Two infinite sums

$\begin{aligned} S_x & = 1 + \frac 1x + \frac 1{x^2} + \frac 1{x^3} + \frac 1{x^4} + \cdots & \small \color{#3D99F6} \text{Multiply both sides by }x \\ xS_x & = x + \color{#3D99F6} 1 + \frac 1x + \frac 1{x^2} + \frac 1{x^3} + \cdots \\ xS_x & = x + \color{#3D99F6} S_x \\ xS_x-S_x & = x \\ \implies S_x & = \frac x{x-1} \\ \therefore S_7 & = \boxed{\dfrac 76} \end{aligned}$

Let's call the infinite sum x, so:

x = 1+ $\frac{1}{7^1}$ + $\frac{1}{7^2}$ + $\frac{1}{7^3}$ + $\frac{1}{7^4}$ ...

Now, divide both sides by 7, that way, 1 turns into $\frac{1}{7^1}$ , $\frac{1}{7^1}$ turns into $\frac{1}{7^2}$ , $\frac{1}{7^2}$ turns into $\frac{1}{7^3}$ and so on. So now:

$\frac{x}{7}$ = $\frac{1}{7^1}$ + $\frac{1}{7^2}$ + $\frac{1}{7^3}$ + $\frac{1}{7^4}$ ...

Here's the important part, you'll notice that the infinite sum is almost identical to x, except, there isn't a 1 in the beginning, so it can now be expressed as: x-1. That way, it turns into:

$\frac{x}{7}$ = x-1

x = 7x-7

-6x = -7

x = $\frac{7}{6}$