Find real-valued sequences $x(n)$ for which $c^{x(n)} = o(1/n )$

Question

For which $x=x(n)$ does it hold that

$$c^x = o\left(\frac{1}{n}\right)$$

where $c\in(0,1)$ is a constant.

So clearly, for $x=n$, this is true. But for which $x =o(n)$ does this hold?

I thought of solving the equation $c^x=\frac{1}{n}$, which yields $$x=-\frac{\log n}{\log c}.$$

But somehow, I am not sure. What does this mean? For $x \gg \log n$, we have that $c^x = o\left(\frac{1}{n}\right)$?

And further: For which $x=x(n)$ does it hold that $$\sum_{i=x}^n c^i = o\left(\frac{1}{n}\right)$$

Answer 1

Edited

1. Since $\log(c)<0$ we have $x=1/o(1/\log(n))$

2. For fixed const $c$ assuming $x$ is integer this sum is $\frac{c^x-c^{n}}{1-c}$. Since $c^n=o(1/n)$ the condition for $x$ remains the same.