Asymptotics for series $\sum_{n \leq x} n/\log n$

Question

What can be said about an equivalent of $$\sum_{n \leqslant x} \frac{n}{\log n}$$

I would like to compare it to $x^2$ which is approximately $\sum_{n \leqslant x} n$. Is it negligible in front of it? I tried partial summation and dyadic cutting, but nothing seems to work...

More generally, given a function $f$, is it true that $\sum \frac{f(n)}{\log(n)^\varepsilon}$ is negligible compared to $\sum f(n)$?

Answer 1

Let $$s_n = \sum^n_{k=2} \frac{k}{\log k}.$$ Then, obviously $s_n\to\infty$ as $n\to\infty$, and $$\lim_{n\to\infty}\frac{s_{n+1}-s_n}{(n+1)^2-n^2}=\lim_{n\to\infty}\frac{n+1}{(2n+1)\log(n+1)}=0.$$ Due to the Stolz–Cesàro theorem, this implies $$\lim_{n\to\infty}\frac{s_n}{n^2}=0.$$
For your more general question, you can just replace $n^2$ by $\sum_{k\le n}f(k)$ and $\log(n)$ by $\log(n)^\varepsilon$.

Answer 2

Wolfram Alpha says

$$\int \frac x{\ln x}dx=\operatorname{Ei}(2\log x).$$

Then it is known that

$$\frac12e^{-x}\log\left(1+\frac2x\right)\le-\operatorname{Ei}(-x)\le e^{-x}\log\left(1+\frac1x\right).$$

This shows that your sum is asymptotic to

$$\frac{cx^2}{\log x}.$$

Answer 3

For an increasing function $f(x)$, we have that $$ \sum_{a \leq n \leq b} f(n) \leq \int_a^{b+1} f(t) dt.$$ So we are led to consider $$ \begin{align} \int_2^X \frac{t}{\log t}\,dt &= \frac{t^2}{\log t} \Big|_2^X - \int_2^X t \left( \frac{t}{\log t}\right)' dt \\ &=\frac{t^2}{\log t} \Big|_2^X - \int_2^X \frac{t(\log t - 1)}{\log^2 t} dt \\ =\frac{t^2}{\log t} \Big|_2^X - \int_2^X \frac{t}{\log t} dt + \int_2^X \frac{t}{\log^2 t} dt. \end{align}$$ Rearranging, we have that $$ \begin{align} 2 \int_2^X \frac{t}{\log t} dt &= \frac{t^2}{\log t} \Big|_2^X + \int_2^X \frac{t}{\log^2 t} dt \\ &= \frac{X^2}{\log X} - 4/\log 4 + \int_2^X \frac{t}{\log^2 t} dt\leq \frac{X^2}{\log X} + \frac{X^2}{\log^2 X}+ c \end{align}$$ for a small constant $c$. Thus $$ \int_2^X \frac{t}{\log t} dt = \frac{X^2}{2 \log X} + \frac{X^2}{2 \log^2 X}+ c.$$

A similar argument gives a similar lower bound, so that the leading order estimate of your sum is $$ \frac{X^2}{2 \log X},$$ as expected from the comments. Performing further integration by parts actually leads to further terms in the expansion, where each term is a logarithmic savings over the previous. This is analogous to one way of proving the asymptotics for the logarithmic integral $\int dt/\log(t)$.

It is clear that this sort of argument will hold whenever $f$ is a well-enough behaved (e.g. monotonic, integrable). For a slightly worse-behaving class of functions $f$, one can replace the naive sum-and-integral bound with the first few terms of an Euler-Maclaurin sum, as long as $f$ is a few-times differentiable.

Answer 4

By summation by parts,

$$\begin{eqnarray*} \sum_{n=2}^{N}\frac{n}{\log n}&=&\frac{N^2+N-2}{2\log N}+\sum_{n=2}^{N-1}\frac{(n^2+n-2)\log\left(1+\frac{1}{n}\right)}{2\log(n)\log(n+1)}\\&=&\frac{N^2}{2\log N}+O(N)+\sum_{n=2}^{N}\frac{n}{2\log^2 n}\\&=&\frac{N^2}{2\log N}\left(1+O\left(\frac{1}{\log N}\right)\right). \end{eqnarray*} $$