Is $2 sum_{n=m+2}^{infty}dfrac{1}{n^{alpha}} geq sum_{n=m+1}^{infty}dfrac{1}{n^{alpha}}$?












1














Working in a project I came across to the following problem, to investigate if the following inequality is true for $alpha>2,$ and $m$ a positive integer:
$$2 sum_{n=m+2}^{infty}dfrac{1}{n^{alpha}} geq sum_{n=m+1}^{infty}dfrac{1}{n^{alpha}}.$$



I have made some numerical experiments in Worfram which indicates that this is true, but I have no idea how to perform ao proof.






real-analysis complex-analysis number-theory numerical-methods zeta-functions







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  • Please link to the numerical experiments. It might help.
    – Shaun
    Nov 30 at 0:18










  • Why are there two sums when you can cancel so many terms from the right?
    – T_M
    Nov 30 at 0:18
















1














Working in a project I came across to the following problem, to investigate if the following inequality is true for $alpha>2,$ and $m$ a positive integer:
$$2 sum_{n=m+2}^{infty}dfrac{1}{n^{alpha}} geq sum_{n=m+1}^{infty}dfrac{1}{n^{alpha}}.$$



I have made some numerical experiments in Worfram which indicates that this is true, but I have no idea how to perform ao proof.






real-analysis complex-analysis number-theory numerical-methods zeta-functions







share|cite|improve this question






















  • Please link to the numerical experiments. It might help.
    – Shaun
    Nov 30 at 0:18










  • Why are there two sums when you can cancel so many terms from the right?
    – T_M
    Nov 30 at 0:18














1












1








1







Working in a project I came across to the following problem, to investigate if the following inequality is true for $alpha>2,$ and $m$ a positive integer:
$$2 sum_{n=m+2}^{infty}dfrac{1}{n^{alpha}} geq sum_{n=m+1}^{infty}dfrac{1}{n^{alpha}}.$$



I have made some numerical experiments in Worfram which indicates that this is true, but I have no idea how to perform ao proof.






real-analysis complex-analysis number-theory numerical-methods zeta-functions







share|cite|improve this question













Working in a project I came across to the following problem, to investigate if the following inequality is true for $alpha>2,$ and $m$ a positive integer:
$$2 sum_{n=m+2}^{infty}dfrac{1}{n^{alpha}} geq sum_{n=m+1}^{infty}dfrac{1}{n^{alpha}}.$$



I have made some numerical experiments in Worfram which indicates that this is true, but I have no idea how to perform ao proof.






real-analysis complex-analysis number-theory numerical-methods zeta-functions




real-analysis complex-analysis number-theory numerical-methods zeta-functions






share|cite|improve this question













share|cite|improve this question











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asked Nov 30 at 0:14









Eduardo

553112




553112












  • Please link to the numerical experiments. It might help.
    – Shaun
    Nov 30 at 0:18










  • Why are there two sums when you can cancel so many terms from the right?
    – T_M
    Nov 30 at 0:18


















  • Please link to the numerical experiments. It might help.
    – Shaun
    Nov 30 at 0:18










  • Why are there two sums when you can cancel so many terms from the right?
    – T_M
    Nov 30 at 0:18
















Please link to the numerical experiments. It might help.
– Shaun
Nov 30 at 0:18




Please link to the numerical experiments. It might help.
– Shaun
Nov 30 at 0:18












Why are there two sums when you can cancel so many terms from the right?
– T_M
Nov 30 at 0:18




Why are there two sums when you can cancel so many terms from the right?
– T_M
Nov 30 at 0:18










2 Answers
2






active

oldest

votes


















3














This is equivalent to just asking the question of whether



$$frac{1}{(m + 1)^{alpha}} le sum_{n = m + 2}^{infty} frac{1}{n^{alpha}}$$



or equivalently, whether



$$sum_{k = 2}^{infty} left(frac{m + 1}{m + k}right)^{alpha} ge 1$$



Now fix an exponent $alpha$. The summands are each increasing in $m$, tending towards $1$ as $m to infty$ (which immediately implies the desired inequality for $m$ large enough, depending on $alpha$); therefore, it is sufficient to study $m = 1$. We need to show that



$$2^{alpha} sum_{j = 3}^{infty} frac{1}{j^{alpha}} ge 1.$$



However, if $alpha$ is sufficiently large this does not hold: in fact,



$$lim_{alpha to infty} 2^{alpha} sum_{j = 3}^{infty} frac{1}{j^{alpha}} = 0.$$




So the punchline is that for each fixed $m$, this works if $alpha$ is small enough. And for each fixed $alpha$, it works if $m$ is large enough.






share|cite|improve this answer





























    3














    When $alpha = 4$ and $m=1$ you can do the sums in closed form:
    $$
    2 sum_{n=3}^infty frac{1}{n^4} - sum_{n=2}^infty frac{1}{n^4}
    = frac{pi^4}{90}-frac98 < 0
    $$






    share|cite|improve this answer























      Your Answer





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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3














      This is equivalent to just asking the question of whether



      $$frac{1}{(m + 1)^{alpha}} le sum_{n = m + 2}^{infty} frac{1}{n^{alpha}}$$



      or equivalently, whether



      $$sum_{k = 2}^{infty} left(frac{m + 1}{m + k}right)^{alpha} ge 1$$



      Now fix an exponent $alpha$. The summands are each increasing in $m$, tending towards $1$ as $m to infty$ (which immediately implies the desired inequality for $m$ large enough, depending on $alpha$); therefore, it is sufficient to study $m = 1$. We need to show that



      $$2^{alpha} sum_{j = 3}^{infty} frac{1}{j^{alpha}} ge 1.$$



      However, if $alpha$ is sufficiently large this does not hold: in fact,



      $$lim_{alpha to infty} 2^{alpha} sum_{j = 3}^{infty} frac{1}{j^{alpha}} = 0.$$




      So the punchline is that for each fixed $m$, this works if $alpha$ is small enough. And for each fixed $alpha$, it works if $m$ is large enough.






      share|cite|improve this answer


























        3














        This is equivalent to just asking the question of whether



        $$frac{1}{(m + 1)^{alpha}} le sum_{n = m + 2}^{infty} frac{1}{n^{alpha}}$$



        or equivalently, whether



        $$sum_{k = 2}^{infty} left(frac{m + 1}{m + k}right)^{alpha} ge 1$$



        Now fix an exponent $alpha$. The summands are each increasing in $m$, tending towards $1$ as $m to infty$ (which immediately implies the desired inequality for $m$ large enough, depending on $alpha$); therefore, it is sufficient to study $m = 1$. We need to show that



        $$2^{alpha} sum_{j = 3}^{infty} frac{1}{j^{alpha}} ge 1.$$



        However, if $alpha$ is sufficiently large this does not hold: in fact,



        $$lim_{alpha to infty} 2^{alpha} sum_{j = 3}^{infty} frac{1}{j^{alpha}} = 0.$$




        So the punchline is that for each fixed $m$, this works if $alpha$ is small enough. And for each fixed $alpha$, it works if $m$ is large enough.






        share|cite|improve this answer
























          3












          3








          3






          This is equivalent to just asking the question of whether



          $$frac{1}{(m + 1)^{alpha}} le sum_{n = m + 2}^{infty} frac{1}{n^{alpha}}$$



          or equivalently, whether



          $$sum_{k = 2}^{infty} left(frac{m + 1}{m + k}right)^{alpha} ge 1$$



          Now fix an exponent $alpha$. The summands are each increasing in $m$, tending towards $1$ as $m to infty$ (which immediately implies the desired inequality for $m$ large enough, depending on $alpha$); therefore, it is sufficient to study $m = 1$. We need to show that



          $$2^{alpha} sum_{j = 3}^{infty} frac{1}{j^{alpha}} ge 1.$$



          However, if $alpha$ is sufficiently large this does not hold: in fact,



          $$lim_{alpha to infty} 2^{alpha} sum_{j = 3}^{infty} frac{1}{j^{alpha}} = 0.$$




          So the punchline is that for each fixed $m$, this works if $alpha$ is small enough. And for each fixed $alpha$, it works if $m$ is large enough.






          share|cite|improve this answer












          This is equivalent to just asking the question of whether



          $$frac{1}{(m + 1)^{alpha}} le sum_{n = m + 2}^{infty} frac{1}{n^{alpha}}$$



          or equivalently, whether



          $$sum_{k = 2}^{infty} left(frac{m + 1}{m + k}right)^{alpha} ge 1$$



          Now fix an exponent $alpha$. The summands are each increasing in $m$, tending towards $1$ as $m to infty$ (which immediately implies the desired inequality for $m$ large enough, depending on $alpha$); therefore, it is sufficient to study $m = 1$. We need to show that



          $$2^{alpha} sum_{j = 3}^{infty} frac{1}{j^{alpha}} ge 1.$$



          However, if $alpha$ is sufficiently large this does not hold: in fact,



          $$lim_{alpha to infty} 2^{alpha} sum_{j = 3}^{infty} frac{1}{j^{alpha}} = 0.$$




          So the punchline is that for each fixed $m$, this works if $alpha$ is small enough. And for each fixed $alpha$, it works if $m$ is large enough.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 30 at 0:25









          T. Bongers

          22.8k54661




          22.8k54661























              3














              When $alpha = 4$ and $m=1$ you can do the sums in closed form:
              $$
              2 sum_{n=3}^infty frac{1}{n^4} - sum_{n=2}^infty frac{1}{n^4}
              = frac{pi^4}{90}-frac98 < 0
              $$






              share|cite|improve this answer




























                3














                When $alpha = 4$ and $m=1$ you can do the sums in closed form:
                $$
                2 sum_{n=3}^infty frac{1}{n^4} - sum_{n=2}^infty frac{1}{n^4}
                = frac{pi^4}{90}-frac98 < 0
                $$






                share|cite|improve this answer


























                  3












                  3








                  3






                  When $alpha = 4$ and $m=1$ you can do the sums in closed form:
                  $$
                  2 sum_{n=3}^infty frac{1}{n^4} - sum_{n=2}^infty frac{1}{n^4}
                  = frac{pi^4}{90}-frac98 < 0
                  $$






                  share|cite|improve this answer














                  When $alpha = 4$ and $m=1$ you can do the sums in closed form:
                  $$
                  2 sum_{n=3}^infty frac{1}{n^4} - sum_{n=2}^infty frac{1}{n^4}
                  = frac{pi^4}{90}-frac98 < 0
                  $$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 30 at 1:13

























                  answered Nov 30 at 0:42









                  Mark Fischler

                  32.3k12250




                  32.3k12250






























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