Logarithm to any base is equal to a constant times logarithm to the base $2$. So let's work with base $2$. For convenience, write $\log n$ for $\log_2 n$.
We use a mild variant of the usual proof that $\sum \frac{1}{n}$ diverges.
Note that $\log 2$ and $\log 3$ are both $\ge 1$; $\log 4$, $\log 5$, $\log 6$, and $\log 7$ are all $\ge 2$; $\log 8$, $\log 9$, and so on up to $\log 15$ are all $\ge 3$, and so on. So we have: $\frac{1}{\log 2}+\frac{1}{\log 3} \le \frac{1}{1}+\frac{1}{1}=2.$ $\frac{1}{\log 4}+\frac{1}{\log 5}+\frac{1}{\log 6}+\frac{1}{\log 7}\le \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2.$
In the same way we can see that for any $m \ge 1$, $\frac{1}{\log (2^{m})}+\frac{1}{\log (2^{m}+1)}+\cdots+\frac{1}{\log (2^{m+1}-1)}\le 2.$ We conclude that $\sum_{k=2}^{2^{m+1}-1} \frac{1}{\log k} <2m,$ since up to $2^{m+1}-1$ we have $m$ chunks each of which has sum $<2$.
Now suppose that $2^{m} \le n \le 2^{m+1}-1$. Then $\sum_{k=2}^n \frac{1}{\log k}\le \sum_{k=2}^{2^{m+1}-1} \frac{1}{\log k} <2m.$ Thus $\frac{1}{n}\sum_{k=2}^n \frac{1}{\log k} <\frac{2m}{2^{m}}.$ Finally, it is well-known that $\displaystyle\lim_{m\to\infty} \frac{m}{2^m}=0$.