Minimum and maximum of this sequence Fixed two integers $n>1$ and $N\geq 2$, consider the (finite) sequence of points
$$
S:=\left\{ \frac{\ln^{k}(2)+\ldots + \ln^{k}(N)}{\ln^{k+1}(2)+\ldots + \ln^{k+1}(N)} : k=0,\ldots, n-1  \right\}
$$
I am interested in computing the minimum and the maximum of $S$, Some suggestions?
Many thanks in advance for your comments.
 A: For $N = 2$, it's straight-forward that the first term (with $k = 0$) and the last term (with $k = n-1$) are the maximum and the minimum of $S$.
We will prove the same result for $N \ge 3$:
$$\frac{\ln^{p}(2)+\ldots + \ln^{p}(N)}{\ln^{p+1}(2)+\ldots + \ln^{p+1}(N)} \ge \frac{\ln^{q}(2)+\ldots + \ln^{q}(N)}{\ln^{q+1}(2)+\ldots + \ln^{q+1}(N)} \qquad \forall 0\le p<q \tag{1}$$
Let's denote
$$f(x) = \ln\left(\sum_{n=2}^N \ln^x(n)  \right) \qquad \forall 0\le p<q $$
$(1)$ is equivalent to
$$f(p)-f(p+1) \ge f(q)-f(q+1) \iff f(p)+f(q+1) \ge f(q)+f(p+1)  \qquad \forall 0\le p<q \tag{2}$$
We prove $(2)$ by applying the Karamata's inequality.
For $N \ge 3$, we have
$$f'(x) = \frac{\sum_{n=2}^N \left(\ln(\ln(n))\ln^x(n)\right)}{\sum_{n=2}^N \ln^x(n)} >0$$
$$f''(x) =  \frac{\left(\sum_{n=2}^N (\ln(\ln(n))^2\ln^x(n))\right)\left(\sum_{n=2}^N \ln^x(n)\right)-\left(\sum_{n=2}^N (\ln(\ln(n))\ln^x(n))\right)^2}
{\left(\sum_{n=2}^N \ln^x(n)\right)^2} \tag{3}$$
By applying the Cauchy Schwarz inequality, the numerator of $(3)$ is positive. Then $f''(x) > 0$.
Hence, the function $f(x)$ is increasing and convexe for $x \in \Bbb R^+$. Applying the Karamata's inequality for $(x_1,x_2,y_1,y_2) = (q+1,p,q,p+1)$ that satisfy
$$x_1 > x_2$$
$$y_1 \ge y_2$$
$$x_1+x_2 = y_1+y_2 = (p+q+1)$$
then $(2)$ holds true.
Conclusion: the first term (with $k = 0$) and the last term (with $k = n-1$) are the maximum and the minimum of $S$.
