Let for $a>0$, $\displaystyle S_n(a)=\sum_{k=1}^{n-1}(1-\frac{k}{n})^{-a}\left(\frac{\log\left(k\right)}{k}-\frac{\log\left(k+1\right)}{k+1}\right)$.

I conjecture that there exist a number $c_a\in\mathbb R$ , such that : $$ \displaystyle S_n(a)\sim c_a \big(\sum_{k=1}^{n-1}{1\over k^a}\big)n^{a-2} \ln n\quad (n\to +\infty).$$

I found by elementary calculus that

When $a=1$, $ \displaystyle S_n(1)\sim \frac 32 \frac{\ln^2 n}n$ , so $c_1=\frac 32$ and when $a=2$, $ \displaystyle S_n(2)\sim \frac{\pi^2}6\ln n$, so $c_2=1$

I need some help for général case

I give this non-rigorous proof $\displaystyle S_n(a) = \sum_{k=1}^{n-1} \Big(1-{k \over n}\Big)^{-a} \Big({\ln k \over k} - {\ln (k+1) \over k+1} \Big). $ with change of index : $\displaystyle S_n (a)= \sum_{k=1}^{n-1} \Big({n \over k}\Big)^a \Big({\ln (n-k) \over n-k} - {\ln (n-k+1) \over n-k+1} \Big).$

I believe we have this asymptotic development : $\displaystyle {\ln (n-k) \over n-k} - {\ln (n-k+1) \over n-k+1} = {\ln n -1\over n^2}\Big(1+ {(2k-1)\over n } {2 \ln n -3 \over 2 \ln n-2} + \big({1\over n^3}\big)\epsilon_{k,n}\Big)$,

Thus ? $\displaystyle S_n (a) \sim c_a \Big(\sum_{k=1}^{n-1}\Big({n \over k}\Big)^a \Big) {\ln n \over n^{2}}$ with a constant $c_a$ to be determined

Explanations at the request of Dr. Wolfgang Hintze

A- Let us establish that Thank's for my friend Lou $\displaystyle \lim_{n\to + \infty}S_n(a)=\left\{\begin{array}{cl} 0 & \text { si } \: 0\leqslant a <2\\+\infty &\:\text {si } \: a\geqslant 2. \end{array}\right.$

Let $f: x\mapsto \dfrac{\log x}x - \dfrac {\log(x+1)}{x+1}.\qquad f(x)\underset{x\to + \infty}\sim \dfrac {\log x}{x^2}.$ $1)\quad\forall a\geqslant 2, \quad S_n(a)\geqslant \dfrac {n^{a}}{(n-1)^{a}}f(1)+\dfrac {n^{a}}{(n-2)^{a}}f(2) +0+0+\dots +0+\dfrac {n^{a}}{(1)^{a}}f(n-1)\underset{n\to + \infty}\sim n^{a-2}\log n.$ $2)\quad $ It's easy to prove that : $\:\:\forall x\in [3;n], \:\: 0<f(x)<\dfrac{ \log n}{x^2} \quad(1).$ If $a\in [0;2[.\quad S_n(a) =-\dfrac{\log n}n+ \displaystyle \sum _{k=1}^{n-1}\left(\dfrac {n^{a}}{(n-k)^{a}} -1\right) f(k).\quad$ write : $\:T_n(a):=\log n\displaystyle \sum _{k=3}^{n-1}\left(\dfrac {n^{a}}{(n-k)^{a}} -1\right) \dfrac 1{k^2}.$ From $ (1) $, it therefore suffices to prove that : $ \displaystyle \lim_{n\to + \infty} T_n(a) =0.$ $\displaystyle T_n(a)=\log n \sum_{k=1}^{n-3}\dfrac{n^{a} - k^{a}}{k^{a}(n-k)^2} =\dfrac{\log n}{n^2} \displaystyle \sum_{k=1}^{n-3}(n^{a}- k^{a})\left( k^{-a}+ \dfrac {2 k^{1-a}}{n-k} + \dfrac {k^{2-a}}{(n-k)^2}\right).\quad $ We also have inequalities: $\:\:\forall a \in \mathbb R^+,\: \forall k\in [1;n]: $ $0 \leqslant a<2 \implies 0\leqslant n^{a}- k^{a}\leqslant an^{a-1}(n-k).\quad (2), \quad 0\leqslant a \leqslant1 \implies 0\leqslant n^{a}- k^{a}\leqslant ak^{a-1}(n-k).\quad (3).$ They lead to: $\bullet \:\text{Si }1\leqslant a<2, \:\text {alors}\:\:0<T_n(a) < \log n \left(n^{a-2}\displaystyle \sum_{k=1}^{n-3}k^{-a} +2a n^{a-3}\sum_{k=1}^{n-3}k^{1-a}+an^{-1}\sum_{k=1}^{n-3} (n-k)^{-1}\right). \qquad (4)$ $\bullet \:\text{Si }0\leqslant a \leqslant1, \:\text {alors}\:\: 0<T_n(a) < \log n \left(n^{a-2}\displaystyle \sum_{k=1}^{n-3}k^{-a} +2a n^{-2}\sum_{k=1}^{n-3}1+an^{-1}\sum_{k=1}^{n-3} (n-k)^{-1}\right).\qquad (5)$ We check that the right-hand side of the inequalities (4) and (5) have a zero limit when $ n \to + \infty.$

B- Show that $\:\:\boxed{S_n(1) \underset{n\to + \infty}\sim \dfrac {3(\log n)^2}{2n}.}\qquad$ Let $g: x\mapsto x^2f(x) -\log x.\quad $ Then : $\:\:\displaystyle \lim_{+\infty} g=-1, \quad g \text { is bounded on } [1;+\infty[\:\: (1).\qquad$ Let $\:\:T_n:= \displaystyle \sum_{k=1}^{n-1} \dfrac k{n-k} f(k). $ Then: $ \:\: S_n(1) = T_n- \dfrac {\log n}n\:\:(2).\:\:\quad T_n=\dfrac 1n \displaystyle \sum_{k=1}^{n-1}\left( \dfrac 1k + \dfrac 1{n-k}\right) k^2f(k) =U_n +V_n +W_n\:\:$ with $U_n,V_n,W_n $ are defined by : $\displaystyle U_n:=\dfrac 1n\sum_{k=1}^{n-1} \dfrac {\log k}k, \quad V_n:= \dfrac 1n\sum_{k=1}^{n-1} \dfrac {\log k}{n-k}, \quad W_n:= \dfrac 1n\sum_{k=1}^{n-1}\left( \dfrac 1k + \dfrac 1{n-k}\right) g(k).\:\:\:$ According to $(1): \quad W_n=\mathcal O\left (\dfrac{\log n}n \right)\:\:(3).\quad \:\:U_n \underset{n\to + \infty}\sim \dfrac{(\log n)^2}{2n}\:\: (4). \qquad V_n =\displaystyle \dfrac {\log n}n\sum_{k=1}^{n-1}\dfrac 1{n-k} +\dfrac 1{n^2}\displaystyle\sum_{k=1}^{n-1}\dfrac {\log(k/n)}{1-(k/n)}. $ The Function $h:x\mapsto \dfrac{\ln x}{1-x}\text{ is continuous, monotonic, integrable on }\:]0;1[,\quad $ so $\:\:\displaystyle \lim_{n\to +\infty}\dfrac 1n\sum_{k=1}^{n-1} \dfrac {\log(k/n)}{1-(k/n)} =\int _0 ^1 h .$ This information, combined with the fact that: $\: \displaystyle \dfrac {\log n}n\sum_{k=1}^{n-1}\dfrac 1{n-k} \underset{n\to + \infty}\sim \dfrac {(\log n)^2}n,\:$ leads to: $\:\:V_n\underset{n\to + \infty}\sim \dfrac {(\log n)^2}n.\quad (5)$ $(2), (3), (4) $ et $(5)$ provide the equivalent of $ S_n (1) $ announced.

I can explain the case a = 2 if needed

  • $\begingroup$ have you tried to split your sum at some point ($k_s=\sqrt n$ or something like that) and estimate both contributions seperatly (the main contribution will be from the part where $n>k>k_s$, the other part should give something $\sim\log(k_s)/k_s$ which is small in comparison). i don't have the time to write something up as an answer, but my premiliminary results look rather promising. $\endgroup$
    – asgeige
    Dec 22, 2021 at 13:55
  • $\begingroup$ I don't know what u mean $\endgroup$
    – Jane
    Dec 22, 2021 at 20:43
  • $\begingroup$ @Jane Thank you very much for your explanation (+1) $\endgroup$ Jan 7 at 22:11

2 Answers 2


I will deal with the case that $a> 3$. We have $$ n^{ - a} S_n (a) = \sum\limits_{k = 1}^{n - 1} {\frac{1}{{k^a }}\left( {\frac{{\log (n - k)}}{{n - k}} - \frac{{\log (n - k + 1)}}{{n - k + 1}}} \right)} . $$ We split the sum at $k=\lfloor \sqrt n \rfloor $. Using the mean value theorem and some trivial estimation, we find \begin{align*} &\left|\sum\limits_{k = \lfloor \sqrt n \rfloor }^{n - 1} {\frac{1}{{k^a }}\left( {\frac{{\log (n - k)}}{{n - k}} - \frac{{\log (n - k + 1)}}{{n - k + 1}}} \right)} \right| \leq \sum\limits_{k = \lfloor \sqrt n \rfloor }^{n - 1} \frac{1}{{k^a }}\left|\frac{{\log (n - t_k ) - 1}}{{(n - t_k )^2 }}\right| \\ & \le C\frac{{\log n}}{{n^2 }}\sum\limits_{k = \left\lfloor {\sqrt n } \right\rfloor }^{n - 1} {\frac{1}{{k^{a - 2} }}} \le C\frac{{\log n}}{{n^2 }}\int_{\left\lfloor {\sqrt n } \right\rfloor - 1}^{ + \infty } {\frac{{dt}}{{t^{a - 2} }}} \le \frac{{C'}}{{a- 3}}\frac{{\log n}}{{n^{(a + 1)/2} }}, \end{align*} where $k-1<t_k<k$ is a suitable number and $C,C'>0$ are appropriate constants. Consequently, $$ \frac{{n^{2 - a} }}{{\log n}}S_n (a) = \sum\limits_{k = 1}^{\lfloor \sqrt n \rfloor - 1} {\frac{1}{{k^a }}\frac{{n^2 }}{{\log n}}\left( {\frac{{\log (n - k)}}{{n - k}} - \frac{{\log (n - k + 1)}}{{n - k + 1}}} \right)} + \mathcal{O}\!\left( {\frac{1}{{n^{(a - 3)/2} }}} \right). $$ If $1\leq k<\sqrt{n}$, then by the mean value theorem and the monotonicity of $x \mapsto (\log x-1)/x^2$, we find \begin{align*} 0<\frac{{n^2 }}{{\log n}}\left( {\frac{{\log (n - k)}}{{n - k}} - \frac{{\log (n - k + 1)}}{{n - k + 1}}} \right) & \le \frac{{n^2 }}{{\log n}}\frac{{\log (n - k) - 1}}{{(n - k)^2 }} \\ & \le \frac{{n^2 }}{{\log n}}\frac{{\log (n - \sqrt n ) - 1}}{{(n - \sqrt n )^2 }} < 2 \end{align*} for large $n$. Therefore, using the asymptotics $$ \frac{{\log (n - k)}}{{n - k}} - \frac{{\log (n - k + 1)}}{{n - k + 1}} \sim \frac{{\log n}}{{n^2 }} $$ for fixed $k$ and large $n$, and Tannery's theorem, we deduce $$ \mathop {\lim }\limits_{n \to + \infty } \frac{{n^{2 - a} }}{{\log n}}S_n (a) = \zeta (a) $$ provided $a>3$.

  • 1
    $\begingroup$ I want to extend my thanks and appreciation for your time and effort. I will try to understand your reasoning $\endgroup$
    – Jane
    Jan 7 at 16:14
  • 2
    $\begingroup$ @Gary. Great proof (1+). As an expert for asymtotic expansions, can you perhaps also solve the problem using my integral representation? $\endgroup$ Jan 7 at 23:05

This is not an answer but an extended comment which provides an integral representation of the sum which might serve as a starting point for further investigations.

We have to study the sum

$$S(a,n) =\sum_{k=1}^{n-1} f(k)\left(c(k) - c(k+1)\right)\tag{1}$$

Here $f(k) = (1-\frac{k}{n})^{-a}$ and $c(k)=\frac{\log(k)}{k}$

The peculiar form of the second factor of the summand suggests immediately to write it as an integral

$$c(k)-c(k+1) = -\int_{k}^{k+1} c'(x)\;dx\tag{2}$$

Then our sum becomes

$$S(a,n) = -\sum_{k=1}^{n-1} f(k) \int_{k}^{k+1} c'(x)\;dx\tag{3}$$

Notice now (this is a useful trick) that we can incorporate the factor $f(k)$ in the integrand writing it with the floor function as $f (\lfloor x\rfloor)$ for $k \lt x \lt k+1$ so that

$$S(a,n) = -\sum_{k=1}^{n-1} \int_{k}^{k+1} f (\lfloor x\rfloor) c'(x)\;dx\tag{4}$$

the sum can be done using $\sum_{k=1}^{n-1}\int_{k}^{k+1}=\int_{1}^{n}$ and using $c'(x) = -\frac{\log(x)-1}{x^2}$ leads to the integral representation of the sum

$$S(a,n) = I(a,n) := \int_{1}^{n} (1-\frac{\lfloor x\rfloor}{n})^{-a} \frac{\log(x)-1}{x^2}\;dx\tag{5}$$

Notice that, due to the floor function, the integrand is a discontinuous but finite function.

I have checked the equivalence $I=S$ symbolically and numerically in Mathematica.

  • $\begingroup$ Thank's for formula 5, but it's not easy with this formula to show that for example $ \int_{1}^{n} (1-\frac{\lfloor x\rfloor}{n})^{-1} \frac{\log(x)-1}{x^2}\;dx \sim \frac 32 \frac{\ln^2 n}n$ or $ \int_{1}^{n} (1-\frac{\lfloor x\rfloor}{n})^{-2} \frac{\log(x)-1}{x^2}\;dx \sim \frac{\pi^2}6\ln n $ $\endgroup$
    – Jane
    Jan 3 at 18:35
  • $\begingroup$ Ok, I agree, but could you please explain how you did the calculation for $a=1$ and $a=2$ $\endgroup$ Jan 4 at 12:53
  • $\begingroup$ I explained the case a = 1, do you also want the case a = 2? $\endgroup$
    – Jane
    Jan 6 at 20:50

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