Big theta notation of harmonic series I want to prove that big theta notation of the harmonic series is $\Theta(\log n)$. I want to work with integral to show that.

I attempted this:

$$\ln(n)=\int^n_1  \frac{dx}x \le \sum _{k=1} ^n  \frac1k \le 1 + \int^n_2 \frac{dx}x = 1 + \ln(n)$$
This approach was not demanded, because I have not proven that $\Theta(\log n)$ is a tight bound for the harmonic series.

How can I show this, and how to overcome this obstacle? Please help.

Thanks.
If we extend the definition of $=$ and $\le$ to $\Theta$s in the obvious way, you have shown that

$$\begin{align}
\Theta \ln (n) &amp;= \Theta \int^n_1  \frac{dx}x \\
&amp;\le \Theta \sum _{k=1} ^n  \frac1k \\
&amp;\le \Theta (1 + \int^n_2 \frac{dx}x) \\
&amp;= \Theta (1 + \ln(n)) \\
\text {which, since } \Theta (1) \lt \Theta (\ln (n))\text {, }\\
&amp;= \Theta \ln (n) \\
\end{align}$$
If we extend the definition of $=$ and $\le$ to $\Theta$s in the obvious way, you have shown that

$$\begin{align}
\Theta \ln (n) &amp;= \Theta \int^n_1  \frac{dx}x \\
&amp;\le \Theta \sum _{k=1} ^n  \frac1k \\
&amp;\le \Theta (1 + \int^n_2 \frac{dx}x) \\
&amp;= \Theta (1 + \ln(n)) \\
\text {which, since } \Theta (1) \lt \Theta (\ln (n))\text {, }\\
&amp;= \Theta \ln (n) \\
\end{align}$$