I was trying to get an asymptotic formula for the sum
$$\sum_{n=2}^{x}\frac{1}{\log n}$$
this looks a bit tricky, but I still tried it. We have
\begin{align}
\sum_{n=2}^{x}\frac{1}{\log n}&=-\sum_{n=2}^{x}\frac{1}{\log(nt)}\Bigg{|}_{t=1}^{t=\infty}\\
&=\sum_{n=2}^{x}\int_{1}^{\infty}\frac1{t\log^2(nt)}dt
\end{align}
Now I don't know what to do further. Any help would be appreciated. Any asymptotic formula or closed form would help.
Note: I will probably update this question to post more of my work.
Update: I proved that this sum is $O(x/\log x)$, and this was also mentioned in the comments. But I want to improve this. How can one improve the error term?
Best Answer
Note that $$ \sum\limits_{n = 2}^x {\frac{1}{{\log n}}} \le \frac{1}{{\log 2}} + \sum\limits_{n = 3}^x {\int_{n - 1}^n {\frac{{dt}}{{\log t}}} } = \frac{1}{{\log 2}} + \int_2^x {\frac{{dt}}{{\log t}}} $$ and $$ \sum\limits_{n = 2}^x {\frac{1}{{\log n}}} \ge \sum\limits_{n = 2}^x {\int_n^{n + 1} {\frac{{dt}}{{\log t}}} } \ge \sum\limits_{n = 2}^{x - 1} {\int_n^{n + 1} {\frac{{dt}}{{\log t}}} } = \int_2^x {\frac{{dt}}{{\log t}}} . $$ Thus, by the known asymptotic expansion of the logarithmic integral, $$ \sum\limits_{n = 2}^x {\frac{1}{{\log n}}} = \int_2^x {\frac{{dt}}{{\log t}}} + \mathcal{O}(1) \sim \frac{x}{{\log x}}\left( {1 + \frac{{1!}}{{\log x}} + \frac{{2!}}{{\log ^2 x}} + \ldots } \right) $$ as $x\to +\infty$.