What is the big-$O$ notation of a summation of a log?
Recall that the sum of $\log$'s is equivalent to the $\log$ of products.

That is:

$$\log(xy) = \log x + \log y$$

Thus we can change your function:

$$
\begin{align}
\sum_{i=n}^{n+m} \log i &amp;= \log \prod_{i = n}^{n + m} i\\
&amp;= \log (n \cdot (n+1) \cdot (n+2) \cdot \ldots \cdot (m-1) \cdot m )\\
&amp;= \log (m!\ /\ (n-1)! )\\
\end{align}
$$

Then we can similarly use the division rule to get these back out:

$$
\log (m!\ /\ (n-1)! ) = \log (m!) - \log((n-1)!)
$$

Then using Stirling's Approximation we can immediately see:

$$
\log (m!) - \log((n-1)!) = O(m \log m)
$$

You might be able to do better though by taking more precise bounds to get:

$$
\begin{align}
\log (m!) - \log((n-1)!) &amp;\leq em^{m + \frac{1}{2}}e^{-m} - \sqrt{2\pi}(n-1)^{n- \frac{1}{2}}e^{1-n}\\
&amp;= \vdots
\end{align}
$$