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Prove for $m,n\in \mathbf N$, $$\frac{\sum_{j=1}^mj^{2n}}{\sum_{j=1}^m\big(j-\frac12\big)^{2n}}>\bigg(\frac{m+\frac12}m\bigg)^{2n+1}.$$

I suspect this is true for any sufficiently large power, so that $2n$ can be replaced by an arbitrary real number above some bound.

This proposition seems to call for using the estimation of summation with integral. $$\frac1{k+1}\bigg(\big(m+\frac12\big)^{k+1}-\frac1{2^{k+1}}\bigg)=\int_{\frac12}^{m+\frac12} x^k dx >\sum_{j=1}^mj^k>\int_0^m x^k{\rm d}x=\frac1{k+1}m^{k+1}.$$ The first inequality comes from the convexity of the $x^k$ for $k>1$. We get a resemblance of the conclusion. However, it does not get us the desired inequality. We need a finer estimate. Perhaps Euler-Maclaurin formula could be of help. One can also resort to the exact evaluation $\sum_{j=1}^m j^k$ in sum of combinatorial numbers. But then one needs to know the lower bound of the combinatorial numbers. It seems this bound is quite tight. I cannot see the way forward right now how.

Hans
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    If there are good asymptotics or bounds for the Faulhaber's formula, then you can write your LHS in terms of $S(m,n) = \sum_{i=1}^m i^n:$ $$LHS = \dfrac{2^{2n}S(m,2n)}{S(2m,2n)-2^{2n}S(m,2n)} $$ – dezdichado Feb 01 '25 at 16:01
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    Function plots indicate that $\frac{\sum_{j=1}^mj^{a}}{\sum_{j=1}^m\left(j-\frac12\right)^{a}}>\left(\frac{m+\frac12}m\right)^{a+1}$ for $a \ge 2$. – Martin R Feb 01 '25 at 18:13
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    I could get very very close just using calculus tricks (i.e. try to prove $\frac{d}{dx} \log(f(x)) \ge \frac{2n+1}{2} (\log(m+1/2) -\log(m))$ over $x\in [-1/2, 0]$, where $f(x) = \sum_{j=1}^m (j+x)^{2n}$). After differentiating and applying the integral inequalities you wrote down, I ended up wanting to prove something like $\frac{(m+x)^{2n}}{2(m+1/2 + x)^{2n+1}} \ge \log(m+1/2) -\log(m)$. These are both $\frac{1}{2m}$ to first order, so it's certainly very close, but I couldn't see how to complete the proof. Maybe someone else will have better luck with this approach. – LVFToENlCx Feb 01 '25 at 22:00
  • @Hans $\log(f(0)) - \log(f(-1/2)) \ge \int_{-1/2}^0 \frac{d}{dx} \log(f(x))~dx \ge \frac{1}{2} \inf_{x \in [-1/2, 0]} \frac{d}{dx} \log(f(x))$. So to prove $\log(f(0)) - \log(f(-1/2)) \ge (2n+1)(\log(m+1/2) - \log(m))$, it suffices to prove the same lower bound for $\frac{1}{2} \frac{d}{dx} \log(f(x))$ for $x \in [-1/2, 0]$. – LVFToENlCx Feb 02 '25 at 00:39
  • @Hans Since $\log(m+1/2) - \log(m)$ is $\frac{1}{2m}$ to first order, we would expect that difference in power, but that doesn't prove that the inequality works. – LVFToENlCx Feb 02 '25 at 00:40
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    @Pranay Good catch, it should have been $\frac{1}{2} \frac{d}{dx} \log(f(x)) \ge (2n+1) (\log(m+1/2) - \log(m))$. I believe I have it right on the rest of the comment. – LVFToENlCx Feb 02 '25 at 00:42
  • @LVFToENlCx: I realized what you were trying to do before you responded. I like your idea. I will try to complete it, later. – Hans Feb 02 '25 at 06:00
  • @LVFToENlCx: Unfortunately, a few numerical trials produced an inequality sign opposite to the desired direction. Try a few examples yourself, in case I erred. – Hans Feb 02 '25 at 07:35
  • Now posted to MO, https://mathoverflow.net/questions/487276/prove-frac-sum-j-1mj2n-sum-j-1m-bigj-frac12-big2n-big-fra – Gerry Myerson Feb 05 '25 at 23:11

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