Proving convergence of series of series

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So this is a question from a test I'm trying to solve: Given the series $a_p=sum_n=2^inftyfrac1n^p$ (I know this series converges), I need to prove that the series $sum_p=2^inftya_p$ also converges.



At first I thought to prove this with induction, since for every $pge2$ each series converges, therefore their partial limit sequences also converge, and from the properties of limits the sum of a finite number of series will also tend to some limit.



But I realized that since I have an infinite amount of series, I probably can't use induction. Any help would be appreciated.




calculus sequences-and-series convergence




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    So this is a question from a test I'm trying to solve: Given the series $a_p=sum_n=2^inftyfrac1n^p$ (I know this series converges), I need to prove that the series $sum_p=2^inftya_p$ also converges.



    At first I thought to prove this with induction, since for every $pge2$ each series converges, therefore their partial limit sequences also converge, and from the properties of limits the sum of a finite number of series will also tend to some limit.



    But I realized that since I have an infinite amount of series, I probably can't use induction. Any help would be appreciated.




    calculus sequences-and-series convergence




    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      So this is a question from a test I'm trying to solve: Given the series $a_p=sum_n=2^inftyfrac1n^p$ (I know this series converges), I need to prove that the series $sum_p=2^inftya_p$ also converges.



      At first I thought to prove this with induction, since for every $pge2$ each series converges, therefore their partial limit sequences also converge, and from the properties of limits the sum of a finite number of series will also tend to some limit.



      But I realized that since I have an infinite amount of series, I probably can't use induction. Any help would be appreciated.




      calculus sequences-and-series convergence




      share|cite|improve this question













      So this is a question from a test I'm trying to solve: Given the series $a_p=sum_n=2^inftyfrac1n^p$ (I know this series converges), I need to prove that the series $sum_p=2^inftya_p$ also converges.



      At first I thought to prove this with induction, since for every $pge2$ each series converges, therefore their partial limit sequences also converge, and from the properties of limits the sum of a finite number of series will also tend to some limit.



      But I realized that since I have an infinite amount of series, I probably can't use induction. Any help would be appreciated.





      calculus sequences-and-series convergence





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      edited 2 days ago









      Arnaud Mortier

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      asked 2 days ago









      A Wombat

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          3 Answers
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          We have that



          $$
          a_p=zeta(p) -1 sim frac12^p text as p to infty,
          $$



          (refer to Growth of Riemann zeta function $zeta(k)$)



          therefore



          $$sum_p=2^inftya_p$$



          converges by limit comparison test.






          share|cite|improve this answer






























            up vote
            1
            down vote













            For an alternative way note that every term in the double series is positive, so we may interchange the order of summation by Tonelli's theorem. We can evaluate the geometric series in $p$ to find
            $$ sum limits_p=2^infty sum limits_n=2^infty frac1n^p = sum limits_n=2^infty sum limits_p=2^infty frac1n^p = sum limits_n=2^infty left(frac11-frac1n - 1 - frac1nright) = sum limits_n=2^infty left(frac1n-1 - frac1nright) , . $$
            Now we either employ the estimate $frac1n-1 - frac1n = frac1(n-1)n < frac1(n-1)^2$ to obtain an upper bound or, even better, we look through a telescope to spot the exact value of the series.






            share|cite|improve this answer




























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              $beginarray\
              a_p
              &=sum_n=2^inftyfrac1n^p\
              &=frac12^psum_n=2^inftyfrac1(n/2)^p\
              &=frac12^p(1+sum_n=3^inftyfrac1(n/2)^p)\
              &lefrac12^p(1+sum_n=3^inftyfrac1(n/2)^2)
              qquadtextsince (n/2)^2 le (n/2)^p\
              &lefrac12^p(1+4sum_n=3^inftyfrac1n^2)\
              &ltfrac12^p(1+4sum_n=3^inftyfrac1n(n-1))
              qquadtextstandard trick\
              &=frac12^p(1+4sum_n=3^infty(frac1n-1-frac1n))\
              &=frac12^p(1+4frac12)\
              &=frac52frac12^p\
              textso\
              sum_p=2^infty a_p
              &< sum_p=2^inftyfrac52frac12^p\
              &=frac52\
              endarray
              $






              share|cite|improve this answer





















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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

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                active

                oldest

                votes






                active

                oldest

                votes








                up vote
                1
                down vote













                We have that



                $$
                a_p=zeta(p) -1 sim frac12^p text as p to infty,
                $$



                (refer to Growth of Riemann zeta function $zeta(k)$)



                therefore



                $$sum_p=2^inftya_p$$



                converges by limit comparison test.






                share|cite|improve this answer



























                  up vote
                  1
                  down vote













                  We have that



                  $$
                  a_p=zeta(p) -1 sim frac12^p text as p to infty,
                  $$



                  (refer to Growth of Riemann zeta function $zeta(k)$)



                  therefore



                  $$sum_p=2^inftya_p$$



                  converges by limit comparison test.






                  share|cite|improve this answer

























                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    We have that



                    $$
                    a_p=zeta(p) -1 sim frac12^p text as p to infty,
                    $$



                    (refer to Growth of Riemann zeta function $zeta(k)$)



                    therefore



                    $$sum_p=2^inftya_p$$



                    converges by limit comparison test.






                    share|cite|improve this answer















                    We have that



                    $$
                    a_p=zeta(p) -1 sim frac12^p text as p to infty,
                    $$



                    (refer to Growth of Riemann zeta function $zeta(k)$)



                    therefore



                    $$sum_p=2^inftya_p$$



                    converges by limit comparison test.







                    share|cite|improve this answer















                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 2 days ago


























                    answered 2 days ago









                    gimusi

                    63.6k73380




                    63.6k73380




















                        up vote
                        1
                        down vote













                        For an alternative way note that every term in the double series is positive, so we may interchange the order of summation by Tonelli's theorem. We can evaluate the geometric series in $p$ to find
                        $$ sum limits_p=2^infty sum limits_n=2^infty frac1n^p = sum limits_n=2^infty sum limits_p=2^infty frac1n^p = sum limits_n=2^infty left(frac11-frac1n - 1 - frac1nright) = sum limits_n=2^infty left(frac1n-1 - frac1nright) , . $$
                        Now we either employ the estimate $frac1n-1 - frac1n = frac1(n-1)n < frac1(n-1)^2$ to obtain an upper bound or, even better, we look through a telescope to spot the exact value of the series.






                        share|cite|improve this answer

























                          up vote
                          1
                          down vote













                          For an alternative way note that every term in the double series is positive, so we may interchange the order of summation by Tonelli's theorem. We can evaluate the geometric series in $p$ to find
                          $$ sum limits_p=2^infty sum limits_n=2^infty frac1n^p = sum limits_n=2^infty sum limits_p=2^infty frac1n^p = sum limits_n=2^infty left(frac11-frac1n - 1 - frac1nright) = sum limits_n=2^infty left(frac1n-1 - frac1nright) , . $$
                          Now we either employ the estimate $frac1n-1 - frac1n = frac1(n-1)n < frac1(n-1)^2$ to obtain an upper bound or, even better, we look through a telescope to spot the exact value of the series.






                          share|cite|improve this answer























                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            For an alternative way note that every term in the double series is positive, so we may interchange the order of summation by Tonelli's theorem. We can evaluate the geometric series in $p$ to find
                            $$ sum limits_p=2^infty sum limits_n=2^infty frac1n^p = sum limits_n=2^infty sum limits_p=2^infty frac1n^p = sum limits_n=2^infty left(frac11-frac1n - 1 - frac1nright) = sum limits_n=2^infty left(frac1n-1 - frac1nright) , . $$
                            Now we either employ the estimate $frac1n-1 - frac1n = frac1(n-1)n < frac1(n-1)^2$ to obtain an upper bound or, even better, we look through a telescope to spot the exact value of the series.






                            share|cite|improve this answer













                            For an alternative way note that every term in the double series is positive, so we may interchange the order of summation by Tonelli's theorem. We can evaluate the geometric series in $p$ to find
                            $$ sum limits_p=2^infty sum limits_n=2^infty frac1n^p = sum limits_n=2^infty sum limits_p=2^infty frac1n^p = sum limits_n=2^infty left(frac11-frac1n - 1 - frac1nright) = sum limits_n=2^infty left(frac1n-1 - frac1nright) , . $$
                            Now we either employ the estimate $frac1n-1 - frac1n = frac1(n-1)n < frac1(n-1)^2$ to obtain an upper bound or, even better, we look through a telescope to spot the exact value of the series.







                            share|cite|improve this answer













                            share|cite|improve this answer



                            share|cite|improve this answer











                            answered yesterday









                            ComplexYetTrivial

                            2,582524




                            2,582524




















                                up vote
                                1
                                down vote













                                $beginarray\
                                a_p
                                &=sum_n=2^inftyfrac1n^p\
                                &=frac12^psum_n=2^inftyfrac1(n/2)^p\
                                &=frac12^p(1+sum_n=3^inftyfrac1(n/2)^p)\
                                &lefrac12^p(1+sum_n=3^inftyfrac1(n/2)^2)
                                qquadtextsince (n/2)^2 le (n/2)^p\
                                &lefrac12^p(1+4sum_n=3^inftyfrac1n^2)\
                                &ltfrac12^p(1+4sum_n=3^inftyfrac1n(n-1))
                                qquadtextstandard trick\
                                &=frac12^p(1+4sum_n=3^infty(frac1n-1-frac1n))\
                                &=frac12^p(1+4frac12)\
                                &=frac52frac12^p\
                                textso\
                                sum_p=2^infty a_p
                                &< sum_p=2^inftyfrac52frac12^p\
                                &=frac52\
                                endarray
                                $






                                share|cite|improve this answer

























                                  up vote
                                  1
                                  down vote













                                  $beginarray\
                                  a_p
                                  &=sum_n=2^inftyfrac1n^p\
                                  &=frac12^psum_n=2^inftyfrac1(n/2)^p\
                                  &=frac12^p(1+sum_n=3^inftyfrac1(n/2)^p)\
                                  &lefrac12^p(1+sum_n=3^inftyfrac1(n/2)^2)
                                  qquadtextsince (n/2)^2 le (n/2)^p\
                                  &lefrac12^p(1+4sum_n=3^inftyfrac1n^2)\
                                  &ltfrac12^p(1+4sum_n=3^inftyfrac1n(n-1))
                                  qquadtextstandard trick\
                                  &=frac12^p(1+4sum_n=3^infty(frac1n-1-frac1n))\
                                  &=frac12^p(1+4frac12)\
                                  &=frac52frac12^p\
                                  textso\
                                  sum_p=2^infty a_p
                                  &< sum_p=2^inftyfrac52frac12^p\
                                  &=frac52\
                                  endarray
                                  $






                                  share|cite|improve this answer























                                    up vote
                                    1
                                    down vote










                                    up vote
                                    1
                                    down vote









                                    $beginarray\
                                    a_p
                                    &=sum_n=2^inftyfrac1n^p\
                                    &=frac12^psum_n=2^inftyfrac1(n/2)^p\
                                    &=frac12^p(1+sum_n=3^inftyfrac1(n/2)^p)\
                                    &lefrac12^p(1+sum_n=3^inftyfrac1(n/2)^2)
                                    qquadtextsince (n/2)^2 le (n/2)^p\
                                    &lefrac12^p(1+4sum_n=3^inftyfrac1n^2)\
                                    &ltfrac12^p(1+4sum_n=3^inftyfrac1n(n-1))
                                    qquadtextstandard trick\
                                    &=frac12^p(1+4sum_n=3^infty(frac1n-1-frac1n))\
                                    &=frac12^p(1+4frac12)\
                                    &=frac52frac12^p\
                                    textso\
                                    sum_p=2^infty a_p
                                    &< sum_p=2^inftyfrac52frac12^p\
                                    &=frac52\
                                    endarray
                                    $






                                    share|cite|improve this answer













                                    $beginarray\
                                    a_p
                                    &=sum_n=2^inftyfrac1n^p\
                                    &=frac12^psum_n=2^inftyfrac1(n/2)^p\
                                    &=frac12^p(1+sum_n=3^inftyfrac1(n/2)^p)\
                                    &lefrac12^p(1+sum_n=3^inftyfrac1(n/2)^2)
                                    qquadtextsince (n/2)^2 le (n/2)^p\
                                    &lefrac12^p(1+4sum_n=3^inftyfrac1n^2)\
                                    &ltfrac12^p(1+4sum_n=3^inftyfrac1n(n-1))
                                    qquadtextstandard trick\
                                    &=frac12^p(1+4sum_n=3^infty(frac1n-1-frac1n))\
                                    &=frac12^p(1+4frac12)\
                                    &=frac52frac12^p\
                                    textso\
                                    sum_p=2^infty a_p
                                    &< sum_p=2^inftyfrac52frac12^p\
                                    &=frac52\
                                    endarray
                                    $







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                                    answered yesterday









                                    marty cohen

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