# Recurrence inequality for Dirichlet's eta function.

I'm studying the following function:

$\theta(p)=\eta(p)\eta(p-2)-\frac{p-1}{p}\eta^2(p-1)$,

where $\eta$ - Dirichlet's eta function. If we build a plot for $p\in [1,150]$, it's easy to see that it's positive and decreasing for $p\in [3,150]$. I have a hypothesis that this function remains positive for all $p\ge 1$.

Are there any analytical results on this subject?

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You can say stronger things. Since $1-2^{-x}<\eta(x)<1$ for real $x>0$ $$1>\eta(p)\eta(p-2)>(1-2^{-p})(1-2^{-p+2})>1-\frac{5}{2^p} \\ 1>\eta(p-1)^2>(1-2^{-p+1})^2>1-\frac{4}{2^p}$$ So $$-\frac{5}{2^p} < \eta(p)\eta(p-2)-\eta^2(p-1) < \frac{4}{2^p}$$ And for a sequence $\epsilon(p)$ taking on values in $[0,1)$ $$\epsilon-\frac{4\epsilon+5}{2^p}<\eta(p)\eta(p-2)-(1-\epsilon)\eta^2(p-1) < \epsilon+\frac{4}{2^p}$$ In your case $\epsilon(p)=1/p$, so $$\frac{1}{p}-\frac{5+4/p}{2^p}<\theta(p)<\frac{1}{p}+\frac{4}{2^p}$$ It follows that $\theta(p)>0$ for $p\ge 5$ and that it converges quickly to $1/p$.
Thank you, this is quite an elegant way. Could you please give a link with a proof for the first line (boundaries for $\eta$)? –  TZakrevskiy May 21 '13 at 18:59
@TZakrevskiy $\eta(x) = 1 - (2^{-x}-3^{-x})-(4^{-x}-5^{-x})\cdots$ each bracketed pair is positive for $x>0$, so $\eta(x)<1$. Likewise $\eta(x) = 1-2^{-x}+(3^{-x}-4^{-x})+(5^{-x}-6^{-x})\cdots$ so $\eta(x)>1-2^{-x}$. It's also described here en.wikipedia.org/wiki/Alternating_series_test#Proof_.5B1.5D –  Zander May 21 '13 at 21:56