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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.

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

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

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