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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?

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    $\begingroup$ For most terms $\ln k$ is very close to $\ln n$, so leading order would be $\frac12n^2/\ln n$ $\endgroup$
    – Empy2
    Dec 22, 2021 at 19:49

2 Answers 2

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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.

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  • $\begingroup$ You can use \left . f \right |_{n=1}^\infty in order to get $\left . f \right |_{n=1}^\infty$. More generally, \left and \right can help make your brackets look nicer. I've edited your answer to have both of these formatting changes, but feel free to undo them if you'd like ^_^ $\endgroup$ Dec 23, 2021 at 5:54
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    $\begingroup$ It may be worth adding that $\sum\nolimits_{k = 2}^n {\frac{k}{{\ln k}}} = I(n) + \mathcal{O}(1)$, so the sum in question and $I(n)$ have the same asymptotic expansion. $\endgroup$
    – Gary
    Dec 23, 2021 at 5:56
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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 $78698.5$.

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  • $\begingroup$ Note that dlmf.nist.gov/6.12.E2 leads directly to the final result without the "intermediate" expansion. $\endgroup$
    – Gary
    Dec 23, 2021 at 3:34
  • $\begingroup$ @Gary. Edited accordingly. $\endgroup$ Dec 23, 2021 at 5:31
  • $\begingroup$ You better use $\sim$ in place of $=$ since these series are divergent asymptotic expansions. $\endgroup$
    – Gary
    Dec 23, 2021 at 5:32

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