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Prove $$\sum_{n=0}^{\infty} a^{n} = \cfrac{1}{1-a}$$ for all a $\in$ $\textbf{R}$ where |a| $<$1 and describe what happens when |a|$\nless$ 1

  • This is a calc two topic
  • I have a start I just need help finishing it
    I started with the partial sums of this series. $s_n=1+a+a^2+a^{3}+...+a^n$
    I also looked at the partial sums of $as_n=a+a^2+a^3+...+a^n+a^{n+1}$
    Then I subtracted $s_n-as_n$ and this is where I got stuck

I eventually want to get to a point where I take the limit to prove this equality is true but I do not know how to proceed

Any help would be appreciated !! Thank you!

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$$ \sum_{n=0}^\infty a^n = \frac{1}{1-a} $$ Let's see what the following leads to $$ (1-a)\sum_{n=0}^\infty a^n = \sum_{n=0}^\infty a^n - \sum_{n=0}^\infty a^{n+1}\\ \sum_{n=0}^\infty a^n - \sum_{k=1}^\infty a^k = \sum_{n=0}^\infty a^n - (\sum_{n=0}^\infty a^n - 1) \\ = \sum_{n=0}^\infty a^n - \sum_{n=0}^\infty a^n - (-1) = 1 $$

so we have $$ (1-a)\sum_{n=0}^\infty a^n = 1 \implies \sum_{n=0}^\infty a^n = \frac{1}{1-a} $$

This approach does not really work when you have the divergent sum when $|a| > 1$ since we have a divergent sum multiplied by a constant does not yield another constant.

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  • $\begingroup$ Everything is correct .. also continue please by answering what happens when $\vert a \vert \geq 1$ as requested by OP. $\endgroup$ – Ahmad Bazzi Sep 26 '18 at 1:31
  • $\begingroup$ thank you so much! where did the k=1 a^k come from $\endgroup$ – Parley Sep 26 '18 at 1:32
  • $\begingroup$ @AhmadBazzi well that case is not really covered in the above expression since $a^n \to 0$ is broken. $\endgroup$ – Chinny84 Sep 26 '18 at 1:34
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    $\begingroup$ I know but please clarify it to OP $\endgroup$ – Ahmad Bazzi Sep 26 '18 at 1:34
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    $\begingroup$ Oh so it was reindexed? $\endgroup$ – Parley Sep 26 '18 at 1:40
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Notice that

$$ \sum_{k=0}^n a^k = \frac{1-a^{n+1}}{1-a} $$

If $|a|<1$, then the sequence $a^{n+1} \to 0$. Therefore, the sequence of partial sums of $\sum a^n $ converges to $\frac{1}{1-a} $. In other words,

$$ \boxed{ \sum_{n=0}^{\infty} a^n = \frac{1}{1-a} } $$

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  • $\begingroup$ is this a proof by induction? $\endgroup$ – Parley Sep 26 '18 at 1:30
  • $\begingroup$ @Parley the first equation is using the expression of geometric series, which could be proved by induction indeed. $\endgroup$ – Ahmad Bazzi Sep 26 '18 at 1:40

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