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.
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. $$ \begin{eqnarray} \sum_{k=0}^\infty a^k &=& \lim_{n\rightarrow \infty}\sum_{k=0}^n a^k\\ &=& \lim_{n\rightarrow \infty} \frac{1-a^{n+1}}{1-a} \\ &=& \frac{1}{1-a} \end{eqnarray} $$
$$(1 + a + ... + a^n)(1-a)=1 + a + ... + a^n-(a + ... + a^{n+1}) =1-a^{n+1}$$ 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 $\frac{1}{1-a}=1+a+a^2+...+a^n$
