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

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?

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

• Answers to your question: no and no. – Yves Daoust Dec 9 '18 at 9:19
• Did you mean to write $D_j = 2D_{j-1}+1$ for $j > 1$? – JimmyK4542 Dec 9 '18 at 9:20
• @JimmyK4542 Yes, I did. – Hussain-Alqatari Dec 9 '18 at 9:23
• @AweKumarJha , $\frac{50}{523}$ is not true, the $50^\text{th}$ term is much less than that. – Hussain-Alqatari Dec 9 '18 at 9:26
• 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. – Awe Kumar Jha Dec 9 '18 at 9:28

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.

• This is helpful. Thank you. What about my second and the third questions? – Hussain-Alqatari Dec 9 '18 at 9:30
• 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. – Anubhab Ghosal Dec 9 '18 at 9:32
• Please use \cdot instead of .. The notation is really confusing. – Kemono Chen Dec 9 '18 at 9:35
• @KemonoChen, fixed it. – Anubhab Ghosal Dec 9 '18 at 9:36
• Appreciate the work, THANKS! – Hussain-Alqatari Dec 9 '18 at 10:36

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 $$\frac{50}{3\cdot2^{49}-1}$$

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

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

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

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

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

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

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

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

• The sum is not less than $4/3$, it is $1.5997809\dots$. – Hussain-Alqatari Dec 9 '18 at 9:58
• My bad, I'll recheck what I wrote. – Shubham Johri Dec 9 '18 at 10:00
• Actually, s_n > n/(3⋅2^n−1) – Ankit Kumar Dec 9 '18 at 10:01
• Yeah, thanks for pointing it out. – Shubham Johri Dec 9 '18 at 10:03