Solve for x, such that: $\sum_{n=0} \frac{x}{4^n} = \sum_{n=0} \frac{1}{x^n}$

Question

The question is stated as the following to solve for x: $$x+\frac{x}{4}+\frac{x}{16}+\frac{x}{64}+...=1+\frac{1}{x}+\frac{1}{x^2}+...$$

I have so far determined their infinite series as shown in the title and have been treating both these series as geometric series. I am pretty confused in terms of next steps, thank you in advance!

Answer 1

I will refer to the question, since the title has a couple of misleading indices I guess... The LHS is equal to $x (1 + \frac{1}{4} + \frac{1}{16} + ...) = x \frac{1}{1 - \frac{1}{4}} = \frac{4x}{3}$. The RHS is equal to $\frac{1}{1 - \frac{1}{x}} = \frac{x}{x-1}$, but it converges only for $|x| > 1$. Now we solve $$ \frac{4x}{3} = \frac{x}{x-1} \implies x = 0, \frac{7}{4}, $$ but $x = \frac{7}{4}$ is the only acceptable solution.

Answer 2

Hint: for $|x|<1$, we have that $$\sum_{n=0}^{+\infty} x^n=\frac{1}{1-x}$$

Answer 3

Indeed you so have

$${\sum_{n=0}^{\infty}\frac{x}{4^n}=\sum_{n=0}^{\infty}\frac{1}{x^n}=\frac{1}{1-\frac{1}{x}}}$$

Where the ${\frac{1}{1-\frac{1}{x}}}$ comes from summing an infinite geometric series with ratio ${\frac{1}{x}}$. Note we are also making an implicit assumption here - that ${|x|>1}$ in order for the geometric series to converge.

Assuming the left hand infinite sum converges, dividing through by ${x}$ yields that

$${\frac{1}{x - 1} = \sum_{n=0}^{\infty}\frac{1}{4^n}=\frac{1}{1-\frac{1}{4}}}$$

(again, the rightmost equality comes from summing an infinite geometric series). This implies

$${x-1 = 1-\frac{1}{4} = \frac{3}{4}}$$

And so

$${x = 1 + \frac{3}{4}=\frac{7}{4}}$$

As required.