Showing that $\sum_{k=1}^n\dfrac{1}{2^{k-1}}$ for $n \geq 2$ is not an integer . Suppose $n\geq2 ,s(n)=\sum_{k=1}^n\dfrac{1}{2^{k-1}} $ 
$$s(2)=1+\frac 12=1.5\\s(3)=1+\frac12+\frac14=1.75 ,\\\vdots$$
Is an elementary proof to $s(n)$ can never be an integer number ?
As honestly as possible : One of my students( k-12) asked this question .I said I will think and answer .But I got stuck ...
I am thankful for your hint,guide or solutions in advanced.
 A: Suppose it were an integer, say $K$.
Then
$$\sum_{k=1}^n \frac{1}{2^k} = K$$
Multiply by $2^n$ on both sides.
Then LHS is $1+2+2^2+...$ which is odd, but RHS is even. Contradiction.
A: $2^{n-1} s(n)$ is odd and $\frac{\text{odd}}{\text{even}\neq 0}$ cannot be an integer.
A: You can compute as sum of geometric progression $\sum_{i={1}}^n \frac{1}{2^{i-1}} = \frac{1 - (1/2)^{n}}{1 - 1/2} = 2 - (1/2)^{n-1} $
A: We can convert all of the fractions in the expansion to the same base, and then add them up. We then get the result $\frac{2^k}{2^k-1}$. Since the denominator is greater than the numerator, it does not divide the numerator, showing that it is not an integer.
A: Since $\sum_{k=1}^1 \frac{1}{2^{k-1}}$ is $1$, showing that $\sum_{k=1}^n\frac{1}{2^{k-1}}$ is not an integer is equivalent to show that $\sum_{k=2}^n \frac{1}{2^{k-1}}$ is not an integer.
$$\begin{align}1&=\sum_{k=2}^\infty \frac{1}{2^{k-1}}\\ 
&=\sum_{k=2}^n \frac{1}{2^{k-1}} + \sum_{k=n+1}^\infty \frac{1}{2^{k-1}}\\
&=S_1+S_2\\
\end{align}$$
Since $1$, the minimum positive integer, cannot be obtained as the sum of two positive integers and both $S_1$ and $S_2$ are positive, neither $S_1$ nor $S_2$ are integers.
