Repeat Fractions

It's intuitive that by adding $\dfrac{1}{256}$ , the sum will telescope and give an answer of $1$ . Therefore, the sum is equal to $1 - \dfrac{1}{256}$ , which evaluates to $\boxed{\dfrac{255}{256}}$ .

The general formula for a sum of a geometric series goes as follows: $\sum_{n=0}^{N} x^n \equiv \frac{x^{N+1}-1}{x-1}$ Here, our series starts at $\dfrac{1}{2}$ so we must subtract 1 from our formula. by plugging in $x=\dfrac{1}{2}$ and $N = \log_{2}(256) = 8$ , se can see that $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots + \frac{1}{256} = \frac{\left(\frac{1}{2}\right)^{8+1}-1}{\frac{1}{2}-1} - 1 = \frac{-\frac{511}{512}}{-\frac{1}{2}}-1 = \frac{255}{256} + 1 - 1 = \boxed{\frac{255}{256}}$

$\frac{1}{2}$ + $\frac{1}{4}$ + $\frac{1}{8}$ + $\cdots$ + $\frac{1}{256}$ = $\frac{128}{256}$ + $\frac{64}{256}$ + $\frac{32}{256}$ + $\cdots$ + $\frac{1}{256}$ = $\boxed{\frac{255}{256}}$

$\begin{aligned} S & = \frac 12 + \frac 14 + \frac 18 + \cdots + \frac 1{256} \\ & = \frac 34 + \frac 18 + \frac 1{16} + \cdots + \frac 1{256} \\ & = \frac 78 + \frac 1{16} + \frac 1{32} + \cdots + \frac 1{256} \\ & = \cdots \quad \cdots \quad \cdots \\ & = \frac {127}{128} + \frac 1{256} \\ & = \boxed{\frac {255}{256}} \end{aligned}$