Substitute or visualise?

Visual approach:

Video link

$\implies x=\dfrac{1}{3}$

$\implies a=1, b=3$

$\implies a+b=\boxed{4}$

Algebraic approach:

$x = \displaystyle \sum_{n=1}^\infty \dfrac{1}{4^n} = \dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3} \cdots$

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

$\implies 4x - 1 = \dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3} \cdots =x$

$\implies 4x-1=x$ $\implies 3x=1$ $\implies x=\dfrac{1}{3}$

$\implies a=1, b=3$

$\implies a+b=\boxed{4}$

$\sum_{n=1}^\infty \left(\frac 14 \right)^n=\sum_{n=0}^\infty \left(\frac 14 \right)^n-1=\dfrac{1}{1-\dfrac{1}{4}}-1=\dfrac{1}{\dfrac{3}{4}}-1=\dfrac{4}{3}-1=\dfrac{1}{3}$

Let $S_n$ be the answer

$S_n=\sum_{n=1}^{∞}=\frac{1}{4}+\frac{1}{16}+\frac{1}{64}...$

$\frac{1}{4}S_n=\frac{1}{16}+\frac{1}{64}+\frac{1}{256}...$

$S_n(1-\frac{1}{4})=\frac{1}{4}$

$S_n(\frac{3}{4})=\frac{1}{4}$

$S_n=\frac{1}{4} \times \frac{4}{3}=\frac{1}{3}$

As $HCF(1,3)=1$ therefore answer is $\boxed{3+1=4}$

Use the geometric series formula:

$a \sum_{n=0}^{\infty} r^n = \frac{a}{1-r}$ .

Here $a = r = \frac{1}{4}$ , so the sum is $(\frac{1}{4} )(\frac{4}{3}) = \frac{1}{3}$ .

Substitution method:

Let $S = \sum_{n=1}^{\infty} \frac{1}{4^n}$ .

Then $4S = \sum_{n=0}^{\infty} \frac{1}{4^n} = 1 + \sum_{n=1}^{\infty} \frac{1}{4^n}$ ,

so $4S - 1 = S$ . I.e., $S = \frac{1}{3}$ .

I'll leave the visual proof for somebody else. It makes a nice picture.