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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 $j>1$, $$ N_j= N_{j-1}+1\qquad D_j= 2D_{j-1}+1$$

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 $\infty$?

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    $\begingroup$ Answers to your question: no and no. $\endgroup$ – Yves Daoust Dec 9 '18 at 9:19
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    $\begingroup$ 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
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The $n^{th}$ term is $\frac{n}{3\cdot 2^{n-1}-1}$, 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 $c\cdot2^n+d$]

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

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.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ Appreciate the work, THANKS! $\endgroup$ – Hussain-Alqatari Dec 9 '18 at 10:36
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If $d_{j+1}=2d_j+1$ then $d_{j+1}+1=2(d_j+1)\Rightarrow d_n+1=(d_1+1)2^{n-1}=3\cdot2^{n-1}$.

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

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

$s_n=N_n/D_n=\displaystyle\frac n{3\cdot2^{n-1}-1}, n\ge1$


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

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

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

$\frac12\sum_1^\infty \frac n{2^n}=0+\frac14+\frac28+\frac3{16}...$

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

$\displaystyle\implies0<\sum_1^\infty s_n<2$

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  • $\begingroup$ The sum is not less than $4/3$, it is $1.5997809\dots$. $\endgroup$ – 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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