Given a decreasing positive function, in your case $f(x) = \frac{1}{\log x},$ it is standard to compare with the integral of the same thing. This is done by, in essence, starting at any integer $i$ and the point $(i,f(i))$ and drawing a horizontal segment of length exactly $1$ to the right. This segment is higher than the graph of the real-valued function. Eventually you get to the right endpoint, your $N,$ and that final segment extends over the graph out to $N+1.$ So $ \sum_{i=2}^N f(i) \; > \; \int_2^{N+1} \; f(x) dx. $ An antiderivative is $\mbox{li} \; x,$ see LOGARITHMIC INTEGRAL A pretty good lower bound is $ \mbox{li} \; (N+1) - \mbox{li} \; 2 $
There is a similar process for an upper bound. Draw segments to the left. In case, as here, $f(1)$ is undefined, just start the process one later and keep the explicit $f(2)$ term. $ \sum_{i=2}^N f(i) \; < \; f(2) + \; \int_2^{N} \; f(x) dx. $ A pretty good upper bound is $ \frac{1}{\log 2} + \mbox{li} \; N - \mbox{li} \; 2 $
There are any number of ways to discuss the size of $\mbox{li} \; x,$ see SUM. One exact sum I like is $ \mbox{li} \; x = \gamma + \log \log x + \sum_{n=1}^\infty \; \frac{(\log x)^n}{n \; n!}. $