It is my calculations about the convergence of a series are correct?












2














Consider $0<q<p$ real numbers. I wana find for which values of $p$ the series $sumfrac{1}{p^n-q^n}$ is convergent.



First note that the following



$$
sumfrac{1}{p^n-q^n}=sumfrac{1}{p^n(1-(frac{q}{p})^n)}
$$



Now we have the inequalities



$$
0leqsqrt[n]{1-left(frac{q}{p}right)^n}leq 1
$$



So we have that $lim_{nrightarrowinfty}sqrt[n]{1-left(frac{q}{p}right)^n}=k$ where $0leq kleq 1$



Then, by the root criteria (if $kneq0$) the serie converge iff $frac{1}{pk}<1$ iff $1<pkleq p$.



My question: As you see, my process depend of $k$ because I assume that $kneq 0$ but I can't prove it. Can someone give me a hint?






calculus sequences-and-series convergence







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    2














    Consider $0<q<p$ real numbers. I wana find for which values of $p$ the series $sumfrac{1}{p^n-q^n}$ is convergent.



    First note that the following



    $$
    sumfrac{1}{p^n-q^n}=sumfrac{1}{p^n(1-(frac{q}{p})^n)}
    $$



    Now we have the inequalities



    $$
    0leqsqrt[n]{1-left(frac{q}{p}right)^n}leq 1
    $$



    So we have that $lim_{nrightarrowinfty}sqrt[n]{1-left(frac{q}{p}right)^n}=k$ where $0leq kleq 1$



    Then, by the root criteria (if $kneq0$) the serie converge iff $frac{1}{pk}<1$ iff $1<pkleq p$.



    My question: As you see, my process depend of $k$ because I assume that $kneq 0$ but I can't prove it. Can someone give me a hint?






    calculus sequences-and-series convergence







    share|cite|improve this question

























      2












      2








      2







      Consider $0<q<p$ real numbers. I wana find for which values of $p$ the series $sumfrac{1}{p^n-q^n}$ is convergent.



      First note that the following



      $$
      sumfrac{1}{p^n-q^n}=sumfrac{1}{p^n(1-(frac{q}{p})^n)}
      $$



      Now we have the inequalities



      $$
      0leqsqrt[n]{1-left(frac{q}{p}right)^n}leq 1
      $$



      So we have that $lim_{nrightarrowinfty}sqrt[n]{1-left(frac{q}{p}right)^n}=k$ where $0leq kleq 1$



      Then, by the root criteria (if $kneq0$) the serie converge iff $frac{1}{pk}<1$ iff $1<pkleq p$.



      My question: As you see, my process depend of $k$ because I assume that $kneq 0$ but I can't prove it. Can someone give me a hint?






      calculus sequences-and-series convergence







      share|cite|improve this question













      Consider $0<q<p$ real numbers. I wana find for which values of $p$ the series $sumfrac{1}{p^n-q^n}$ is convergent.



      First note that the following



      $$
      sumfrac{1}{p^n-q^n}=sumfrac{1}{p^n(1-(frac{q}{p})^n)}
      $$



      Now we have the inequalities



      $$
      0leqsqrt[n]{1-left(frac{q}{p}right)^n}leq 1
      $$



      So we have that $lim_{nrightarrowinfty}sqrt[n]{1-left(frac{q}{p}right)^n}=k$ where $0leq kleq 1$



      Then, by the root criteria (if $kneq0$) the serie converge iff $frac{1}{pk}<1$ iff $1<pkleq p$.



      My question: As you see, my process depend of $k$ because I assume that $kneq 0$ but I can't prove it. Can someone give me a hint?






      calculus sequences-and-series convergence




      calculus sequences-and-series convergence






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 1 at 23:14









      Gödel

      1,413319




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          2 Answers
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          $(frac q p)^{n} to 0$ because $q<p$. Hence $1-(frac q p)^{n} to 1$. Comparing the given series with $sum frac 1 {p^{n}}$ we see that the series converges iff $p>1$.






          share|cite|improve this answer































            1














            You could better compute $k$, but it's not the best method. Note also that the simple fact that $0le a_nle1$ doesn't guarantee the existence of $lim_{ntoinfty}a_n$, so you don't really have $k$.



            First method.



            Note that
            $$
            lim_{ntoinfty}frac{dfrac{1}{p^n}}{dfrac{1}{p^n-q^n}}=
            lim_{ntoinfty}left(1-left(dfrac{q}{p}right)^{!n}right)=1
            $$
            so the two series $sum_nfrac{1}{p^n-q^n}$ and $sum_{n}frac{1}{p^n}$ are either both convergent or both divergent.



            Second method.



            The ratio test, setting $r=q/p<1$,
            $$
            lim_{ntoinfty}frac{dfrac{1}{p^{n+1}-q^{n+1}}}{dfrac{1}{p^n-q^n}}=
            lim_{ntoinfty}frac{p^{n}-q^n}{p^{n+1}-q^{n+1}}=
            lim_{ntoinfty}frac{1-r^n}{p-qr^n}=frac{1}{p}
            $$
            would only leave the case $p=1$ to be determined. Can you do the case $p=1$?






            share|cite|improve this answer





















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              2 Answers
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              2 Answers
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              3














              $(frac q p)^{n} to 0$ because $q<p$. Hence $1-(frac q p)^{n} to 1$. Comparing the given series with $sum frac 1 {p^{n}}$ we see that the series converges iff $p>1$.






              share|cite|improve this answer




























                3














                $(frac q p)^{n} to 0$ because $q<p$. Hence $1-(frac q p)^{n} to 1$. Comparing the given series with $sum frac 1 {p^{n}}$ we see that the series converges iff $p>1$.






                share|cite|improve this answer


























                  3












                  3








                  3






                  $(frac q p)^{n} to 0$ because $q<p$. Hence $1-(frac q p)^{n} to 1$. Comparing the given series with $sum frac 1 {p^{n}}$ we see that the series converges iff $p>1$.






                  share|cite|improve this answer














                  $(frac q p)^{n} to 0$ because $q<p$. Hence $1-(frac q p)^{n} to 1$. Comparing the given series with $sum frac 1 {p^{n}}$ we see that the series converges iff $p>1$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 2 at 4:21

























                  answered Dec 1 at 23:24









                  Kavi Rama Murthy

                  50.1k31854




                  50.1k31854























                      1














                      You could better compute $k$, but it's not the best method. Note also that the simple fact that $0le a_nle1$ doesn't guarantee the existence of $lim_{ntoinfty}a_n$, so you don't really have $k$.



                      First method.



                      Note that
                      $$
                      lim_{ntoinfty}frac{dfrac{1}{p^n}}{dfrac{1}{p^n-q^n}}=
                      lim_{ntoinfty}left(1-left(dfrac{q}{p}right)^{!n}right)=1
                      $$
                      so the two series $sum_nfrac{1}{p^n-q^n}$ and $sum_{n}frac{1}{p^n}$ are either both convergent or both divergent.



                      Second method.



                      The ratio test, setting $r=q/p<1$,
                      $$
                      lim_{ntoinfty}frac{dfrac{1}{p^{n+1}-q^{n+1}}}{dfrac{1}{p^n-q^n}}=
                      lim_{ntoinfty}frac{p^{n}-q^n}{p^{n+1}-q^{n+1}}=
                      lim_{ntoinfty}frac{1-r^n}{p-qr^n}=frac{1}{p}
                      $$
                      would only leave the case $p=1$ to be determined. Can you do the case $p=1$?






                      share|cite|improve this answer


























                        1














                        You could better compute $k$, but it's not the best method. Note also that the simple fact that $0le a_nle1$ doesn't guarantee the existence of $lim_{ntoinfty}a_n$, so you don't really have $k$.



                        First method.



                        Note that
                        $$
                        lim_{ntoinfty}frac{dfrac{1}{p^n}}{dfrac{1}{p^n-q^n}}=
                        lim_{ntoinfty}left(1-left(dfrac{q}{p}right)^{!n}right)=1
                        $$
                        so the two series $sum_nfrac{1}{p^n-q^n}$ and $sum_{n}frac{1}{p^n}$ are either both convergent or both divergent.



                        Second method.



                        The ratio test, setting $r=q/p<1$,
                        $$
                        lim_{ntoinfty}frac{dfrac{1}{p^{n+1}-q^{n+1}}}{dfrac{1}{p^n-q^n}}=
                        lim_{ntoinfty}frac{p^{n}-q^n}{p^{n+1}-q^{n+1}}=
                        lim_{ntoinfty}frac{1-r^n}{p-qr^n}=frac{1}{p}
                        $$
                        would only leave the case $p=1$ to be determined. Can you do the case $p=1$?






                        share|cite|improve this answer
























                          1












                          1








                          1






                          You could better compute $k$, but it's not the best method. Note also that the simple fact that $0le a_nle1$ doesn't guarantee the existence of $lim_{ntoinfty}a_n$, so you don't really have $k$.



                          First method.



                          Note that
                          $$
                          lim_{ntoinfty}frac{dfrac{1}{p^n}}{dfrac{1}{p^n-q^n}}=
                          lim_{ntoinfty}left(1-left(dfrac{q}{p}right)^{!n}right)=1
                          $$
                          so the two series $sum_nfrac{1}{p^n-q^n}$ and $sum_{n}frac{1}{p^n}$ are either both convergent or both divergent.



                          Second method.



                          The ratio test, setting $r=q/p<1$,
                          $$
                          lim_{ntoinfty}frac{dfrac{1}{p^{n+1}-q^{n+1}}}{dfrac{1}{p^n-q^n}}=
                          lim_{ntoinfty}frac{p^{n}-q^n}{p^{n+1}-q^{n+1}}=
                          lim_{ntoinfty}frac{1-r^n}{p-qr^n}=frac{1}{p}
                          $$
                          would only leave the case $p=1$ to be determined. Can you do the case $p=1$?






                          share|cite|improve this answer












                          You could better compute $k$, but it's not the best method. Note also that the simple fact that $0le a_nle1$ doesn't guarantee the existence of $lim_{ntoinfty}a_n$, so you don't really have $k$.



                          First method.



                          Note that
                          $$
                          lim_{ntoinfty}frac{dfrac{1}{p^n}}{dfrac{1}{p^n-q^n}}=
                          lim_{ntoinfty}left(1-left(dfrac{q}{p}right)^{!n}right)=1
                          $$
                          so the two series $sum_nfrac{1}{p^n-q^n}$ and $sum_{n}frac{1}{p^n}$ are either both convergent or both divergent.



                          Second method.



                          The ratio test, setting $r=q/p<1$,
                          $$
                          lim_{ntoinfty}frac{dfrac{1}{p^{n+1}-q^{n+1}}}{dfrac{1}{p^n-q^n}}=
                          lim_{ntoinfty}frac{p^{n}-q^n}{p^{n+1}-q^{n+1}}=
                          lim_{ntoinfty}frac{1-r^n}{p-qr^n}=frac{1}{p}
                          $$
                          would only leave the case $p=1$ to be determined. Can you do the case $p=1$?







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 1 at 23:58









                          egreg

                          178k1484201




                          178k1484201






























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