Analyzing $sum^infty _{n=1}{frac{1}{p^n - q^n}}, 0lt qlt p$ [on hold]












1














I am told to analyze the convergence of $sum^infty _{n=1}{frac{1}{p^n - q^n}}, 0lt qlt p$ depending on the values of $p$ and $q$. The only thing I have concluded is that ovbiously, the series is a positive term series. I do not know which criteria to use in this case. Thanks in advance.










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put on hold as off-topic by Did, Strants, rtybase, amWhy, José Carlos Santos 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Strants, rtybase, amWhy, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.













  • Faced with such a problem, I would experiment. Try different values of p, q, and n (e.g. p=3 or 5, q = 2, n =5. Look for patterns in the behavior of the sums of the fractions. Form hypotheses around the observed behavior. Prove the hypotheses. Form hypotheses around convergence. Try to prove these hypostheses. Then, if you still can't complete the problem, edit your post to show all of your work.
    – user2661923
    2 days ago


















1














I am told to analyze the convergence of $sum^infty _{n=1}{frac{1}{p^n - q^n}}, 0lt qlt p$ depending on the values of $p$ and $q$. The only thing I have concluded is that ovbiously, the series is a positive term series. I do not know which criteria to use in this case. Thanks in advance.










share|cite|improve this question















put on hold as off-topic by Did, Strants, rtybase, amWhy, José Carlos Santos 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Strants, rtybase, amWhy, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.













  • Faced with such a problem, I would experiment. Try different values of p, q, and n (e.g. p=3 or 5, q = 2, n =5. Look for patterns in the behavior of the sums of the fractions. Form hypotheses around the observed behavior. Prove the hypotheses. Form hypotheses around convergence. Try to prove these hypostheses. Then, if you still can't complete the problem, edit your post to show all of your work.
    – user2661923
    2 days ago
















1












1








1







I am told to analyze the convergence of $sum^infty _{n=1}{frac{1}{p^n - q^n}}, 0lt qlt p$ depending on the values of $p$ and $q$. The only thing I have concluded is that ovbiously, the series is a positive term series. I do not know which criteria to use in this case. Thanks in advance.










share|cite|improve this question















I am told to analyze the convergence of $sum^infty _{n=1}{frac{1}{p^n - q^n}}, 0lt qlt p$ depending on the values of $p$ and $q$. The only thing I have concluded is that ovbiously, the series is a positive term series. I do not know which criteria to use in this case. Thanks in advance.







real-analysis calculus sequences-and-series






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edited 2 days ago









Foobaz John

21.4k41351




21.4k41351










asked 2 days ago









Andarrkor

446




446




put on hold as off-topic by Did, Strants, rtybase, amWhy, José Carlos Santos 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Strants, rtybase, amWhy, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by Did, Strants, rtybase, amWhy, José Carlos Santos 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Strants, rtybase, amWhy, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.












  • Faced with such a problem, I would experiment. Try different values of p, q, and n (e.g. p=3 or 5, q = 2, n =5. Look for patterns in the behavior of the sums of the fractions. Form hypotheses around the observed behavior. Prove the hypotheses. Form hypotheses around convergence. Try to prove these hypostheses. Then, if you still can't complete the problem, edit your post to show all of your work.
    – user2661923
    2 days ago




















  • Faced with such a problem, I would experiment. Try different values of p, q, and n (e.g. p=3 or 5, q = 2, n =5. Look for patterns in the behavior of the sums of the fractions. Form hypotheses around the observed behavior. Prove the hypotheses. Form hypotheses around convergence. Try to prove these hypostheses. Then, if you still can't complete the problem, edit your post to show all of your work.
    – user2661923
    2 days ago


















Faced with such a problem, I would experiment. Try different values of p, q, and n (e.g. p=3 or 5, q = 2, n =5. Look for patterns in the behavior of the sums of the fractions. Form hypotheses around the observed behavior. Prove the hypotheses. Form hypotheses around convergence. Try to prove these hypostheses. Then, if you still can't complete the problem, edit your post to show all of your work.
– user2661923
2 days ago






Faced with such a problem, I would experiment. Try different values of p, q, and n (e.g. p=3 or 5, q = 2, n =5. Look for patterns in the behavior of the sums of the fractions. Form hypotheses around the observed behavior. Prove the hypotheses. Form hypotheses around convergence. Try to prove these hypostheses. Then, if you still can't complete the problem, edit your post to show all of your work.
– user2661923
2 days ago












2 Answers
2






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1














Observe that
$$sum_{n=1}^inftyfrac{1}{p^n-q^n}=frac{1}{p-q}sum_{n=1}^inftyfrac{1}{p^{n-1}+p^{n-2}q+cdots+pq^{n-2}+q^{n-1}}$$
We can apply the comparison test to $dfrac{1}{p^{n-1}}$ and $dfrac{1}{q^{n-1}}$ to see that the series converges whenever $p>1$ or $q>1$.



If both $pleq 1$ and $qleq 1$, then it is not the case that $dfrac 1{(p^n-q^n)}to 0$, hence the series diverges.






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    0














    Write the general term $a_n$ of the series as
    $$
    0lefrac{m}{p^n}leq a_n=frac{1}{p^n(1-(q/p)^n)}leq frac{M}{p^n}
    $$

    since $frac{1}{1-(q/p)^n}to 1$ for some $m, M>0$. The above inequality should allow you to deduce convergence depending on whether $p>1$ or $p<1$.



    For the case with $p=1$, the general term of the series is
    $$
    a_n=frac{1}{1-q^n}
    $$

    What can we say about $lim_{nto infty} a_n$ if $q<1$.






    share|cite|improve this answer





















    • How did you delimit the series? Thanks for answering.
      – Andarrkor
      2 days ago


















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    Observe that
    $$sum_{n=1}^inftyfrac{1}{p^n-q^n}=frac{1}{p-q}sum_{n=1}^inftyfrac{1}{p^{n-1}+p^{n-2}q+cdots+pq^{n-2}+q^{n-1}}$$
    We can apply the comparison test to $dfrac{1}{p^{n-1}}$ and $dfrac{1}{q^{n-1}}$ to see that the series converges whenever $p>1$ or $q>1$.



    If both $pleq 1$ and $qleq 1$, then it is not the case that $dfrac 1{(p^n-q^n)}to 0$, hence the series diverges.






    share|cite|improve this answer




























      1














      Observe that
      $$sum_{n=1}^inftyfrac{1}{p^n-q^n}=frac{1}{p-q}sum_{n=1}^inftyfrac{1}{p^{n-1}+p^{n-2}q+cdots+pq^{n-2}+q^{n-1}}$$
      We can apply the comparison test to $dfrac{1}{p^{n-1}}$ and $dfrac{1}{q^{n-1}}$ to see that the series converges whenever $p>1$ or $q>1$.



      If both $pleq 1$ and $qleq 1$, then it is not the case that $dfrac 1{(p^n-q^n)}to 0$, hence the series diverges.






      share|cite|improve this answer


























        1












        1








        1






        Observe that
        $$sum_{n=1}^inftyfrac{1}{p^n-q^n}=frac{1}{p-q}sum_{n=1}^inftyfrac{1}{p^{n-1}+p^{n-2}q+cdots+pq^{n-2}+q^{n-1}}$$
        We can apply the comparison test to $dfrac{1}{p^{n-1}}$ and $dfrac{1}{q^{n-1}}$ to see that the series converges whenever $p>1$ or $q>1$.



        If both $pleq 1$ and $qleq 1$, then it is not the case that $dfrac 1{(p^n-q^n)}to 0$, hence the series diverges.






        share|cite|improve this answer














        Observe that
        $$sum_{n=1}^inftyfrac{1}{p^n-q^n}=frac{1}{p-q}sum_{n=1}^inftyfrac{1}{p^{n-1}+p^{n-2}q+cdots+pq^{n-2}+q^{n-1}}$$
        We can apply the comparison test to $dfrac{1}{p^{n-1}}$ and $dfrac{1}{q^{n-1}}$ to see that the series converges whenever $p>1$ or $q>1$.



        If both $pleq 1$ and $qleq 1$, then it is not the case that $dfrac 1{(p^n-q^n)}to 0$, hence the series diverges.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 days ago









        amWhy

        192k28224439




        192k28224439










        answered 2 days ago









        Ben W

        1,825514




        1,825514























            0














            Write the general term $a_n$ of the series as
            $$
            0lefrac{m}{p^n}leq a_n=frac{1}{p^n(1-(q/p)^n)}leq frac{M}{p^n}
            $$

            since $frac{1}{1-(q/p)^n}to 1$ for some $m, M>0$. The above inequality should allow you to deduce convergence depending on whether $p>1$ or $p<1$.



            For the case with $p=1$, the general term of the series is
            $$
            a_n=frac{1}{1-q^n}
            $$

            What can we say about $lim_{nto infty} a_n$ if $q<1$.






            share|cite|improve this answer





















            • How did you delimit the series? Thanks for answering.
              – Andarrkor
              2 days ago
















            0














            Write the general term $a_n$ of the series as
            $$
            0lefrac{m}{p^n}leq a_n=frac{1}{p^n(1-(q/p)^n)}leq frac{M}{p^n}
            $$

            since $frac{1}{1-(q/p)^n}to 1$ for some $m, M>0$. The above inequality should allow you to deduce convergence depending on whether $p>1$ or $p<1$.



            For the case with $p=1$, the general term of the series is
            $$
            a_n=frac{1}{1-q^n}
            $$

            What can we say about $lim_{nto infty} a_n$ if $q<1$.






            share|cite|improve this answer





















            • How did you delimit the series? Thanks for answering.
              – Andarrkor
              2 days ago














            0












            0








            0






            Write the general term $a_n$ of the series as
            $$
            0lefrac{m}{p^n}leq a_n=frac{1}{p^n(1-(q/p)^n)}leq frac{M}{p^n}
            $$

            since $frac{1}{1-(q/p)^n}to 1$ for some $m, M>0$. The above inequality should allow you to deduce convergence depending on whether $p>1$ or $p<1$.



            For the case with $p=1$, the general term of the series is
            $$
            a_n=frac{1}{1-q^n}
            $$

            What can we say about $lim_{nto infty} a_n$ if $q<1$.






            share|cite|improve this answer












            Write the general term $a_n$ of the series as
            $$
            0lefrac{m}{p^n}leq a_n=frac{1}{p^n(1-(q/p)^n)}leq frac{M}{p^n}
            $$

            since $frac{1}{1-(q/p)^n}to 1$ for some $m, M>0$. The above inequality should allow you to deduce convergence depending on whether $p>1$ or $p<1$.



            For the case with $p=1$, the general term of the series is
            $$
            a_n=frac{1}{1-q^n}
            $$

            What can we say about $lim_{nto infty} a_n$ if $q<1$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 days ago









            Foobaz John

            21.4k41351




            21.4k41351












            • How did you delimit the series? Thanks for answering.
              – Andarrkor
              2 days ago


















            • How did you delimit the series? Thanks for answering.
              – Andarrkor
              2 days ago
















            How did you delimit the series? Thanks for answering.
            – Andarrkor
            2 days ago




            How did you delimit the series? Thanks for answering.
            – Andarrkor
            2 days ago



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