Note that $\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n}>\frac{1}{2n}+\ldots+\frac{1}{2n}=\frac{1}{2}$. Therefore, if we apply this to $n=1,2,4,8,16,32,64,128$, we get $\left(\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\ldots+\left(\frac{1}{129}+\ldots+\frac{1}{256}\right)>\frac{1}{2}+\ldots+\frac{1}{2}=4$ so, adding in $\frac{1}{1}$, we get $\sum_{n=1}^{256} \frac{1}{n}>5$ so $k=256$ will suffice.