# sum $\sum_{k=1}^{n-1}(1-\frac{k}{n})^{-a}\left(\frac{\log\left(k\right)}{k}-\frac{\log\left(k+1\right)}{k+1}\right)$

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 : $$\:\:\forall x\in [3;n], \:\: 0 If $$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 $$\bullet \:\text{Si }0\leqslant a \leqslant1, \:\text {alors}\:\: 0 We check that the right-hand side of the inequalities (4) and (5) have a zero limit when $$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

• 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. Dec 22, 2021 at 13:55
• I don't know what u mean
– Jane
Dec 22, 2021 at 20:43
• @Jane Thank you very much for your explanation (+1) Jan 7 at 22:11

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 $$k-1 is a suitable number and $$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$$.

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

• 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 $\int_{1}^{n} (1-\frac{\lfloor x\rfloor}{n})^{-2} \frac{\log(x)-1}{x^2}\;dx \sim \frac{\pi^2}6\ln n$
– Jane
Jan 3 at 18:35
• Ok, I agree, but could you please explain how you did the calculation for $a=1$ and $a=2$ Jan 4 at 12:53
• I explained the case a = 1, do you also want the case a = 2?
– Jane
Jan 6 at 20:50