Mathematical Induction & Sequencing












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Let $q_1$ be an arbitrary positive real number and define the sequence {$ q_{n};$} by $ q_{n+1};$ = $ q_{n};$ + $frac{1}{q_n}$



Must there be an index k for which $q_k > 5^{50}$?






sequences-and-series combinatorics







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Let $q_1$ be an arbitrary positive real number and define the sequence {$ q_{n};$} by $ q_{n+1};$ = $ q_{n};$ + $frac{1}{q_n}$



    Must there be an index k for which $q_k > 5^{50}$?






    sequences-and-series combinatorics







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let $q_1$ be an arbitrary positive real number and define the sequence {$ q_{n};$} by $ q_{n+1};$ = $ q_{n};$ + $frac{1}{q_n}$



      Must there be an index k for which $q_k > 5^{50}$?






      sequences-and-series combinatorics







      share|cite|improve this question











      $endgroup$




      Let $q_1$ be an arbitrary positive real number and define the sequence {$ q_{n};$} by $ q_{n+1};$ = $ q_{n};$ + $frac{1}{q_n}$



      Must there be an index k for which $q_k > 5^{50}$?






      sequences-and-series combinatorics




      sequences-and-series combinatorics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 23 '18 at 22:32









      Scientifica

      6,37641335




      6,37641335










      asked Apr 2 '17 at 14:24







      user431696





























          3 Answers
          3






          active

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          3












          $begingroup$

          It is easy to show $q_n$ is a strictly increasing sequence of positive numbers. Thus if it is bounded, it converges. Let us assume $q_n$ is bounded. If $q_n$ converges, then $lim_{nrightarrowinfty}(q_n) =lim_{nrightarrowinfty}(q_{n+1}) =lim_{nrightarrowinfty}(q_n + frac{1}{q_n}) iff lim_{nrightarrowinfty} (frac{1}{q_n} =0)$ $ iff lim_{nrightarrowinfty}{q_n} = infty$



          But then $q_n$ is not bounded, which is a contradiction.
          Thus $q_n$ is unbounded, and for all M we can find an n for which $q_n >M$






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            Let's assume the sequence were bounded. Since it is a strictly increasing sequence, it follows that is must converge to some finite limit. Let's call that limit $q$. Note that since for $q_nle 1$ we have $q_{n+1} = q_n + 1/q_n > 1/q_n ge 1$, therefore $q>1$.



            Now since $q_n$ converges to $q$, for every $epsilon>0$ there's an $N$ so that $q_n>q-epsilon$ for all $nge N$. Let's specifically take $epsilon=1/q$. Note that $epsilon>0$ because of $q>1$. Then you've got some $N$ so that $q-epsilon<q_N<q$. But then, $q_{N+1}=q_N+1/q_N > q-epsilon +1/q_N = q - 1/q + 1/q_N > q -1/q + 1/q = q$, in contradiction to the assumption that $q$ is an upper bound. Therefore the assumption that the sequence is bounded must be wrong.



            Since the sequence is unbounded and strictly monotonous growing, it will grow above any bound, including the bound $5^{50}$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              $1/q$ isn't better?
              $endgroup$
              – Rafa Budría
              Apr 2 '17 at 15:25










            • $begingroup$
              @RafaBudría: Yes; I originally named the limit $a$ and then renamed it to $q$, but then erroneously continued to use $a$ in the text added after doing the change.
              $endgroup$
              – celtschk
              Apr 2 '17 at 15:28










            • $begingroup$
              I supposed that. Still remain some a's :)
              $endgroup$
              – Rafa Budría
              Apr 2 '17 at 15:29










            • $begingroup$
              @RafaBudría: I hope now I missed no further one; thank you.
              $endgroup$
              – celtschk
              Apr 2 '17 at 15:30



















            1












            $begingroup$

            Since $q_1$ is positive and greater $0$ and $frac{1}{q_1}$ is positive and greater $0$, $q_2=q_1+frac{1}{q_1}$ will be positive as well. Furthermore since $frac{1}{q_1}$ is greater 0, $q_2=q_1+frac{1}{q_1}$ will be greater than $q_1$ and by induction this means that the sequence is strictly increasing.



            If the sequence were bounded, that would mean it would converge to a certain value. If that were the case, the difference between two consecutive numbers in the sequence would need to tend to $0$ as $n$ tends to infinity, $limlimits_{nrightarrowinfty}(q_{n+1}-q_n)=0$. Substituting $q_{n+1}=q_n+frac{1}{q_n}$, we get $0=limlimits_{nrightarrowinfty}(q_{n+1}-q_n)=limlimits_{nrightarrowinfty}(q_n+frac{1}{q_n}-q_n)=limlimits_{nrightarrowinfty}(frac{1}{q_n})$, yet $limlimits_{nrightarrowinfty}(frac{1}{q_n})=0Leftrightarrow limlimits_{nrightarrowinfty}q_n=infty$, which would contradicts the series being bounded.



            Therefore the series is not bounded and will eventually reach any arbitrarily large number.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Doesn't follow. You proved it's increasing, not that it's unbounded. I think in fact it's unbounded, but proof is needed.
              $endgroup$
              – Rafa Budría
              Apr 2 '17 at 14:34













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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            It is easy to show $q_n$ is a strictly increasing sequence of positive numbers. Thus if it is bounded, it converges. Let us assume $q_n$ is bounded. If $q_n$ converges, then $lim_{nrightarrowinfty}(q_n) =lim_{nrightarrowinfty}(q_{n+1}) =lim_{nrightarrowinfty}(q_n + frac{1}{q_n}) iff lim_{nrightarrowinfty} (frac{1}{q_n} =0)$ $ iff lim_{nrightarrowinfty}{q_n} = infty$



            But then $q_n$ is not bounded, which is a contradiction.
            Thus $q_n$ is unbounded, and for all M we can find an n for which $q_n >M$






            share|cite|improve this answer











            $endgroup$


















              3












              $begingroup$

              It is easy to show $q_n$ is a strictly increasing sequence of positive numbers. Thus if it is bounded, it converges. Let us assume $q_n$ is bounded. If $q_n$ converges, then $lim_{nrightarrowinfty}(q_n) =lim_{nrightarrowinfty}(q_{n+1}) =lim_{nrightarrowinfty}(q_n + frac{1}{q_n}) iff lim_{nrightarrowinfty} (frac{1}{q_n} =0)$ $ iff lim_{nrightarrowinfty}{q_n} = infty$



              But then $q_n$ is not bounded, which is a contradiction.
              Thus $q_n$ is unbounded, and for all M we can find an n for which $q_n >M$






              share|cite|improve this answer











              $endgroup$
















                3












                3








                3





                $begingroup$

                It is easy to show $q_n$ is a strictly increasing sequence of positive numbers. Thus if it is bounded, it converges. Let us assume $q_n$ is bounded. If $q_n$ converges, then $lim_{nrightarrowinfty}(q_n) =lim_{nrightarrowinfty}(q_{n+1}) =lim_{nrightarrowinfty}(q_n + frac{1}{q_n}) iff lim_{nrightarrowinfty} (frac{1}{q_n} =0)$ $ iff lim_{nrightarrowinfty}{q_n} = infty$



                But then $q_n$ is not bounded, which is a contradiction.
                Thus $q_n$ is unbounded, and for all M we can find an n for which $q_n >M$






                share|cite|improve this answer











                $endgroup$



                It is easy to show $q_n$ is a strictly increasing sequence of positive numbers. Thus if it is bounded, it converges. Let us assume $q_n$ is bounded. If $q_n$ converges, then $lim_{nrightarrowinfty}(q_n) =lim_{nrightarrowinfty}(q_{n+1}) =lim_{nrightarrowinfty}(q_n + frac{1}{q_n}) iff lim_{nrightarrowinfty} (frac{1}{q_n} =0)$ $ iff lim_{nrightarrowinfty}{q_n} = infty$



                But then $q_n$ is not bounded, which is a contradiction.
                Thus $q_n$ is unbounded, and for all M we can find an n for which $q_n >M$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Apr 2 '17 at 15:25

























                answered Apr 2 '17 at 15:12









                Daphna KeidarDaphna Keidar

                1996




                1996























                    1












                    $begingroup$

                    Let's assume the sequence were bounded. Since it is a strictly increasing sequence, it follows that is must converge to some finite limit. Let's call that limit $q$. Note that since for $q_nle 1$ we have $q_{n+1} = q_n + 1/q_n > 1/q_n ge 1$, therefore $q>1$.



                    Now since $q_n$ converges to $q$, for every $epsilon>0$ there's an $N$ so that $q_n>q-epsilon$ for all $nge N$. Let's specifically take $epsilon=1/q$. Note that $epsilon>0$ because of $q>1$. Then you've got some $N$ so that $q-epsilon<q_N<q$. But then, $q_{N+1}=q_N+1/q_N > q-epsilon +1/q_N = q - 1/q + 1/q_N > q -1/q + 1/q = q$, in contradiction to the assumption that $q$ is an upper bound. Therefore the assumption that the sequence is bounded must be wrong.



                    Since the sequence is unbounded and strictly monotonous growing, it will grow above any bound, including the bound $5^{50}$.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      $1/q$ isn't better?
                      $endgroup$
                      – Rafa Budría
                      Apr 2 '17 at 15:25










                    • $begingroup$
                      @RafaBudría: Yes; I originally named the limit $a$ and then renamed it to $q$, but then erroneously continued to use $a$ in the text added after doing the change.
                      $endgroup$
                      – celtschk
                      Apr 2 '17 at 15:28










                    • $begingroup$
                      I supposed that. Still remain some a's :)
                      $endgroup$
                      – Rafa Budría
                      Apr 2 '17 at 15:29










                    • $begingroup$
                      @RafaBudría: I hope now I missed no further one; thank you.
                      $endgroup$
                      – celtschk
                      Apr 2 '17 at 15:30
















                    1












                    $begingroup$

                    Let's assume the sequence were bounded. Since it is a strictly increasing sequence, it follows that is must converge to some finite limit. Let's call that limit $q$. Note that since for $q_nle 1$ we have $q_{n+1} = q_n + 1/q_n > 1/q_n ge 1$, therefore $q>1$.



                    Now since $q_n$ converges to $q$, for every $epsilon>0$ there's an $N$ so that $q_n>q-epsilon$ for all $nge N$. Let's specifically take $epsilon=1/q$. Note that $epsilon>0$ because of $q>1$. Then you've got some $N$ so that $q-epsilon<q_N<q$. But then, $q_{N+1}=q_N+1/q_N > q-epsilon +1/q_N = q - 1/q + 1/q_N > q -1/q + 1/q = q$, in contradiction to the assumption that $q$ is an upper bound. Therefore the assumption that the sequence is bounded must be wrong.



                    Since the sequence is unbounded and strictly monotonous growing, it will grow above any bound, including the bound $5^{50}$.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      $1/q$ isn't better?
                      $endgroup$
                      – Rafa Budría
                      Apr 2 '17 at 15:25










                    • $begingroup$
                      @RafaBudría: Yes; I originally named the limit $a$ and then renamed it to $q$, but then erroneously continued to use $a$ in the text added after doing the change.
                      $endgroup$
                      – celtschk
                      Apr 2 '17 at 15:28










                    • $begingroup$
                      I supposed that. Still remain some a's :)
                      $endgroup$
                      – Rafa Budría
                      Apr 2 '17 at 15:29










                    • $begingroup$
                      @RafaBudría: I hope now I missed no further one; thank you.
                      $endgroup$
                      – celtschk
                      Apr 2 '17 at 15:30














                    1












                    1








                    1





                    $begingroup$

                    Let's assume the sequence were bounded. Since it is a strictly increasing sequence, it follows that is must converge to some finite limit. Let's call that limit $q$. Note that since for $q_nle 1$ we have $q_{n+1} = q_n + 1/q_n > 1/q_n ge 1$, therefore $q>1$.



                    Now since $q_n$ converges to $q$, for every $epsilon>0$ there's an $N$ so that $q_n>q-epsilon$ for all $nge N$. Let's specifically take $epsilon=1/q$. Note that $epsilon>0$ because of $q>1$. Then you've got some $N$ so that $q-epsilon<q_N<q$. But then, $q_{N+1}=q_N+1/q_N > q-epsilon +1/q_N = q - 1/q + 1/q_N > q -1/q + 1/q = q$, in contradiction to the assumption that $q$ is an upper bound. Therefore the assumption that the sequence is bounded must be wrong.



                    Since the sequence is unbounded and strictly monotonous growing, it will grow above any bound, including the bound $5^{50}$.






                    share|cite|improve this answer











                    $endgroup$



                    Let's assume the sequence were bounded. Since it is a strictly increasing sequence, it follows that is must converge to some finite limit. Let's call that limit $q$. Note that since for $q_nle 1$ we have $q_{n+1} = q_n + 1/q_n > 1/q_n ge 1$, therefore $q>1$.



                    Now since $q_n$ converges to $q$, for every $epsilon>0$ there's an $N$ so that $q_n>q-epsilon$ for all $nge N$. Let's specifically take $epsilon=1/q$. Note that $epsilon>0$ because of $q>1$. Then you've got some $N$ so that $q-epsilon<q_N<q$. But then, $q_{N+1}=q_N+1/q_N > q-epsilon +1/q_N = q - 1/q + 1/q_N > q -1/q + 1/q = q$, in contradiction to the assumption that $q$ is an upper bound. Therefore the assumption that the sequence is bounded must be wrong.



                    Since the sequence is unbounded and strictly monotonous growing, it will grow above any bound, including the bound $5^{50}$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Apr 2 '17 at 15:30

























                    answered Apr 2 '17 at 15:21









                    celtschkceltschk

                    29.8k75599




                    29.8k75599












                    • $begingroup$
                      $1/q$ isn't better?
                      $endgroup$
                      – Rafa Budría
                      Apr 2 '17 at 15:25










                    • $begingroup$
                      @RafaBudría: Yes; I originally named the limit $a$ and then renamed it to $q$, but then erroneously continued to use $a$ in the text added after doing the change.
                      $endgroup$
                      – celtschk
                      Apr 2 '17 at 15:28










                    • $begingroup$
                      I supposed that. Still remain some a's :)
                      $endgroup$
                      – Rafa Budría
                      Apr 2 '17 at 15:29










                    • $begingroup$
                      @RafaBudría: I hope now I missed no further one; thank you.
                      $endgroup$
                      – celtschk
                      Apr 2 '17 at 15:30


















                    • $begingroup$
                      $1/q$ isn't better?
                      $endgroup$
                      – Rafa Budría
                      Apr 2 '17 at 15:25










                    • $begingroup$
                      @RafaBudría: Yes; I originally named the limit $a$ and then renamed it to $q$, but then erroneously continued to use $a$ in the text added after doing the change.
                      $endgroup$
                      – celtschk
                      Apr 2 '17 at 15:28










                    • $begingroup$
                      I supposed that. Still remain some a's :)
                      $endgroup$
                      – Rafa Budría
                      Apr 2 '17 at 15:29










                    • $begingroup$
                      @RafaBudría: I hope now I missed no further one; thank you.
                      $endgroup$
                      – celtschk
                      Apr 2 '17 at 15:30
















                    $begingroup$
                    $1/q$ isn't better?
                    $endgroup$
                    – Rafa Budría
                    Apr 2 '17 at 15:25




                    $begingroup$
                    $1/q$ isn't better?
                    $endgroup$
                    – Rafa Budría
                    Apr 2 '17 at 15:25












                    $begingroup$
                    @RafaBudría: Yes; I originally named the limit $a$ and then renamed it to $q$, but then erroneously continued to use $a$ in the text added after doing the change.
                    $endgroup$
                    – celtschk
                    Apr 2 '17 at 15:28




                    $begingroup$
                    @RafaBudría: Yes; I originally named the limit $a$ and then renamed it to $q$, but then erroneously continued to use $a$ in the text added after doing the change.
                    $endgroup$
                    – celtschk
                    Apr 2 '17 at 15:28












                    $begingroup$
                    I supposed that. Still remain some a's :)
                    $endgroup$
                    – Rafa Budría
                    Apr 2 '17 at 15:29




                    $begingroup$
                    I supposed that. Still remain some a's :)
                    $endgroup$
                    – Rafa Budría
                    Apr 2 '17 at 15:29












                    $begingroup$
                    @RafaBudría: I hope now I missed no further one; thank you.
                    $endgroup$
                    – celtschk
                    Apr 2 '17 at 15:30




                    $begingroup$
                    @RafaBudría: I hope now I missed no further one; thank you.
                    $endgroup$
                    – celtschk
                    Apr 2 '17 at 15:30











                    1












                    $begingroup$

                    Since $q_1$ is positive and greater $0$ and $frac{1}{q_1}$ is positive and greater $0$, $q_2=q_1+frac{1}{q_1}$ will be positive as well. Furthermore since $frac{1}{q_1}$ is greater 0, $q_2=q_1+frac{1}{q_1}$ will be greater than $q_1$ and by induction this means that the sequence is strictly increasing.



                    If the sequence were bounded, that would mean it would converge to a certain value. If that were the case, the difference between two consecutive numbers in the sequence would need to tend to $0$ as $n$ tends to infinity, $limlimits_{nrightarrowinfty}(q_{n+1}-q_n)=0$. Substituting $q_{n+1}=q_n+frac{1}{q_n}$, we get $0=limlimits_{nrightarrowinfty}(q_{n+1}-q_n)=limlimits_{nrightarrowinfty}(q_n+frac{1}{q_n}-q_n)=limlimits_{nrightarrowinfty}(frac{1}{q_n})$, yet $limlimits_{nrightarrowinfty}(frac{1}{q_n})=0Leftrightarrow limlimits_{nrightarrowinfty}q_n=infty$, which would contradicts the series being bounded.



                    Therefore the series is not bounded and will eventually reach any arbitrarily large number.






                    share|cite|improve this answer











                    $endgroup$









                    • 1




                      $begingroup$
                      Doesn't follow. You proved it's increasing, not that it's unbounded. I think in fact it's unbounded, but proof is needed.
                      $endgroup$
                      – Rafa Budría
                      Apr 2 '17 at 14:34


















                    1












                    $begingroup$

                    Since $q_1$ is positive and greater $0$ and $frac{1}{q_1}$ is positive and greater $0$, $q_2=q_1+frac{1}{q_1}$ will be positive as well. Furthermore since $frac{1}{q_1}$ is greater 0, $q_2=q_1+frac{1}{q_1}$ will be greater than $q_1$ and by induction this means that the sequence is strictly increasing.



                    If the sequence were bounded, that would mean it would converge to a certain value. If that were the case, the difference between two consecutive numbers in the sequence would need to tend to $0$ as $n$ tends to infinity, $limlimits_{nrightarrowinfty}(q_{n+1}-q_n)=0$. Substituting $q_{n+1}=q_n+frac{1}{q_n}$, we get $0=limlimits_{nrightarrowinfty}(q_{n+1}-q_n)=limlimits_{nrightarrowinfty}(q_n+frac{1}{q_n}-q_n)=limlimits_{nrightarrowinfty}(frac{1}{q_n})$, yet $limlimits_{nrightarrowinfty}(frac{1}{q_n})=0Leftrightarrow limlimits_{nrightarrowinfty}q_n=infty$, which would contradicts the series being bounded.



                    Therefore the series is not bounded and will eventually reach any arbitrarily large number.






                    share|cite|improve this answer











                    $endgroup$









                    • 1




                      $begingroup$
                      Doesn't follow. You proved it's increasing, not that it's unbounded. I think in fact it's unbounded, but proof is needed.
                      $endgroup$
                      – Rafa Budría
                      Apr 2 '17 at 14:34
















                    1












                    1








                    1





                    $begingroup$

                    Since $q_1$ is positive and greater $0$ and $frac{1}{q_1}$ is positive and greater $0$, $q_2=q_1+frac{1}{q_1}$ will be positive as well. Furthermore since $frac{1}{q_1}$ is greater 0, $q_2=q_1+frac{1}{q_1}$ will be greater than $q_1$ and by induction this means that the sequence is strictly increasing.



                    If the sequence were bounded, that would mean it would converge to a certain value. If that were the case, the difference between two consecutive numbers in the sequence would need to tend to $0$ as $n$ tends to infinity, $limlimits_{nrightarrowinfty}(q_{n+1}-q_n)=0$. Substituting $q_{n+1}=q_n+frac{1}{q_n}$, we get $0=limlimits_{nrightarrowinfty}(q_{n+1}-q_n)=limlimits_{nrightarrowinfty}(q_n+frac{1}{q_n}-q_n)=limlimits_{nrightarrowinfty}(frac{1}{q_n})$, yet $limlimits_{nrightarrowinfty}(frac{1}{q_n})=0Leftrightarrow limlimits_{nrightarrowinfty}q_n=infty$, which would contradicts the series being bounded.



                    Therefore the series is not bounded and will eventually reach any arbitrarily large number.






                    share|cite|improve this answer











                    $endgroup$



                    Since $q_1$ is positive and greater $0$ and $frac{1}{q_1}$ is positive and greater $0$, $q_2=q_1+frac{1}{q_1}$ will be positive as well. Furthermore since $frac{1}{q_1}$ is greater 0, $q_2=q_1+frac{1}{q_1}$ will be greater than $q_1$ and by induction this means that the sequence is strictly increasing.



                    If the sequence were bounded, that would mean it would converge to a certain value. If that were the case, the difference between two consecutive numbers in the sequence would need to tend to $0$ as $n$ tends to infinity, $limlimits_{nrightarrowinfty}(q_{n+1}-q_n)=0$. Substituting $q_{n+1}=q_n+frac{1}{q_n}$, we get $0=limlimits_{nrightarrowinfty}(q_{n+1}-q_n)=limlimits_{nrightarrowinfty}(q_n+frac{1}{q_n}-q_n)=limlimits_{nrightarrowinfty}(frac{1}{q_n})$, yet $limlimits_{nrightarrowinfty}(frac{1}{q_n})=0Leftrightarrow limlimits_{nrightarrowinfty}q_n=infty$, which would contradicts the series being bounded.



                    Therefore the series is not bounded and will eventually reach any arbitrarily large number.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Apr 2 '17 at 15:35

























                    answered Apr 2 '17 at 14:29









                    ThorgottThorgott

                    566314




                    566314








                    • 1




                      $begingroup$
                      Doesn't follow. You proved it's increasing, not that it's unbounded. I think in fact it's unbounded, but proof is needed.
                      $endgroup$
                      – Rafa Budría
                      Apr 2 '17 at 14:34
















                    • 1




                      $begingroup$
                      Doesn't follow. You proved it's increasing, not that it's unbounded. I think in fact it's unbounded, but proof is needed.
                      $endgroup$
                      – Rafa Budría
                      Apr 2 '17 at 14:34










                    1




                    1




                    $begingroup$
                    Doesn't follow. You proved it's increasing, not that it's unbounded. I think in fact it's unbounded, but proof is needed.
                    $endgroup$
                    – Rafa Budría
                    Apr 2 '17 at 14:34






                    $begingroup$
                    Doesn't follow. You proved it's increasing, not that it's unbounded. I think in fact it's unbounded, but proof is needed.
                    $endgroup$
                    – Rafa Budría
                    Apr 2 '17 at 14:34




















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