Possible Duplicate:
Value of $\sum\limits_n x^n$
In my lecture notes:
$$a_n = \frac{1}{2^n}, \qquad \sum_{n-1}^{\infty} a_n = \lim_{n\to\infty} (1-\frac{1}{2^n}) = 1$$
How do I get $(1-\frac{1}{2^n})$?
A similar example given is
$$a_n = 2^{n-1}, \qquad \sum_{n-1}^{\infty} a_n = \lim_{n\to\infty} (2^n - 1)$$
How do I get these?
Those are from the standard formula for the sum of a finite geometric series:
$$\sum_{k=0}^nar^k=\frac{a-ar^{n+1}}{1-r}=\frac{ar^{n+1}-a}{r-1}\;.\tag{1}$$
Write $$S=\sum_{k=0}^nar^k\;;$$ then
$$\begin{align*}S&=a\color{red}{+ar+ar^2+\dots+ar^{n-1}+ar^n}\\ rS&=\quad\;\;\,\color{red}{ar+ar^2+\dots+ar^{n-1}+ar^n}+ar^{n+1}\;, \end{align*}$$
and when you subtract the second equation from the first, the red terms cancel out to leave $(1-r)S=a+ar^{n+1}$, from which $(1)$ follows immediately.
$$a_n = \frac{1}{2^n}, \qquad \sum_{n=1}^{\infty} a_n = \lim_{n\to\infty} (1-\frac{1}{2^n}) = 1$$
Depends how you interpret it. First let us look at the solution to
First we will be using (for $r<1$)
$$ \sum_{k=0}^{n} ar^{k} = \frac{a- ar^{n+1}}{1-r} $$ $$ \sum_{k=0}^\infty ar^k = \frac{a}{1-r} $$
Now in the problem above
$$a_n = \frac{1}{2^n}$$ and therefore
$$ \begin{align*} \sum_{n=1}^{\infty} a_n &= \sum_{n=1}^{\infty}\frac{1}{2^n}\\ &= {\frac{1}{2}} ( 1 + \frac{1}{2^{1}} + \frac{1}{2^{2}} + \dots )\\ &= {\frac{1}{2}} (\frac{1}{1-\frac{1}{2}}) &= 1 \end{align*} $$
And
$$\lim_{n\to\infty} (1-\frac{1}{2^n}) = 1 - \lim_{n\to\infty} \frac{1}{2^n} = 1$$
It is just a way to rewrite the series which is also explained on the relevant wikipedia article.
Basically all you have to know is that
$$(1+r+r^2+r^3+\ldots+r^n)(1-r)=1-r^{n+1}$$
