# Upper bound of the sum $\sum_{i=2}^{N}{\frac{1}{\log(i)}}$

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?

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

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