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?

  • $\begingroup$ Probably big-oh of $x/\log x$. $\endgroup$ Dec 19, 2020 at 8:48
  • $\begingroup$ @GerryMyerson how can you say "probably"? $\endgroup$ Dec 19, 2020 at 8:53
  • $\begingroup$ @GerryMyerson oh proved this. But, this is not much good. I want an even better bound. $\endgroup$ Dec 19, 2020 at 9:04
  • $\begingroup$ en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula $\endgroup$ Dec 19, 2020 at 9:07
  • $\begingroup$ You've written $x\log x$, but you mean $x/\log x$, right? $\endgroup$ Dec 19, 2020 at 10:04

2 Answers 2


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 $x\to +\infty$.

  • $\begingroup$ And $\lim_{x\to \infty} Li(x)-\sum_{n=1}^x 1/\log n= \sum_n O(\frac{1}{n\log^2 n})$ converges $\endgroup$
    – reuns
    Dec 19, 2020 at 9:23

By Riemann-Stieltjes integration we have

$$ \begin{aligned} \sum_{2\le n\le x}{1\over\log n} &=\sum_{2<n\le x}{1\over\log n}+{1\over\log2} \\ &=\int_2^x{\mathrm d(t-\{t\})\over\log t}+{1\over\log2} \\ &=\int_2^x{\mathrm dt\over\log t}+{1\over\log2}-\int_2^x{\mathrm d\{t\}\over\log t} \\ &=\operatorname{Li}(x)+{1\over\log2}-{\{x\}\over\log x}-\int_2^x{\{t\}\over t\log^2t}\mathrm dt \\ &=\operatorname{Li}(x)+A+\mathcal O\left(1\over\log x\right) \end{aligned} $$

in which

$$ A={1\over\log2}-\int_2^\infty{\{t\}\over t\log^2t}\mathrm dt $$

and $\operatorname{Li}(x)=\operatorname{li}(x)-\operatorname{li}(2)$ is the logarithmic integral function satisfying

$$ \operatorname{Li}(x)\sim{x\over\log x} $$


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