Asymptotic analysis of $\sum_{k=2}^n\frac{k}{\ln k}$ 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?
 A: 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.
A: 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 $78698.5$.
