
$\frac 12 + \frac 1{2^2} + .... + \frac 1{2^{n-1}} + \frac 1{2^n} = k$
$\frac 12 + \frac 1{2^2} + .... + \frac 1{2^{n-1}} + \frac 1{2^n} + \frac 1{2^n}= k + \frac 1{2^n}$
$\frac 12 + \frac 1{2^2} + .... + \frac 1{2^{n-1}} + \frac 2{2^n}= k + \frac 1{2^n}$
$\frac 12 + \frac 1{2^2} + .... + \frac 1{2^{n-1}} + \frac 1{2^{n-1}}= k + \frac 1{2^n}$
$\frac 12 + \frac 1{2^2} + .... + \frac 2{2^{n-1}} = k + \frac 1{2^n}$
$\frac 12 + \frac 1{2^2} + .... + \frac 1{2^{n-2}} + \frac 1{2^{n-2}}= k + \frac 1{2^n}$
$\frac 12 + \frac 1{2^2} + .... + \frac 2{2^{n-2}} = k + \frac 1{2^n}$
$.....$
$\frac 12 + \frac 1{2^2} + \frac 1{2^3} + \frac 1{2^3} = k + \frac 1{2^n}$
$\frac 12 + \frac 1{2^2} + \frac 2{2^3} = k + \frac 1{2^n}$
$\frac 12 + \frac 1{2^2} + \frac 1{2^2} = k + \frac 1{2^n}$
$\frac 12 + \frac 2{2^2} = k + \frac 1{2^n}$
$\frac 12 + \frac 12 = k + \frac 1{2^n}$
$1 = k + \frac 1{2^n}$
$k = 1 - \frac 1{2^n}$.
.... or ....
$\frac 12 + \frac 1{2^2} + .... + \frac 1{2^{n-1}} + \frac 1{2^n} = k$
$2^n(\frac 12 + \frac 1{2^2} + .... + \frac 1{2^{n-1}} + \frac 1{2^n}) = 2^n*k$
$2^{n-1} + 2^{n-2} + ...... + 2^2 + 2 + 1 = 2^n*k$
$2^{n-1} + 2^{n-2} + ...... + 2^2 + 2 + 1 + 1 =2^n*k+1$
$2^{n-1} + 2^{n-2} + ...... + 2^2 + 2 + 2 =2^n*k+1$
$2^{n-1} + 2^{n-2} + ...... + 2^2 + 4 =2^n*k+1$
$2^{n-1} + 2^{n-2} + ...... + 8 =2^n*k+1$
$....$
$2^{n-1} + 2^{n-2} + 2^{n-2} = 2^n*k + 1$
$2^{n-1} + 2^{n-1} = 2^n*k + 1$
$2^n = 2^n*k + 1 > 2^n*k$
$1 = \frac {2^n}{2^n} = \frac {2^n}{2^n}*k + \frac 1{2^n} = k + \frac 1{2^n} > k$.
..... or ......
$k = \sum_{i=1}^n \frac 1{2^i}$
$2k = 2\sum_{i=1}^n \frac 1{2^i} = \sum_{i=1}^n \frac 2{2^{i}}=\sum_{i=1}^{n}\frac 1{2^{i-1}} = \sum_{i=0}^{n-1}\frac 1{2^i}$.
$2k = \sum_{i=0}^{n-1} \frac 1{2^i} = \frac 1{2^0} + \sum_{i=1}^n\frac 1{2^i} - \frac 1{2^n} = 1 + k - \frac 1{2^n}$
$k = 1-\frac 1{2^n} < 1$.