in the present case).
                      $endgroup$
                      – Did
                      Dec 4 '18 at 22:16
















                    1












                    1








                    1





                    $begingroup$

                    You can use also the ratio test to get the radius of convergence, I mean, if $a_nneq 0$ for all $nge N$ for enough large $NinBbb N$ and the limit



                    $$rho=lim_{ntoinfty}frac{|a_n|}{|a_{n+1}|}$$



                    exists then $rho$ is the radius of convergence of the power series. In this case we have that



                    $$frac{|a_n|}{|a_{n+1}|}=frac{frac{4^n+3^n}{4^n 3^n}}{frac{4^{n+1}+3^{n+1}}{4^{n+1}3^{n+1}}}=12frac{4^n+3^n}{4^{n+1}+3^{n+1}}=12frac{1+(3/4)^n}{4+3(3/4)^n}tofrac{12}{4}=3$$






                    share|cite|improve this answer











                    $endgroup$



                    You can use also the ratio test to get the radius of convergence, I mean, if $a_nneq 0$ for all $nge N$ for enough large $NinBbb N$ and the limit



                    $$rho=lim_{ntoinfty}frac{|a_n|}{|a_{n+1}|}$$



                    exists then $rho$ is the radius of convergence of the power series. In this case we have that



                    $$frac{|a_n|}{|a_{n+1}|}=frac{frac{4^n+3^n}{4^n 3^n}}{frac{4^{n+1}+3^{n+1}}{4^{n+1}3^{n+1}}}=12frac{4^n+3^n}{4^{n+1}+3^{n+1}}=12frac{1+(3/4)^n}{4+3(3/4)^n}tofrac{12}{4}=3$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 4 '18 at 22:16

























                    answered Dec 4 '18 at 22:16









                    MasacrosoMasacroso

                    13k41746




                    13k41746












                    • $begingroup$
                      This is the ratio test, not the root test (and yes, it works smoothly in the present case).
                      $endgroup$
                      – Did
                      Dec 4 '18 at 22:16




















                    • $begingroup$
                      This is the ratio test, not the root test (and yes, it works smoothly in the present case).
                      $endgroup$
                      – Did
                      Dec 4 '18 at 22:16


















                    $begingroup$
                    This is the ratio test, not the root test (and yes, it works smoothly in the present case).
                    $endgroup$
                    – Did
                    Dec 4 '18 at 22:16






                    $begingroup$
                    This is the ratio test, not the root test (and yes, it works smoothly in the present case).
                    $endgroup$
                    – Did
                    Dec 4 '18 at 22:16













                    1












                    $begingroup$

                    We have that



                    $$|x|cdotsqrt[n]{left(frac{1}{4}right)^n+left(frac{1}{3}right)^n}=|x|cdotfrac13sqrt[n]{left(frac{3}{4}right)^n+1} to|x|cdotfrac13cdot 1 = |x|cdotfrac13<1$$



                    and therefore the radius of convergence is $3$.



                    Moreover note that




                    • for $x=3 implies left(left(frac{1}{4}right)^n+left(frac{1}{3}right)^nright)x^n=left(frac{3}{4}right)^n+1$


                    • for $x=-3 implies left(left(frac{1}{4}right)^n+left(frac{1}{3}right)^nright)x^n=left(-frac{3}{4}right)^n+(-1)^n$







                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Why the downvote, something wrong?
                      $endgroup$
                      – gimusi
                      Dec 4 '18 at 22:21
















                    1












                    $begingroup$

                    We have that



                    $$|x|cdotsqrt[n]{left(frac{1}{4}right)^n+left(frac{1}{3}right)^n}=|x|cdotfrac13sqrt[n]{left(frac{3}{4}right)^n+1} to|x|cdotfrac13cdot 1 = |x|cdotfrac13<1$$



                    and therefore the radius of convergence is $3$.



                    Moreover note that




                    • for $x=3 implies left(left(frac{1}{4}right)^n+left(frac{1}{3}right)^nright)x^n=left(frac{3}{4}right)^n+1$


                    • for $x=-3 implies left(left(frac{1}{4}right)^n+left(frac{1}{3}right)^nright)x^n=left(-frac{3}{4}right)^n+(-1)^n$







                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Why the downvote, something wrong?
                      $endgroup$
                      – gimusi
                      Dec 4 '18 at 22:21














                    1












                    1








                    1





                    $begingroup$

                    We have that



                    $$|x|cdotsqrt[n]{left(frac{1}{4}right)^n+left(frac{1}{3}right)^n}=|x|cdotfrac13sqrt[n]{left(frac{3}{4}right)^n+1} to|x|cdotfrac13cdot 1 = |x|cdotfrac13<1$$



                    and therefore the radius of convergence is $3$.



                    Moreover note that




                    • for $x=3 implies left(left(frac{1}{4}right)^n+left(frac{1}{3}right)^nright)x^n=left(frac{3}{4}right)^n+1$


                    • for $x=-3 implies left(left(frac{1}{4}right)^n+left(frac{1}{3}right)^nright)x^n=left(-frac{3}{4}right)^n+(-1)^n$







                    share|cite|improve this answer











                    $endgroup$



                    We have that



                    $$|x|cdotsqrt[n]{left(frac{1}{4}right)^n+left(frac{1}{3}right)^n}=|x|cdotfrac13sqrt[n]{left(frac{3}{4}right)^n+1} to|x|cdotfrac13cdot 1 = |x|cdotfrac13<1$$



                    and therefore the radius of convergence is $3$.



                    Moreover note that




                    • for $x=3 implies left(left(frac{1}{4}right)^n+left(frac{1}{3}right)^nright)x^n=left(frac{3}{4}right)^n+1$


                    • for $x=-3 implies left(left(frac{1}{4}right)^n+left(frac{1}{3}right)^nright)x^n=left(-frac{3}{4}right)^n+(-1)^n$








                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 4 '18 at 22:18

























                    answered Dec 4 '18 at 22:11









                    gimusigimusi

                    1




                    1












                    • $begingroup$
                      Why the downvote, something wrong?
                      $endgroup$
                      – gimusi
                      Dec 4 '18 at 22:21


















                    • $begingroup$
                      Why the downvote, something wrong?
                      $endgroup$
                      – gimusi
                      Dec 4 '18 at 22:21
















                    $begingroup$
                    Why the downvote, something wrong?
                    $endgroup$
                    – gimusi
                    Dec 4 '18 at 22:21




                    $begingroup$
                    Why the downvote, something wrong?
                    $endgroup$
                    – gimusi
                    Dec 4 '18 at 22:21


















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