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## Is $Q_n(x)=\sigma_{n+1}(x)/\sigma_n(x)$ logarithmically convex on $\mathbf{R}$?

In 1975 J. van de Lune considered the monotony properties of the canonical Riemann Upper and Lower sums for $\int_0^1 t^xdt$, with $x>0$.
Writing $\sigma_n(x) := 1^x+2^x+\cdots+n^x$ these sums are
$$U_n:=U_n(x)=\sigma_n(x)/n^{x+1} \quad \text{and}\quad L_n:= L_n(x)=\sigma_{n-1}(x)/n^{x+1}.$$ He proved (by mathematical induction) that $U_n>U_{n+1}$ and $L_n<L_{n+1}$.

A later proof of $U_n>U_{n+1}$ was obtained by showing that the function
$$h(x):= h_n(x)=(\sigma_{n+1}/\sigma_n(x))(n/(n+1))^x$$ is strictly decreasing on all of $\mathbf{R}$.
Soon afterwards he came to realize that the monotonicity of $h(x)$ would be a consequence of the logarithmic convexity of $Q(x):=Q_n(x)=\sigma_{n+1}(x)/\sigma_n(x)$ on all of $\mathbf{R}$.
Various numerical tests on $Q(x)$ were performed, but no proof was found.

Another application: From the logarithmic convexity of $Q(x)$ one may also obtain a simple proof of a conjecture made by H. Alzer and A. A. Jagers: $$f(x):=f_n(x)=\Bigl(\frac{\frac{1}{n+1}\sigma_{n+1}(x)}{\frac1n \sigma_n(x)}\Bigr)^{1/x}$$ is strictly increasing for $x>0$.

Question: Is for every (fixed) integer $n\ge2$, the function
$$x\in\mathbf{R}\mapsto Q_n(x):=(1^x+2^x+\cdots+n^x+(n+1)^x)/(1^x+2^x+\cdots+n^x)$$ logarithmically convex on all of $\mathbf{R}$?

-
 btw. google reveals a claim of van de Lune that he has a proof of the above for $n \in [2,10]$, and he leaves the general case a conjecture. a very nice conjecture though! – suVRit Sep 19 at 19:55 @Suvrit Much of the work of van de Lune about his conjecture is described in the CWI Report PNA - R0502, May 2005: [J. van de Lune, H. J. J. te Riele, "On some conjectural inequalities and their consequences"](oai.cwi.nl/oai/asset/10943/10943D.pdf) – juan Sep 20 at 19:14 @Suvrit Related material can be found in the paper S. Abramovich, J. Baric, M. Matic, J. Pecaric, "On van de Lune-Alzer's Inequality" J. of Math. Inequalities, 1, (2007) 563-587.