How to solve $T(n)= T(n - 1) + \frac{1}{n\log n}$?
We have
$$
T(n) = \sum_{m=3}^n \frac{1}{m\log m} + T(2).
$$
We can bound
$$
\int_3^{n+1} \frac{dx}{x\log x} \leq \sum_{m=3}^n \frac{1}{m\log m} \leq \int_2^n \frac{dx}{x\log x}.
$$
Since $(\log\log x)' = \frac{1}{x\log x}$, this shows that
$$
\log\log(n+1) - \log\log 3 \leq \sum_{m=3}^n \frac{1}{m\log m} \leq \log\log n - \log\log 2,
$$
and so $T(n) = \log\log n + \Theta(1)$.