On the interval 0 to 1 vertical lines are drawn at the points

x = \frac{k}{2^{n}}

where

k

is an odd positive integer. The height of the lines at 0 and 1 are 1 unit, the height of the line at

x = \frac{1}{2}

\frac{1}{2}

unit, the height of the line at

x = \frac{1}{4}

and

x = \frac{3}{4}

\frac{1}{4}

unit and the height of the line at

\frac{k}{2^{n}}

\frac{1}{2^{n}}

for all values of

k

Now consider the line $y = h$ where $\frac{1}{2^{n}} > h > \frac{1}{2^{n + 1}}$ . How many of the vertical lines from $x = 0$ to $x = 1$ does it cut?

What happens to the heights of these lines as $n$ gets larger? What happens to the number of lines cut as $n$ gets larger?