Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm studying the following function:


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?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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 $1/p$.

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.