We must solve the recurrence $T(n)=T(n-3)+ \frac{1}{\log {n}}$ using substitution. I've got up to here $$T(n)=T(n-3-3)+\frac1{\log(n-3)} + \frac1{\log n} \implies T(n-3-3-3) + \frac1{\log(n-3-3)} + \frac1{\log(n-3)} + \frac1{\log n}$$
And after $k$ steps
$$\begin{split} T(n)&=T(n-3k)+\frac1{\log(n-3(k-1))}+\cdots+\frac1{\log(n-3(0))}\\ &=T(n-3k)+\sum_{i=0}^{k-1}\frac{1}{\log{(n-3i)}}\\ \end{split}$$
But I don't know how to approximate or solve the $\sum_{i=0}^{k-1}\frac{1}{\log{(n-3i)}}$, what can I do next?