Terminating Both Algorithms

Let the expression be $f(x)$ , then we have:

$\begin{aligned} f(x) & = \frac{1+x+x^2+x^3+...}{1+2x+3x^2+4x^3+...} = \frac{1+x+x^2+x^3+...}{\frac{d}{dx}(1+x+x^2+x^3+...)} \\ & = \frac{\frac{1}{1-x}}{\frac{d}{dx}\left(\frac{1}{1-x}\right)} = \frac{\frac{1}{1-x}}{\frac{1}{(1-x)^2}} = \frac{(1-x)^2}{1-x} = \boxed{1-x} \end{aligned}$

$Required\quad value,\quad S=\cfrac { 1+x+{ x }^{ 2 }{ +x }^{ 3 }+... }{ 1+2x+{ 3x }^{ 2 }{ +4x }^{ 3 }+... } \\ \\ The\quad denominator\quad of\quad S\quad is\quad an\quad arithmetic-geometric\quad series.\quad \\ \\ Let\quad y=1+2x+{ 3x }^{ 2 }{ +4x }^{ 3 }+...\\ \\ \quad \Rightarrow xy=\quad \quad x+{ 2x }^{ 2 }{ +3x }^{ 3 }+...\\ \\ Subtracting\quad the\quad two,\quad we\quad get\\ y(1-x)=1+(2-1)x+(3-2){ x }^{ 2 }+(4-3){ x }^{ 3 }+...\\ \\ \Rightarrow y=\cfrac { 1+x+{ x }^{ 2 }{ +x }^{ 3 }+... }{ 1-x } \\ \\ Hence,\quad S=\cfrac { 1+x+{ x }^{ 2 }{ +x }^{ 3 }+... }{ \cfrac { 1+x+{ x }^{ 2 }{ +x }^{ 3 }+... }{ 1-x } } \quad =\quad 1-x.\\ \\$

Let

$1+x+x^{2}+x^{3}+\cdots = S_{1}$

$1+2x+3x^{2}+4x^{3}+\cdots = S_{2}$

Now,

$~~~~~~ S_{2} = 1+2x+3x^{2}+4x^{3}+\cdots \\ - S_{2}x = ~~~~~~~~~x+2x^{2}+3x^{3}+4x^{4}+\cdots \\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_} \\ S_{2}(1-x) = 1+x+x^{2}+x^{3}+\cdots \\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_} \\ \implies S_{2}(1-x) = S_{1} \\ \implies S_{2} = \dfrac{S_{1}}{(1-x)} \\$

Hence,

$\dfrac{S_{1}}{S_{2}} \iff \dfrac{S_{1}}{{S_{1}}/{(1-x)}} \implies \boxed{(1-x)}$