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I'm trying to prove the positivity of the function $(n+2)\zeta(n+3)-\zeta(n+2)-n-1$ on $\mathbb{N}$ using the inequalities \begin{equation*} \frac{s+1}{s}\zeta(s)\zeta(s+2)\geq \zeta^2(s+1),\quad s>1 \end{equation*} and \begin{equation*} \zeta(s)\zeta(s+2)> \zeta^2(s+1),\quad s>1 \end{equation*} and $$ \zeta^2(s + 1)< \left(\frac{s + 1}{s}\right) \left( \frac{(1 - 2^{-s})(1 - 2^{-(s+2)})}{(1 - 2^{-(s+1)})^2} \right) \zeta(s)\zeta(s + 2), \quad s > 1. $$

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    $\begingroup$ You may simply use the definition of $\zeta(s)=\sum1/n^s$, after cancellation the remaining positive terms are much larger in absolute values then negative $\endgroup$
    – Fedor Petrov
    Commented Oct 19 at 11:38
  • $\begingroup$ Is this sum $\sum_{k=2}^{\infty}\left(\frac{n+2}{k^{n+3}}-\frac{1}{k^{n+2}}\right) $ positive for $n\in \mathbb{N}?$ $\endgroup$
    – user90533
    Commented Oct 20 at 5:01
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    $\begingroup$ Certainly: negative terms start with $k=n+3$, and they are way too small than the $k=2$ term, by the integral bound or whatever. $\endgroup$
    – Fedor Petrov
    Commented Oct 20 at 5:05

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