Asymptotic analysis of $\sum_{k=2}^n\frac{k}{\ln k}$

Question

I know that by integrating, we can show $$\sum_{k=1}^n \ln(k)\sim n\ln n-n$$

Now what can we do for $$\sum_{k=2}^n\frac{k}{\ln k}$$ which doesn't admit an integral approximation in analytic form?

(Or perhaps there is a way to evaluate $\int \frac{x}{\ln x}dx$? I am not sure how to do it.)

Since, for large $x$, $\ln x\sim~x^\epsilon$ for arbitrarily small $\epsilon$ (I know this is not a mathematically rigorous expression, but it is quite handy in analyzing asymptotic behavior), the sum will, in some sense, be of the order $n^{2-\epsilon}$ for arbitrarily small $\epsilon$. Thus, I guess it could be approximated by something like $n^2/\ln n$ or maybe $n^2/\ln(\ln n)$ that could have this order $n^{2-\epsilon}$.

What is then the correct approximation?

Answer 1

I will treat the main term $I(n) = \int_{x = *}^n \frac x{\ln x}dx$ when $n$ tends to infinity, where $*$ is any fixed constant.

By integration by parts, we can write \begin{eqnarray} I(n) &=& \left . \frac{x^2}{2\ln x} \right |_{x = *}^n - \int_{x = *}^n \left ( \frac 1{\ln x} \right )'\frac{x^2}2 dx\\ &\approx& \frac{n^2}{2\ln n} + \frac 12\int_{x = *}^n\frac x{(\ln x)^2}dx \end{eqnarray} where $\approx$ means up to $O(1)$. It is clear that the integral $\int_{x = *}^n\frac x{(\ln x)^2}dx$ is of size $O \left (\frac n{(\ln n)^2} \right )$.

This procedure can continue: \begin{eqnarray} \int_{x = *}^n\frac x{(\ln x)^2}dx &=& \left . \frac{x^2}{2(\ln x)^2} \right |_{x = *}^n - \int_{x = *}^n \left ( \frac 1{(\ln x)^2} \right )'\frac{x^2}2 dx\\ &\approx& \frac{n^2}{2(\ln n)^2} + \int_{x = *}^n\frac x{(\ln x)^3}dx \end{eqnarray} and so on.

The result is an asymptotic expansion: $I(n) \sim n^2 \left (\frac 1{2\ln n} + \frac 1{4 (\ln n)^2} + \frac 1{4(\ln n)^3} + \frac 3{8(\ln n)^4} + \dots \right )$

Note that this is not a power series expansion, as it is not convergent for any $n$. It should be understood as many asymptotic formulas such as $$I(n) = n^2 \left ( \frac 1{2\ln n} + \frac 1{4 (\ln n)^2} + \frac 1{4(\ln n)^3} \right ) + O \left (\frac{n^2}{(\ln n)^4} \right )$$ by taking any finite number of starting terms.

Answer 2

If you consider $$I=\int_2^n \frac{x}{\log (x)}\,dx$$ let $x=e^y$ to make $$I=\int_{\log(2)}^{\log(n)}\frac {e^{2y}} y \,dy=\int_{\log(2)}^{\log(n)}\frac {e^{2y}} {2y} \,d(2y)=\text{Ei}(2 \log (n))-\text{Ei}(2 \log (2))$$ where appears the exponential integral function.

Now, for large values of $n$, as @Gary commented (have a look here) a series expansion is $$\text{Ei}(x)\sim\frac {e^x} x \sum_{k=0}^\infty \frac {k!} {x^k}$$

$$\text{Ei}(\log (n^2))\sim\frac{n^2} 2 \sum_{k=1}^\infty \frac {a_k}{\log^k(n)}$$ where the first $a_k$ form the sequence $$\left\{1,\frac{1}{2},\frac{1}{2},\frac{3}{4},\frac{3}{2},\frac{15}{4},\frac{45}{4},\frac{315}{8},\frac{315}{2},\cdots\right\}$$ already given by @WhatsUp.

The numerators correspond to largest odd divisor of $k!$ (see sequence $A049606$ in $OEIS$) and the denominator correspond to Dress's sequence (see sequence $A001316$ in $OEIS$).

For $n=1000$, the above truncated expression would give $78623.2$ while the corresponding summation gives $78698.5$.