# An Inequality Involving The Riemann Zeta Function

I'm having trouble proving the following inequality for $2<r<3$: $$(1+2^{-r})\frac{(3^r+1)^2}{3^{2r}+1}>\frac{\zeta(r)}{\zeta(2r)}.$$ I can easily plot the graph, and the inequality clearly holds. I just don't seem to know how I could give a rigorous proof. I require $r<3$ because, when $r\geq3$, I can prove the inequality using the trivial bounds $1+2^{-r}<\zeta(r)<1+2^{-r}+\int_{2}^{\infty}x^{-r}dx$.

• – user153012 Oct 24 '14 at 12:09

I'm not able to prove it for little $r$, but I hope this can help you, maybe improving some constant. Let $r\geq2$. For the Euler product you have $$\frac{\zeta\left(r\right)}{\zeta\left(2r\right)}=\underset{p}{\prod}\left(1+\frac{1}{p^{r}}\right)$$ so your inequality is equivalet to $$\underset{p}{\prod}\left(1+\frac{1}{p^{r}}\right)<\left(1+\frac{1}{2^{r}}\right)\frac{\left(3^{r}+1\right)^{2}}{3^{2r}+1}=\left(1+\frac{1}{2^{r}}\right)\frac{\left(1+\frac{1}{3^{r}}\right)^{2}}{\left(1+\frac{1}{3^{2r}}\right)}$$ hence$$\underset{p\geq5}{\prod}\left(1+\frac{1}{p^{r}}\right)<\frac{\left(1+\frac{1}{3^{r}}\right)}{\left(1+\frac{1}{3^{2r}}\right)}$$ and again it's equivalent to $$\underset{p\geq5}{\sum}\log\left(1+\frac{1}{p^{r}}\right)<\log\left(1+\frac{1}{3^{r}}\right)-\log\left(1+\frac{1}{3^{2r}}\right).$$ Let $T>5$ an integer. You have, for partial summation $$\underset{5\leq p\leq T}{\sum}\log\left(1+\frac{1}{p^{r}}\right) \leq$$ $$\leq \pi\left(T\right)\,\log\left(1+\frac{1}{T^{r}}\right)+r\int_{5}^{T}\frac{\pi\left(t\right)}{t\left(t^{r}+1\right)}dt\,\overrightarrow{T\rightarrow\infty}r\int_{5}^{\infty}\frac{\pi\left(t\right)}{t\left(t^{r}+1\right)}dt.$$ So, because $\pi\left(n\right)<1.25506\,\frac{n}{\log\left(n\right)}$ for $n>1$, we have $$\underset{p\geq5}{\sum}\log\left(1+\frac{1}{p^{r}}\right)<1.25506\, r\int_{5}^{\infty}\frac{1}{\log\left(t\right)t^{r}}dt=1.25506\, r\int_{\log\left(5\right)\left(r-1\right)}^{\infty}\frac{e^{-t}}{t}dt=1.25506\, r\, E_{1}\left(\log\left(5\right)\left(r-1\right)\right).$$ Using the bound $E_{1}\left(x\right)\leq\log\left(1+\frac{1}{x}\right)e^{-x}$ we have $$\underset{p\geq5}{\sum}\log\left(1+\frac{1}{p^{r}}\right)<1.25506\, r\left(\log\left(1+\frac{1}{\log\left(5\right)\left(r-1\right)}\right)\frac{1}{5^{r-1}}\right).$$ On the other side we have $$\log\left(1+\frac{1}{3^{r}}\right)-\log\left(1+\frac{1}{3^{2r}}\right)=r\int_{3}^{3^{2}}\frac{1}{t\left(t^{r}+1\right)}dt>\frac{r}{2}\int_{3}^{3^{2}}\frac{1}{t^{r+1}}dt= \frac{3^{r}-1}{2\,3^{2r}}$$ so your inequality holds if $$1.25506\, r\left(\log\left(1+\frac{1}{\log\left(5\right)\left(r-1\right)}\right)\frac{1}{5^{r-1}}\right)<\frac{3^{r}-1}{2\,3^{2r}}$$ and this is true for $r$ large enough.