Unlike a decreasing geometric series, the sum of the harmonic series $1,1 / 2,1 / 3,1 / 4,1 / 5, \ldots$ diverges; that is,
$$\sum_{i=1}^{\infty} \frac{1}{i}=\infty$$
It turns out that, for large $n$, the sum of the first $n$ terms of this series can be well approximated as
$$\sum_{i=1}^{n} \frac{1}{i} \approx \ln n+\gamma$$
where ln is natural logarithm (log base $e=2.718 \ldots$ ) and $\gamma$ is a particular constant $0.57721 \ldots$ Show that
$$\sum_{i=1}^{n} \frac{1}{i}=\Theta(\log n)$$
(Hint: To show an upper bound, decrease each denominator to the next power of two. For a lower bound, increase each denominator to the next power of $2 . .$ )