# Prove that the summation from n=0 to infinity of a^n is 1/(1-a)

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!

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

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

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,

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

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