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?

Must we add them one-by-one?

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

 A: 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.
A: 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 $\frac{50}{3\cdot2^{49}-1}$
A: $D_j=2D_{j-1}+1\\\ \ \ \ =2(2D_{j-2}+1)+1\\\ \ \ \ =4D_{j-2}+1+2\\\ \ \ \ =4(2D_{j-3}+1)+1+2\\\ \ \ \ \ \ \ \ \vdots\\\ \ \ \ =2^kD_{j-k}+2^k-1\\\ \ \ \ =2^{j-1}D_1+2^{j-1}-1\\\ \ \ \ =3\cdot2^{j-1}-1$
$s_n=N_n/D_n=\displaystyle\frac n{3\cdot2^{n-1}-1}, n\ge1$

$s_n<\displaystyle\frac n{3\cdot2^{n-1}-2^{n-1}}=\frac n{2^n}$
$\displaystyle\sum_1^\infty s_n<\sum_1^\infty \frac n{2^n}$ which is an AP-GP series
$\sum_1^\infty \frac n{2^n}=\frac12+\frac24+\frac38...$
$\frac12\sum_1^\infty \frac n{2^n}=0+\frac14+\frac28+\frac3{16}...$
$\sum_1^\infty \frac n{2^n}-\frac12\sum_1^\infty \frac n{2^n}=\frac12\sum_1^\infty \frac n{2^n}=\frac12+\frac14+\frac18...=1$
$\displaystyle\implies0<\sum_1^\infty s_n<2$
