5. An aside: the sum of all the unit fractions
It is possible to show that, given any real $\mathrm{M}$, there exists a positive integer $k$ large enough that:
$$
\sum_{n=1}^{k} \frac{1}{n}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3} \ldots>M
$$

Note that this statement means that the infinite harmonic series, $\sum_{n=1}^{\infty} \frac{1}{n}$, grows without bound, or diverges. For the specific example $\mathrm{M}=5$, find a value of $k$, not necessarily the smallest, such that the inequality holds. Justify your answer.
Solution: Note that $\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2 n}>\frac{1}{2 n}+\ldots+\frac{1}{2 n}=\frac{1}{2}$. 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.