Solving a sum similar to geometric series

Question

How do I solve the sum $$\sum_{k=1}^y \left( 1-\frac{1}{\ln x} \right)^k \hspace{0.5cm} $$ for $x>0$ and $y$ a positive integer greater than one?

Despite resembling a geometric series, it does not appear that the sum can be solved as one. Furthermore, even though the summand is asymptotic to $1$ as $x \to \infty$, the sum itself does not behave as $(y^2+y)/2$ for such $x$. In fact, when graphing it, the sum is $o(\ln x)$ ($x \to \infty$).

Answer 1

I think you mean finding an explicit expression for it :

For it by geometric sum :

$$ \sum_{k=1}^y \left(1-\dfrac{1}{\ln(x)}\right)^k=\dfrac{1-(1-\dfrac{1}{\ln(x)})^{y+1}}{1-(1-\dfrac{1}{\ln(x)})} $$

And after I think you can end from now.