One of the questions in Sierpinski's book on number theory lead to finding a tight upper bound for the following sum:
$$\sum_{i=2}^N {\frac{1}{\log(i)}}$$

The trivial upper bound like $\frac{N-1}{\log(2)}$ wouldn't work. Can someone please suggest a stronger bound?


1 Answer 1


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 $$ \mbox{li} \; x = \gamma + \log \log x + \sum_{n=1}^\infty \; \frac{(\log x)^n}{n \; n!}. $$

  • $\begingroup$ Thanks! Asymptotic value of li(x) seems sufficient for me :) $\endgroup$
    – Five
    Commented Nov 17, 2012 at 7:15

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