The series $frac{1}{2}+frac{2}{5}+frac{3}{11}+frac{4}{23}+…$












1












$begingroup$


Consider the expression $frac{1}{2}+frac{2}{5}+frac{3}{11}+frac{4}{23}+...$



Denote the numerator and the denominator of the $j^text{th}$ term by $N_{j}$ and $D_{j}$, respectively. Then, $N_1=1$, $D_1=2$, and, for every $j>1$, $$
N_j=
N_{j-1}+1qquad D_j=
2D_{j-1}+1$$




What is the $50^text{th}$ term?




Must we evaluate that term-by-term until we reach the $50^text{th}$ term?




What is the sum of the first $25$ term?




Must we add them one-by-one?




What is the exact value of the sum to $infty$?







sequences-and-series convergence summation recurrence-relations







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Answers to your question: no and no.
    $endgroup$
    – Yves Daoust
    Dec 9 '18 at 9:19








  • 1




    $begingroup$
    Did you mean to write $D_j = 2D_{j-1}+1$ for $j > 1$?
    $endgroup$
    – JimmyK4542
    Dec 9 '18 at 9:20










  • $begingroup$
    @JimmyK4542 Yes, I did.
    $endgroup$
    – Hussain-Alqatari
    Dec 9 '18 at 9:23










  • $begingroup$
    @AweKumarJha , $frac{50}{523}$ is not true, the $50^text{th}$ term is much less than that.
    $endgroup$
    – Hussain-Alqatari
    Dec 9 '18 at 9:26










  • $begingroup$
    Then it just looks like a very similar series involving primes. Try using the method of generating functions to convert the recursion into explicit formula.
    $endgroup$
    – Awe Kumar Jha
    Dec 9 '18 at 9:28


















1












$begingroup$


Consider the expression $frac{1}{2}+frac{2}{5}+frac{3}{11}+frac{4}{23}+...$



Denote the numerator and the denominator of the $j^text{th}$ term by $N_{j}$ and $D_{j}$, respectively. Then, $N_1=1$, $D_1=2$, and, for every $j>1$, $$
N_j=
N_{j-1}+1qquad D_j=
2D_{j-1}+1$$




What is the $50^text{th}$ term?




Must we evaluate that term-by-term until we reach the $50^text{th}$ term?




What is the sum of the first $25$ term?




Must we add them one-by-one?




What is the exact value of the sum to $infty$?







sequences-and-series convergence summation recurrence-relations







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Answers to your question: no and no.
    $endgroup$
    – Yves Daoust
    Dec 9 '18 at 9:19








  • 1




    $begingroup$
    Did you mean to write $D_j = 2D_{j-1}+1$ for $j > 1$?
    $endgroup$
    – JimmyK4542
    Dec 9 '18 at 9:20










  • $begingroup$
    @JimmyK4542 Yes, I did.
    $endgroup$
    – Hussain-Alqatari
    Dec 9 '18 at 9:23










  • $begingroup$
    @AweKumarJha , $frac{50}{523}$ is not true, the $50^text{th}$ term is much less than that.
    $endgroup$
    – Hussain-Alqatari
    Dec 9 '18 at 9:26










  • $begingroup$
    Then it just looks like a very similar series involving primes. Try using the method of generating functions to convert the recursion into explicit formula.
    $endgroup$
    – Awe Kumar Jha
    Dec 9 '18 at 9:28
















1












1








1


1



$begingroup$


Consider the expression $frac{1}{2}+frac{2}{5}+frac{3}{11}+frac{4}{23}+...$



Denote the numerator and the denominator of the $j^text{th}$ term by $N_{j}$ and $D_{j}$, respectively. Then, $N_1=1$, $D_1=2$, and, for every $j>1$, $$
N_j=
N_{j-1}+1qquad D_j=
2D_{j-1}+1$$




What is the $50^text{th}$ term?




Must we evaluate that term-by-term until we reach the $50^text{th}$ term?




What is the sum of the first $25$ term?




Must we add them one-by-one?




What is the exact value of the sum to $infty$?







sequences-and-series convergence summation recurrence-relations







share|cite|improve this question











$endgroup$




Consider the expression $frac{1}{2}+frac{2}{5}+frac{3}{11}+frac{4}{23}+...$



Denote the numerator and the denominator of the $j^text{th}$ term by $N_{j}$ and $D_{j}$, respectively. Then, $N_1=1$, $D_1=2$, and, for every $j>1$, $$
N_j=
N_{j-1}+1qquad D_j=
2D_{j-1}+1$$




What is the $50^text{th}$ term?




Must we evaluate that term-by-term until we reach the $50^text{th}$ term?




What is the sum of the first $25$ term?




Must we add them one-by-one?




What is the exact value of the sum to $infty$?







sequences-and-series convergence summation recurrence-relations




sequences-and-series convergence summation recurrence-relations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 9:56









Did

247k23222458




247k23222458










asked Dec 9 '18 at 9:14









Hussain-AlqatariHussain-Alqatari

3187




3187








  • 2




    $begingroup$
    Answers to your question: no and no.
    $endgroup$
    – Yves Daoust
    Dec 9 '18 at 9:19








  • 1




    $begingroup$
    Did you mean to write $D_j = 2D_{j-1}+1$ for $j > 1$?
    $endgroup$
    – JimmyK4542
    Dec 9 '18 at 9:20










  • $begingroup$
    @JimmyK4542 Yes, I did.
    $endgroup$
    – Hussain-Alqatari
    Dec 9 '18 at 9:23










  • $begingroup$
    @AweKumarJha , $frac{50}{523}$ is not true, the $50^text{th}$ term is much less than that.
    $endgroup$
    – Hussain-Alqatari
    Dec 9 '18 at 9:26










  • $begingroup$
    Then it just looks like a very similar series involving primes. Try using the method of generating functions to convert the recursion into explicit formula.
    $endgroup$
    – Awe Kumar Jha
    Dec 9 '18 at 9:28
















  • 2




    $begingroup$
    Answers to your question: no and no.
    $endgroup$
    – Yves Daoust
    Dec 9 '18 at 9:19








  • 1




    $begingroup$
    Did you mean to write $D_j = 2D_{j-1}+1$ for $j > 1$?
    $endgroup$
    – JimmyK4542
    Dec 9 '18 at 9:20










  • $begingroup$
    @JimmyK4542 Yes, I did.
    $endgroup$
    – Hussain-Alqatari
    Dec 9 '18 at 9:23










  • $begingroup$
    @AweKumarJha , $frac{50}{523}$ is not true, the $50^text{th}$ term is much less than that.
    $endgroup$
    – Hussain-Alqatari
    Dec 9 '18 at 9:26










  • $begingroup$
    Then it just looks like a very similar series involving primes. Try using the method of generating functions to convert the recursion into explicit formula.
    $endgroup$
    – Awe Kumar Jha
    Dec 9 '18 at 9:28










2




2




$begingroup$
Answers to your question: no and no.
$endgroup$
– Yves Daoust
Dec 9 '18 at 9:19






$begingroup$
Answers to your question: no and no.
$endgroup$
– Yves Daoust
Dec 9 '18 at 9:19






1




1




$begingroup$
Did you mean to write $D_j = 2D_{j-1}+1$ for $j > 1$?
$endgroup$
– JimmyK4542
Dec 9 '18 at 9:20




$begingroup$
Did you mean to write $D_j = 2D_{j-1}+1$ for $j > 1$?
$endgroup$
– JimmyK4542
Dec 9 '18 at 9:20












$begingroup$
@JimmyK4542 Yes, I did.
$endgroup$
– Hussain-Alqatari
Dec 9 '18 at 9:23




$begingroup$
@JimmyK4542 Yes, I did.
$endgroup$
– Hussain-Alqatari
Dec 9 '18 at 9:23












$begingroup$
@AweKumarJha , $frac{50}{523}$ is not true, the $50^text{th}$ term is much less than that.
$endgroup$
– Hussain-Alqatari
Dec 9 '18 at 9:26




$begingroup$
@AweKumarJha , $frac{50}{523}$ is not true, the $50^text{th}$ term is much less than that.
$endgroup$
– Hussain-Alqatari
Dec 9 '18 at 9:26












$begingroup$
Then it just looks like a very similar series involving primes. Try using the method of generating functions to convert the recursion into explicit formula.
$endgroup$
– Awe Kumar Jha
Dec 9 '18 at 9:28






$begingroup$
Then it just looks like a very similar series involving primes. Try using the method of generating functions to convert the recursion into explicit formula.
$endgroup$
– Awe Kumar Jha
Dec 9 '18 at 9:28












3 Answers
3






active

oldest

votes


















3












$begingroup$

The $n^{th}$ term is $frac{n}{3cdot 2^{n-1}-1}$, as can easily be proven by induction. [You can guess this by simply looking at the first two terms as you know it has to be of the form $ccdot2^n+d$]



$therefore 50^{th}$ term$=frac{50}{3cdot 2^{49}-1}$.



I tried all the methods I know(which includes generating functions, bruteforce calculation, CAS). The sum does not have a closed form formula.(As is the case with most nontrivial rapidly converging sums). You may however calculate the sum with arbitrary precision pretty easily.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is helpful. Thank you. What about my second and the third questions?
    $endgroup$
    – Hussain-Alqatari
    Dec 9 '18 at 9:30










  • $begingroup$
    I am trying those parts. Will edit as soon as I get something useful. As of now, I think there is no closed form for the summation.
    $endgroup$
    – Anubhab Ghosal
    Dec 9 '18 at 9:32






  • 1




    $begingroup$
    Please use cdot instead of .. The notation is really confusing.
    $endgroup$
    – Kemono Chen
    Dec 9 '18 at 9:35










  • $begingroup$
    @KemonoChen, fixed it.
    $endgroup$
    – Anubhab Ghosal
    Dec 9 '18 at 9:36






  • 1




    $begingroup$
    Appreciate the work, THANKS!
    $endgroup$
    – Hussain-Alqatari
    Dec 9 '18 at 10:36



















1












$begingroup$

If $d_{j+1}=2d_j+1$ then $d_{j+1}+1=2(d_j+1)Rightarrow d_n+1=(d_1+1)2^{n-1}=3cdot2^{n-1}$.



So 50th term is $frac{50}{3cdot2^{49}-1}$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    $D_j=2D_{j-1}+1\ =2(2D_{j-2}+1)+1\ =4D_{j-2}+1+2\ =4(2D_{j-3}+1)+1+2\ vdots\ =2^kD_{j-k}+2^k-1\ =2^{j-1}D_1+2^{j-1}-1\ =3cdot2^{j-1}-1$



    $s_n=N_n/D_n=displaystylefrac n{3cdot2^{n-1}-1}, nge1$




    $s_n<displaystylefrac n{3cdot2^{n-1}-2^{n-1}}=frac n{2^n}$



    $displaystylesum_1^infty s_n<sum_1^infty frac n{2^n}$ which is an AP-GP series



    $sum_1^infty frac n{2^n}=frac12+frac24+frac38...$



    $frac12sum_1^infty frac n{2^n}=0+frac14+frac28+frac3{16}...$



    $sum_1^infty frac n{2^n}-frac12sum_1^infty frac n{2^n}=frac12sum_1^infty frac n{2^n}=frac12+frac14+frac18...=1$



    $displaystyleimplies0<sum_1^infty s_n<2$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      The sum is not less than $4/3$, it is $1.5997809dots$.
      $endgroup$
      – Hussain-Alqatari
      Dec 9 '18 at 9:58










    • $begingroup$
      My bad, I'll recheck what I wrote.
      $endgroup$
      – Shubham Johri
      Dec 9 '18 at 10:00










    • $begingroup$
      Actually, s_n > n/(3⋅2^n−1)
      $endgroup$
      – Ankit Kumar
      Dec 9 '18 at 10:01










    • $begingroup$
      Yeah, thanks for pointing it out.
      $endgroup$
      – Shubham Johri
      Dec 9 '18 at 10:03











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    The $n^{th}$ term is $frac{n}{3cdot 2^{n-1}-1}$, as can easily be proven by induction. [You can guess this by simply looking at the first two terms as you know it has to be of the form $ccdot2^n+d$]



    $therefore 50^{th}$ term$=frac{50}{3cdot 2^{49}-1}$.



    I tried all the methods I know(which includes generating functions, bruteforce calculation, CAS). The sum does not have a closed form formula.(As is the case with most nontrivial rapidly converging sums). You may however calculate the sum with arbitrary precision pretty easily.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      This is helpful. Thank you. What about my second and the third questions?
      $endgroup$
      – Hussain-Alqatari
      Dec 9 '18 at 9:30










    • $begingroup$
      I am trying those parts. Will edit as soon as I get something useful. As of now, I think there is no closed form for the summation.
      $endgroup$
      – Anubhab Ghosal
      Dec 9 '18 at 9:32






    • 1




      $begingroup$
      Please use cdot instead of .. The notation is really confusing.
      $endgroup$
      – Kemono Chen
      Dec 9 '18 at 9:35










    • $begingroup$
      @KemonoChen, fixed it.
      $endgroup$
      – Anubhab Ghosal
      Dec 9 '18 at 9:36






    • 1




      $begingroup$
      Appreciate the work, THANKS!
      $endgroup$
      – Hussain-Alqatari
      Dec 9 '18 at 10:36
















    3












    $begingroup$

    The $n^{th}$ term is $frac{n}{3cdot 2^{n-1}-1}$, as can easily be proven by induction. [You can guess this by simply looking at the first two terms as you know it has to be of the form $ccdot2^n+d$]



    $therefore 50^{th}$ term$=frac{50}{3cdot 2^{49}-1}$.



    I tried all the methods I know(which includes generating functions, bruteforce calculation, CAS). The sum does not have a closed form formula.(As is the case with most nontrivial rapidly converging sums). You may however calculate the sum with arbitrary precision pretty easily.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      This is helpful. Thank you. What about my second and the third questions?
      $endgroup$
      – Hussain-Alqatari
      Dec 9 '18 at 9:30










    • $begingroup$
      I am trying those parts. Will edit as soon as I get something useful. As of now, I think there is no closed form for the summation.
      $endgroup$
      – Anubhab Ghosal
      Dec 9 '18 at 9:32






    • 1




      $begingroup$
      Please use cdot instead of .. The notation is really confusing.
      $endgroup$
      – Kemono Chen
      Dec 9 '18 at 9:35










    • $begingroup$
      @KemonoChen, fixed it.
      $endgroup$
      – Anubhab Ghosal
      Dec 9 '18 at 9:36






    • 1




      $begingroup$
      Appreciate the work, THANKS!
      $endgroup$
      – Hussain-Alqatari
      Dec 9 '18 at 10:36














    3












    3








    3





    $begingroup$

    The $n^{th}$ term is $frac{n}{3cdot 2^{n-1}-1}$, as can easily be proven by induction. [You can guess this by simply looking at the first two terms as you know it has to be of the form $ccdot2^n+d$]



    $therefore 50^{th}$ term$=frac{50}{3cdot 2^{49}-1}$.



    I tried all the methods I know(which includes generating functions, bruteforce calculation, CAS). The sum does not have a closed form formula.(As is the case with most nontrivial rapidly converging sums). You may however calculate the sum with arbitrary precision pretty easily.






    share|cite|improve this answer











    $endgroup$



    The $n^{th}$ term is $frac{n}{3cdot 2^{n-1}-1}$, as can easily be proven by induction. [You can guess this by simply looking at the first two terms as you know it has to be of the form $ccdot2^n+d$]



    $therefore 50^{th}$ term$=frac{50}{3cdot 2^{49}-1}$.



    I tried all the methods I know(which includes generating functions, bruteforce calculation, CAS). The sum does not have a closed form formula.(As is the case with most nontrivial rapidly converging sums). You may however calculate the sum with arbitrary precision pretty easily.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 9 '18 at 10:22

























    answered Dec 9 '18 at 9:28









    Anubhab GhosalAnubhab Ghosal

    82518




    82518












    • $begingroup$
      This is helpful. Thank you. What about my second and the third questions?
      $endgroup$
      – Hussain-Alqatari
      Dec 9 '18 at 9:30










    • $begingroup$
      I am trying those parts. Will edit as soon as I get something useful. As of now, I think there is no closed form for the summation.
      $endgroup$
      – Anubhab Ghosal
      Dec 9 '18 at 9:32






    • 1




      $begingroup$
      Please use cdot instead of .. The notation is really confusing.
      $endgroup$
      – Kemono Chen
      Dec 9 '18 at 9:35










    • $begingroup$
      @KemonoChen, fixed it.
      $endgroup$
      – Anubhab Ghosal
      Dec 9 '18 at 9:36






    • 1




      $begingroup$
      Appreciate the work, THANKS!
      $endgroup$
      – Hussain-Alqatari
      Dec 9 '18 at 10:36


















    • $begingroup$
      This is helpful. Thank you. What about my second and the third questions?
      $endgroup$
      – Hussain-Alqatari
      Dec 9 '18 at 9:30










    • $begingroup$
      I am trying those parts. Will edit as soon as I get something useful. As of now, I think there is no closed form for the summation.
      $endgroup$
      – Anubhab Ghosal
      Dec 9 '18 at 9:32






    • 1




      $begingroup$
      Please use cdot instead of .. The notation is really confusing.
      $endgroup$
      – Kemono Chen
      Dec 9 '18 at 9:35










    • $begingroup$
      @KemonoChen, fixed it.
      $endgroup$
      – Anubhab Ghosal
      Dec 9 '18 at 9:36






    • 1




      $begingroup$
      Appreciate the work, THANKS!
      $endgroup$
      – Hussain-Alqatari
      Dec 9 '18 at 10:36
















    $begingroup$
    This is helpful. Thank you. What about my second and the third questions?
    $endgroup$
    – Hussain-Alqatari
    Dec 9 '18 at 9:30




    $begingroup$
    This is helpful. Thank you. What about my second and the third questions?
    $endgroup$
    – Hussain-Alqatari
    Dec 9 '18 at 9:30












    $begingroup$
    I am trying those parts. Will edit as soon as I get something useful. As of now, I think there is no closed form for the summation.
    $endgroup$
    – Anubhab Ghosal
    Dec 9 '18 at 9:32




    $begingroup$
    I am trying those parts. Will edit as soon as I get something useful. As of now, I think there is no closed form for the summation.
    $endgroup$
    – Anubhab Ghosal
    Dec 9 '18 at 9:32




    1




    1




    $begingroup$
    Please use cdot instead of .. The notation is really confusing.
    $endgroup$
    – Kemono Chen
    Dec 9 '18 at 9:35




    $begingroup$
    Please use cdot instead of .. The notation is really confusing.
    $endgroup$
    – Kemono Chen
    Dec 9 '18 at 9:35












    $begingroup$
    @KemonoChen, fixed it.
    $endgroup$
    – Anubhab Ghosal
    Dec 9 '18 at 9:36




    $begingroup$
    @KemonoChen, fixed it.
    $endgroup$
    – Anubhab Ghosal
    Dec 9 '18 at 9:36




    1




    1




    $begingroup$
    Appreciate the work, THANKS!
    $endgroup$
    – Hussain-Alqatari
    Dec 9 '18 at 10:36




    $begingroup$
    Appreciate the work, THANKS!
    $endgroup$
    – Hussain-Alqatari
    Dec 9 '18 at 10:36











    1












    $begingroup$

    If $d_{j+1}=2d_j+1$ then $d_{j+1}+1=2(d_j+1)Rightarrow d_n+1=(d_1+1)2^{n-1}=3cdot2^{n-1}$.



    So 50th term is $frac{50}{3cdot2^{49}-1}$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      If $d_{j+1}=2d_j+1$ then $d_{j+1}+1=2(d_j+1)Rightarrow d_n+1=(d_1+1)2^{n-1}=3cdot2^{n-1}$.



      So 50th term is $frac{50}{3cdot2^{49}-1}$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        If $d_{j+1}=2d_j+1$ then $d_{j+1}+1=2(d_j+1)Rightarrow d_n+1=(d_1+1)2^{n-1}=3cdot2^{n-1}$.



        So 50th term is $frac{50}{3cdot2^{49}-1}$






        share|cite|improve this answer









        $endgroup$



        If $d_{j+1}=2d_j+1$ then $d_{j+1}+1=2(d_j+1)Rightarrow d_n+1=(d_1+1)2^{n-1}=3cdot2^{n-1}$.



        So 50th term is $frac{50}{3cdot2^{49}-1}$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 9 '18 at 9:39









        MinzMinz

        951127




        951127























            1












            $begingroup$

            $D_j=2D_{j-1}+1\ =2(2D_{j-2}+1)+1\ =4D_{j-2}+1+2\ =4(2D_{j-3}+1)+1+2\ vdots\ =2^kD_{j-k}+2^k-1\ =2^{j-1}D_1+2^{j-1}-1\ =3cdot2^{j-1}-1$



            $s_n=N_n/D_n=displaystylefrac n{3cdot2^{n-1}-1}, nge1$




            $s_n<displaystylefrac n{3cdot2^{n-1}-2^{n-1}}=frac n{2^n}$



            $displaystylesum_1^infty s_n<sum_1^infty frac n{2^n}$ which is an AP-GP series



            $sum_1^infty frac n{2^n}=frac12+frac24+frac38...$



            $frac12sum_1^infty frac n{2^n}=0+frac14+frac28+frac3{16}...$



            $sum_1^infty frac n{2^n}-frac12sum_1^infty frac n{2^n}=frac12sum_1^infty frac n{2^n}=frac12+frac14+frac18...=1$



            $displaystyleimplies0<sum_1^infty s_n<2$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              The sum is not less than $4/3$, it is $1.5997809dots$.
              $endgroup$
              – Hussain-Alqatari
              Dec 9 '18 at 9:58










            • $begingroup$
              My bad, I'll recheck what I wrote.
              $endgroup$
              – Shubham Johri
              Dec 9 '18 at 10:00










            • $begingroup$
              Actually, s_n > n/(3⋅2^n−1)
              $endgroup$
              – Ankit Kumar
              Dec 9 '18 at 10:01










            • $begingroup$
              Yeah, thanks for pointing it out.
              $endgroup$
              – Shubham Johri
              Dec 9 '18 at 10:03
















            1












            $begingroup$

            $D_j=2D_{j-1}+1\ =2(2D_{j-2}+1)+1\ =4D_{j-2}+1+2\ =4(2D_{j-3}+1)+1+2\ vdots\ =2^kD_{j-k}+2^k-1\ =2^{j-1}D_1+2^{j-1}-1\ =3cdot2^{j-1}-1$



            $s_n=N_n/D_n=displaystylefrac n{3cdot2^{n-1}-1}, nge1$




            $s_n<displaystylefrac n{3cdot2^{n-1}-2^{n-1}}=frac n{2^n}$



            $displaystylesum_1^infty s_n<sum_1^infty frac n{2^n}$ which is an AP-GP series



            $sum_1^infty frac n{2^n}=frac12+frac24+frac38...$



            $frac12sum_1^infty frac n{2^n}=0+frac14+frac28+frac3{16}...$



            $sum_1^infty frac n{2^n}-frac12sum_1^infty frac n{2^n}=frac12sum_1^infty frac n{2^n}=frac12+frac14+frac18...=1$



            $displaystyleimplies0<sum_1^infty s_n<2$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              The sum is not less than $4/3$, it is $1.5997809dots$.
              $endgroup$
              – Hussain-Alqatari
              Dec 9 '18 at 9:58










            • $begingroup$
              My bad, I'll recheck what I wrote.
              $endgroup$
              – Shubham Johri
              Dec 9 '18 at 10:00










            • $begingroup$
              Actually, s_n > n/(3⋅2^n−1)
              $endgroup$
              – Ankit Kumar
              Dec 9 '18 at 10:01










            • $begingroup$
              Yeah, thanks for pointing it out.
              $endgroup$
              – Shubham Johri
              Dec 9 '18 at 10:03














            1












            1








            1





            $begingroup$

            $D_j=2D_{j-1}+1\ =2(2D_{j-2}+1)+1\ =4D_{j-2}+1+2\ =4(2D_{j-3}+1)+1+2\ vdots\ =2^kD_{j-k}+2^k-1\ =2^{j-1}D_1+2^{j-1}-1\ =3cdot2^{j-1}-1$



            $s_n=N_n/D_n=displaystylefrac n{3cdot2^{n-1}-1}, nge1$




            $s_n<displaystylefrac n{3cdot2^{n-1}-2^{n-1}}=frac n{2^n}$



            $displaystylesum_1^infty s_n<sum_1^infty frac n{2^n}$ which is an AP-GP series



            $sum_1^infty frac n{2^n}=frac12+frac24+frac38...$



            $frac12sum_1^infty frac n{2^n}=0+frac14+frac28+frac3{16}...$



            $sum_1^infty frac n{2^n}-frac12sum_1^infty frac n{2^n}=frac12sum_1^infty frac n{2^n}=frac12+frac14+frac18...=1$



            $displaystyleimplies0<sum_1^infty s_n<2$






            share|cite|improve this answer











            $endgroup$



            $D_j=2D_{j-1}+1\ =2(2D_{j-2}+1)+1\ =4D_{j-2}+1+2\ =4(2D_{j-3}+1)+1+2\ vdots\ =2^kD_{j-k}+2^k-1\ =2^{j-1}D_1+2^{j-1}-1\ =3cdot2^{j-1}-1$



            $s_n=N_n/D_n=displaystylefrac n{3cdot2^{n-1}-1}, nge1$




            $s_n<displaystylefrac n{3cdot2^{n-1}-2^{n-1}}=frac n{2^n}$



            $displaystylesum_1^infty s_n<sum_1^infty frac n{2^n}$ which is an AP-GP series



            $sum_1^infty frac n{2^n}=frac12+frac24+frac38...$



            $frac12sum_1^infty frac n{2^n}=0+frac14+frac28+frac3{16}...$



            $sum_1^infty frac n{2^n}-frac12sum_1^infty frac n{2^n}=frac12sum_1^infty frac n{2^n}=frac12+frac14+frac18...=1$



            $displaystyleimplies0<sum_1^infty s_n<2$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 9 '18 at 10:18

























            answered Dec 9 '18 at 9:52









            Shubham JohriShubham Johri

            4,885717




            4,885717












            • $begingroup$
              The sum is not less than $4/3$, it is $1.5997809dots$.
              $endgroup$
              – Hussain-Alqatari
              Dec 9 '18 at 9:58










            • $begingroup$
              My bad, I'll recheck what I wrote.
              $endgroup$
              – Shubham Johri
              Dec 9 '18 at 10:00










            • $begingroup$
              Actually, s_n > n/(3⋅2^n−1)
              $endgroup$
              – Ankit Kumar
              Dec 9 '18 at 10:01










            • $begingroup$
              Yeah, thanks for pointing it out.
              $endgroup$
              – Shubham Johri
              Dec 9 '18 at 10:03


















            • $begingroup$
              The sum is not less than $4/3$, it is $1.5997809dots$.
              $endgroup$
              – Hussain-Alqatari
              Dec 9 '18 at 9:58










            • $begingroup$
              My bad, I'll recheck what I wrote.
              $endgroup$
              – Shubham Johri
              Dec 9 '18 at 10:00










            • $begingroup$
              Actually, s_n > n/(3⋅2^n−1)
              $endgroup$
              – Ankit Kumar
              Dec 9 '18 at 10:01










            • $begingroup$
              Yeah, thanks for pointing it out.
              $endgroup$
              – Shubham Johri
              Dec 9 '18 at 10:03
















            $begingroup$
            The sum is not less than $4/3$, it is $1.5997809dots$.
            $endgroup$
            – Hussain-Alqatari
            Dec 9 '18 at 9:58




            $begingroup$
            The sum is not less than $4/3$, it is $1.5997809dots$.
            $endgroup$
            – Hussain-Alqatari
            Dec 9 '18 at 9:58












            $begingroup$
            My bad, I'll recheck what I wrote.
            $endgroup$
            – Shubham Johri
            Dec 9 '18 at 10:00




            $begingroup$
            My bad, I'll recheck what I wrote.
            $endgroup$
            – Shubham Johri
            Dec 9 '18 at 10:00












            $begingroup$
            Actually, s_n > n/(3⋅2^n−1)
            $endgroup$
            – Ankit Kumar
            Dec 9 '18 at 10:01




            $begingroup$
            Actually, s_n > n/(3⋅2^n−1)
            $endgroup$
            – Ankit Kumar
            Dec 9 '18 at 10:01












            $begingroup$
            Yeah, thanks for pointing it out.
            $endgroup$
            – Shubham Johri
            Dec 9 '18 at 10:03




            $begingroup$
            Yeah, thanks for pointing it out.
            $endgroup$
            – Shubham Johri
            Dec 9 '18 at 10:03


















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