TITLE: sum $\sum_{k=1}^{n-1}(1-\frac{k}{n})^{-a}\left(\frac{\log\left(k\right)}{k}-\frac{\log\left(k+1\right)}{k+1}\right)$
QUESTION [10 upvotes]: 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

REPLY [3 votes]: 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 $a>3$.