How do I prove that $ \sum_{k=0}^{\infty} (1-x)^k=1/x, 0This is a step in a derivation I'm looking into for my high school Econ class. Is it even true?
 A: $$0<x<1 \implies 0<1-x<1$$
we know that for $q\in (-1,1):$
$$\sum_{n=0}^{+\infty}q^n=\frac{1}{1-q}$$
which gives if $q=1-x$,
$$\frac{1}{1-1+x}=\frac{1}{x}.$$
A: First, note that
$$
(1-y)\sum_{k=0}^{K}y^{k}=\sum_{k=0}^{K}y^{k}-\sum_{k=0}^{K}y^{k+1}=\sum_{k=0}^{K}y^{k}-\sum_{k=1}^{K+1}y^{k}=1-y^{K+1}.
$$
If $y\in[0,1)$, $y^{K+1}\rightarrow 0$ as $K\rightarrow \infty$. Therefore, we can take limits in the above to get
$$
\lim_{K\rightarrow\infty}(1-y)\sum_{k=0}^{K}y^{k}=1\text{ whenever }y\in[0,1).
$$
Dividing both sides by $1-y$, we get the familiar identity
$$
\lim_{K\rightarrow\infty}\sum_{k=0}^{K}y^{k}=\frac{1}{1-y}\text{ whenever }y\in[0,1).
$$
Now, making the substitution $x=1-y$, we get
$$
\lim_{K\rightarrow\infty}\sum_{k=0}^{K}(1-x)^{k}=\frac{1}{x}\text{ whenever }x\in(0,1],
$$
as desired.
A: Recall that for $$\sum_{n=0}^{\infty}y^n = \frac{1}{1-y}$$ for $|y|<1$, i.e. also for $0<y<1$.
Now substitute $y$ with $1-x$ and you'll get the same.
Recall that you found the above formula by looking at $$1+y+y^2+..+y^n = S$$ for some total sum $S$. Then you multiply by $(1-y)$ both sides yielding $$(1-y^{n+1}) = S(1-y)$$ and finally $$S=\frac{1-y^{n+1}}{1-y}$$. Now the sequence tends to a limit if this sum tends to a limit as $n\to\infty$, and for $|y|<1$ you get that the limit is exactly the above-stated $\frac{1}{1-y}$
A: It is true. Assume that it converges, and call $S$ that sum. Then
$$
(1-x)S=S-1 \implies S=\frac{1}{x}.
$$
