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 $1+2^{-r}<\zeta(r)<1+2^{-r}+\int_{2}^{\infty}x^{-r}dx$.

• Oct 24, 2014 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.

Some thoughts.

Remark. It remains to give a rigorous proof of (2). I have a very complicated approach: We have $$F''(r) = \sum_{k=1}^\infty f(k,r)$$. We can prove that $$f(k,r) \le 0$$ for all $$k\ge 3$$ and $$r\in [2, 3]$$. It suffices to prove that $$\sum_{k=1}^{20} f(k,r) \le 0$$. It seems true by plotting it.

It suffices to prove that $$(1+2^{-r})\frac{(3^r+1)^2}{3^{2r}+1}> \frac{12}{23} + \frac{2}{r} >\frac{\zeta(r)}{\zeta(2r)}. \tag{1}$$

(i) For the right inequality in (1).

Let $$F(r) := \left(\frac{12}{23} + \frac{2}{r}\right) \zeta(2r) - \zeta(r) = \left(\frac{12}{23} + \frac{2}{r}\right) \sum_{k=1}^\infty \frac{1}{k^{2r}} - \sum_{k=1}^\infty \frac{1}{k^r}.$$ We have \begin{align*} F''(r) &= \sum_{k=1}^\infty \left( \frac{4}{r^3 k^{2r}} + \frac{8\ln k}{r^2 k^{2r}} + 4\left(\frac{12}{23} + \frac{2}{r}\right)\frac{\ln^2 k}{k^{2r}}\right) - \sum_{k=1}^\infty \frac{\ln^2 k}{k^r} \le 0. \tag{2} \end{align*} Also, we have $$F(2) = \frac{7}{414}\pi^4 - \frac{1}{6}\pi^2 > 0$$, and $$F(3) = \frac{82}{65205}\pi^6 - \zeta(3) > 0$$. Thus, $$F(r) > 0$$ on $$[2, 3]$$.

$$\phantom{2}$$

(ii) For the left inequality in (1).

Fact 1. It holds that, for all $$r \in [2, 3]$$, $$\frac92(3-r)(r-2)^2c^2 + 9(r-1)(r-2)(3-r)c + 18(r-2)^3 + 9\ge 3^r$$ where $$c = \ln 3$$. (Note: The proof is not difficult by taking logarithm and then taking derivative.)

Fact 2. It holds that, for all $$r\in [2,3]$$, $$2(3-r)(r-2)^2d^2 + 4(r-1)(r-2)(3-r)d + 4(r-2)^3+4 \ge 2^r$$ where $$d = \ln 2$$. (Note: The proof is not difficult by taking logarithm and then taking derivative.)

By Fact 1, using $$\ln 3 < 11/10$$, we have, for all $$r\in [2, 3]$$, $$\frac92(3-r)(r-2)^2(11/10)^2 + 9(r-1)(r-2)(3-r)(11/10) + 18(r-2)^3 + 9\ge 3^r.$$

By Fact 2, using $$\ln 2 < 7/10$$, we have, for all $$r \in [2, 3]$$, $$2(3-r)(r-2)^2(7/10)^2 + 4(r-1)(r-2)(3-r)(7/10) + 4(r-2)^3+4 \ge 2^r.$$

Let $$a := 2(3-r)(r-2)^2(7/10)^2 + 4(r-1)(r-2)(3-r)(7/10) + 4(r-2)^3+4,$$ and $$b := \frac92(3-r)(r-2)^2(11/10)^2 + 9(r-1)(r-2)(3-r)(11/10) + 18(r-2)^3 + 9.$$ By Facts 1-2, we have $$a \ge 2^r$$ and $$b \ge 3^r$$. Thus, we have (easy) $$(1+2^{-r})\frac{(3^r+1)^2}{3^{2r}+1} \ge (1+1/a)\frac{(b+1)^2}{b^2+1}.$$

Thus, it suffices to prove that $$(1+1/a)\frac{(b+1)^2}{b^2+1} > \frac{12}{23} + \frac{2}{r},$$ or (equivalently) \begin{align*} &34117281\,{r}^{10}-464867667\,{r}^{9}+3045114081\,{r}^{8} -12404084751 \,{r}^{7}\\ &\qquad +33959433510\,{r}^{6}-63295263486\,{r}^{5}+77189075224\,{r}^{ 4}\\ &\qquad -53559438912\,{r}^{3}+10081316512\,{r}^{2}+10987214240\,r-5474906752\\ >{}& 0 \end{align*} which is true for all $$r\in [2, 3]$$.