# Prove that $1+a+a^2+…+a^n$ = $\frac{1-a^{n+1}}{1-a}$ for all $a\ne 1$. [duplicate]

I've started by saying $$a_n$$=$$1+a+a^2+...+a^n$$=$$\frac{1-a^{n+1}}{1-a}$$

Now I think I need to do: $$lim_{n\to \infty}\frac{1-a^{n+1}}{1-a}$$

Should let $$a=2$$ and then solve the limit, then write a proof for it using $$\epsilon$$?

I've also considered that $$1+a+a^2+...+a^n$$=$$1+a^n$$. But I'm not sure what to do with this information.

I'd like to understand how to start this problem properly.

Hint: You can multiply out $$(1+a+a^2+…+a^n)(1-a)=…$$

• OH! That's way easier than I was trying to make it. So: – Elizabeth Austin Griffith Oct 15 '18 at 18:56
• That is nice, i hope you can solve your problem now! – Dr. Sonnhard Graubner Oct 15 '18 at 18:58
• And a hint again: $$\lim_{n\to \infty}a^n=0$$ if $$|a|<1$$ – Dr. Sonnhard Graubner Oct 15 '18 at 18:59

Or, a standard proof by induction.

$$1+a+a^2+...+a^n =\dfrac{1-a^{n+1}}{1-a}$$ is true for $$n=0$$ and $$n=1$$.

If true for $$n$$, then

$$\begin{array}\\ 1+a+a^2+...+a^n+a^{n+1} &=(1+a+a^2+...+a^n)+a^{n+1}\\ &=\dfrac{1-a^{n+1}}{1-a}+a^{n+1}\\ &=\dfrac{1-a^{n+1}+a^{n+1}(1-a)}{1-a}\\ &=\dfrac{1-a^{n+1}+a^{n+1}-a^{n+2}}{1-a}\\ &=\dfrac{1-a^{n+2}}{1-a}\\ \end{array}$$

which completes the proof.