Equivalent of $\sum_{k=1}^{n-1}\frac1{\ln(n/k)}$ when $n\to+\infty$? I'm looking for an equivalent of $$\displaystyle\sum_{k=1}^{n-1}\frac1{\ln(n/k)}\quad n\to+\infty$$ It must be $n\ln n$, but I can't prove it.
It looks like a Riemann sum but the function is not Riemann-integrable...
Thanks.
 A: $$\sum_{k=1}^{n-1}\frac{1}{\log(n/k)}=\sum_{k=1}^{n-1}\frac{1}{-\log\left(1-\frac{k}{n}\right)}=\sum_{k=1}^{n-1}\frac{1}{\frac{k}{n}+\frac{k^2}{2n^2}+\frac{k^3}{3n^3}+\ldots} $$
and $\sum_{k=1}^{n-1}\frac{1}{\frac{k}{n}}= n H_{n-1} = n\log(n)+\gamma n+o(n)$, while
$$ \sum_{k=1}^{n-1}\left(\frac{1}{\frac{k}{n}}-\frac{1}{\frac{k}{n}+\frac{k^2}{2n^2}+\frac{k^3}{3n^3}+\ldots}\right)=\sum_{k=1}^{n-1}\frac{\frac{k}{2n}+\frac{k^2}{3n^2}+\ldots}{\frac{k}{n}+\frac{k^2}{2n^2}+\frac{k^3}{3n^3}+\ldots} $$
is easily proved to be $O(n)$, since $\frac{1}{x}+\frac{1}{\log(1-x)}$ is improperly Riemann-integrable over $(0,1)$ and its integral equals $\gamma$. Gregory coefficients are involved in a subtle way.
A: Since $x\mapsto \frac{1}{\ln(\frac{n}{x})}$ is increasing, $$\frac{1}{\ln n}+\int_1^{n-1} \frac{1}{\ln(\frac{n}{t})}dt \leq \sum_{k=1}^{n-1}\frac1{\ln(n/k)}\leq \int_1^{n-1} \frac{1}{\ln(\frac{n}{t})}dt + \frac{1}{\ln\left(\frac{n}{n-1} \right)} $$
By the substitution $u=\frac nt$, $$\int_1^{n-1} \frac{1}{\ln(\frac{n}{t})}dt = n\int_{\frac{n}{n-1}}^n\frac{du}{u^2\ln u}$$
This integral diverges solely because of the contribution at $\frac{n}{n-1}\approx 1$, that is  $$\int_{\frac{n}{n-1}}^n\frac{du}{u^2\ln u} = \int_{\frac{n}{n-1}}^2\frac{du}{u^2\ln u} + O(1)$$
Since $ \frac{1}{u^2\ln u}\sim_{u\to 1} \frac{1}{u\ln u}$ and $\int_1^2 \frac{1}{u\ln u}=\infty$, one has the estimate $\displaystyle \int_{\frac{n}{n-1}}^2\frac{du}{u^2\ln u}\sim \int_{\frac{n}{n-1}}^2\frac{du}{u\ln u}$
But $\displaystyle \int_{\frac{n}{n-1}}^2\frac{du}{u\ln u} = \bigg[\ln \ln u \bigg]_\frac{n}{n-1}^2 =\ldots= \ln n+O(1)$
Thus $\displaystyle \int_1^{n-1} \frac{1}{\ln(\frac{n}{t})}dt = n\ln n+O(n)$. It remains to see that $\frac{1}{\ln\left(\frac{n}{n-1} \right)} =O(n)$ and the chain of inequalities above proves that $$\sum_{k=1}^{n-1}\frac1{\ln(n/k)} = n\ln n+ O(n)$$
