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

Question

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!

Answer 1

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,

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

Answer 2

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