Limit of smallest real solution to equation

Question

I think if you solve $$x \sum_{i=0}^{n} (1-x)^i = 0.085$$ for x, the equation always has one real solution such that $0 \leq x \leq 1$, and as $n \rightarrow \infty$, this solution converges towards 0.085 (EDIT: sorry, this was wrong in the original question ... I mean converges to $\frac{0.085}{n}$). (Presumably 0.085 could be replaced by any number between 0 and 1 ...)

Could anyone point me into some direction if I wanted to prove that there is always one such solution (real between 0 and 1) and that it converges towards $\frac{0.085}{n}$? I have basically no education in the subject.

(This is also not a homework question ... it came up as part of my work)

Answer 1

You can simplify the sum $$\sum_{i=0}^n (1-x)^i = \frac{1-(1-x)^{n+1}}{x}$$

So your equation is equivalent to $$1-(1-x)^{n+1} = 0.085$$ i.e. $$(1-x)^{n+1} = 0.915$$

i.e., if $0 \leq x \leq 1$, $$1-x = \sqrt[n+1]{0.915}$$

i.e. $$x = 1-\sqrt[n+1]{0.915}$$

This tends to $1$ as $n \rightarrow + \infty$.

Answer 2

If we write $t=1-x$ then we have $$(1-t)\sum_{i=0}^{n}t^i = 0.085$$

so $$\sum_{i=0}^{n}t^i-\sum_{i=0}^{n}t^{i+1} = 0.085$$

so $${t^{n+1}-1\over t-1}-{t^{n+2}-1\over t-1} = 0.085$$

so $$ t^{n+1}-t^{n+2} = 0.085(t-1)$$

so $$1-t^{n+1} =0.085$$ so $$1-x = \sqrt[n]{0.915}\implies x = 1-\sqrt[n]{0.915}\to 1$$