# How to prove $1 + a + \cdots + a^n + \cdots = \frac{1}{1-a}$? [duplicate]

Possible Duplicate:
Value of $\sum\limits_n x^n$

Let $\lvert a \rvert < 1$.

How would I show that the infinite sum $$1 + a + \cdots + a^n + \cdots = \frac{1}{1-a}$$

Update: Thanks everyone. Having different solutions from different people really helps to see the problem from a different perspective.

-

## marked as duplicate by user7530, N. S., leonbloy, Fabian, The Chaz 2.0Feb 1 '13 at 0:27

Also this is true for any complex number $a$, provided $a \not= 1$. –  ncmathsadist Jan 31 '13 at 23:58
@AndréNicolas fixed my question. –  Blue Pony Inc. Jan 31 '13 at 23:59
Consider the finite geometric series André Nicolas mentioned and then take the limit as $n \to \infty$. Since $\vert a \vert < 1$ it becomes arbitrarily small, negligible. –  Robert Cardona Feb 1 '13 at 0:01

As already pointed out, in general you have $$1+ a + \cdots + a^n = \frac{1-a^{n+1}}{1-a}\ .$$ Taking the limit to infinity and using that if $|a|<1$, then $\lim_{n\rightarrow\infty}a^n=0$ you get the result. $$\begin{eqnarray} \sum_{k=0}^\infty a^k &=& \lim_{n\rightarrow \infty}\sum_{k=0}^n a^k\\ &=& \lim_{n\rightarrow \infty} \frac{1-a^{n+1}}{1-a} \\ &=& \frac{1}{1-a} \end{eqnarray}$$

-
$\vert a \vert < 1$ correct? –  Robert Cardona Feb 1 '13 at 0:02
I think you mean $k=0$, not $i=k$ in the bounds of the sum. –  John Moeller Feb 1 '13 at 0:02

$$(1 + a + ... + a^n)(1-a)=1 + a + ... + a^n-(a + ... + a^{n+1}) =1-a^{n+1}$$ since $a\neq1$ divide by $1-a$ to get $$(1 + a + ... + a^n)=\frac{a^n-1}{a-1}$$ because $$\lim_{n\rightarrow \infty} \frac{a^n-1}{a-1}=\frac{1}{1-a},|a|<1$$ so $$1 + a + ... + a^n+...=\frac{1}{1-a}$$

-

i think it's just using MacLaurin's series

suppose $f(a)=\frac{1}{1-a}$ then we get $f(0)=1$

$f'(a)=\frac{1}{(1-a)^2}$ then we get $f'(0)=1$

$f"(a)=\frac{2}{(1-a)^3}$ then we get $f"(0)=2$

$f'''(a)=\frac{2.3}{(1-a)^2}$ then we get $f'(0)=3!$

do this till $n$-th factor then we get the MacLaurin series of $f(a)$

$$f(a)=f(0)+\frac{f'(0)}{1!}a+\frac{f"(0)}{2!}a^2+...+\frac{f^{(n)}}{n!}a^n$$

$$f(a)=1+a+a^2+...+a^n$$

Finally we get $\frac{1}{1-a}=1+a+a^2+...+a^n$

-