If you want to stop at $$k = \infty$$ you have three problems. $$\infty$$ is not a counting number (not any type of number in standard practice), the expression you are quoting doesn't have an upper bound on where to stop, so you want to talk about $$S_{\infty}$$ if $$\infty$$ was a counting number, and since if $$\infty$$ was a counting number then $$S_{\infty} = 1 - 10^{-\infty}$$ which it stands to reason that: $$1 - \frac{1}{\infty} \lt 1 - \frac{1}{\infty^3}\lt S_{\infty} \lt S_{\infty+1} \lt S$$ (because this is true for all counting numbers) but the precise value of $$S_{\infty}$$ and how it relates to $$S_{\infty-1}$$, $$S_{\infty+1}$$ and $$S$$ requires you to do the heavy lifting. Specifically, I do not know that the hyperreals really have a good handle on exponentials of infinity quantities because non-standard analysis is a topic for math graduate students for the most part, but I thought this was one of the problems that they had.
In the hyperreals, $$\infty$$ is not the largest possible number -- there is no such number. In the hyperreals, $$\infty-1 \lt \infty \lt \infty + 1 \lt \infty^2 \lt \infty^3 \lt e^{\infty} \lt 10^{\infty}$$ because $$e^{\infty} = 1 + \infty + \frac{\infty^2}{2!} + \frac{\infty^3}{3!} + \dots $$ but I don't know how they get detailed answers from that.
In the extended real numbers $$\infty$$ doesn't have any properties of a counting number, such that $$\infty-1 \not\lt \infty \not\lt \infty + 1$$ so specifically $$S_{\infty} = S = 1$$.
Because the concepts of infinity are different, if you want answers you have to do the heavy lifting.
