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Does this reduce to values of a known special function for arbitrary real (or complex) $s$? Answered by Johannes Trost in a comment: it's also known as a "Roman harmonic number". But this …

The factor $\exp(-n\lambda/(2n+2))$ probably precludes significant simplification of $f(N,\lambda)$, but one can still construct an integral representation that shows $f(N,\lambda) > 0$ for all positi …

The desired inequality should be true iff $$ c < c_0 := (r - \sqrt{r^2-1})^2 \quad\ \text{where} \quad\ r = \frac{|a|}{2b} $$ (NB the hypotheses $b>0$ and $a < -2b$ imply $r>1$, so $0 < c_0 < 1$). Num …

Since you know already that the optimal $a_i$ have $2a_i + 1 = x_i = \pm 1$ and the roots of $P'_{n-1}$, the calculation of $V_n$ comes down to the discriminant of $P'_{n-1}$, its leading coefficient, …

[Revised and expanded to give the answer for all $k>1$ and incorporate further terms of an asymptotic expansion as $n \rightarrow \infty$] Fix $k>1$, and write $a_1=f(1,k)=1$ and $$ a_n = f(n,k) = \f …