<s>[INST] Proving for all n that $\sum_{i=0}^n \frac1{2^{i}} < 2$ Proving for all n $\in \mathbb N$, $$\sum_{i=0}^n \frac1{2^{i}} < 2$$
Hint. First prove that the left hand side can be expressed in closed form, i.e. without
using the summation operator.
This is one of my review questions and I do not know how to start even with the hint. Any help would be appreciated.
 [/INST] Notice for $|x| < 1 $
$$ 1 + x + x^2 + x^3 + .... + x^r = \frac{1 - x^{r+1}}{1-x} $$
Hence
$$ \sum_{i=1}^n \frac{1}{2^i} = 1 + \left( \frac{1}{2}\right)^1 + \left( \frac{1}{2}\right)^2 + .... + \left( \frac{1}{2}\right)^n = \frac{1 - \left( \frac{1}{2}\right)^{n+1}}{1 - \frac{1}{2}} = 2 - \frac{1}{2^n} $$
And since $\frac{1}{2^n} > 0$  for all $n>0$, then
$$ 2 + \frac{1}{2^n} > 2 \iff2 > 1 -\frac{1}{2^n} = \sum_{i=1}^n \frac{1}{2^i}$$
</s>