# On evaluating the Riemann zeta function, including that $\zeta(2)\gt \varphi$ where $\varphi$ is the golden ratio

A week ago, I got the following :

For a positive integer $k$, using Cauchy–Schwarz inequality, $$\left(\sum_{n=1}^{\infty}\frac{1}{n^k(n+1)^k}\right)^2\lt \left(\sum_{n=1}^{\infty}\frac{1}{n^{2k}}\right)\left(\sum_{n=1}^{\infty}\frac{1}{(n+1)^{2k}}\right)=\zeta(2k)(\zeta(2k)-1),$$ i.e. $$\zeta(2k)^2-\zeta(2k)-\left(\sum_{n=1}^{\infty}\frac{1}{n^k(n+1)^k}\right)^2\gt 0.$$ So, $$\zeta(2k)\gt \dfrac{1+\sqrt{1+4\left(\sum_{n=1}^{\infty}\dfrac{1}{n^k(n+1)^k}\right)^2}}{2}.$$

From this, we can have the followings :

$$\zeta(2)\gt \varphi,\qquad\zeta(4)\gt \dfrac{1+\sqrt{1+4\left(\dfrac{\pi^2}{3}-3\right)^2}}{2},\qquad \zeta(6)\gt \dfrac{1+\sqrt{1+4\left(10-\pi^2\right)^2}}{2}$$where $\varphi=\frac{1+\sqrt 5}{2}$ is the golden ratio.

Now let us define a sequence $\{a_k\}$ as $$a_k=\zeta(2k)-\dfrac{1+\sqrt{1+4\left(\sum_{n=1}^{\infty}\dfrac{1}{n^k(n+1)^k}\right)^2}}{2}$$

Then, it seems that $\{a_k\}$ is decreasing : $$a_1\approx 0.0269,\quad a_2\approx 0.0044,\quad a_3\approx 0.0006.$$ But I've been facing difficulty in proving that.

Question : Is $\{a_k\}$ decreasing? If so, how can we prove that?

• Those inequalities for $\zeta(2k)$ are beautiful... Commented Jun 21, 2018 at 14:56

Let's look at the first few terms of each side of

$$\zeta(2k) \gt \dfrac{1+\sqrt{1+4\left(\sum_{n=1}^{\infty}\dfrac{1}{n^k(n+1)^k}\right)^2}}{2}.$$

$\zeta(2k) \approx 1+\frac1{4^k}+\frac1{9^k}$ and $\sum_{n=1}^{\infty}\dfrac{1}{n^k(n+1)^k} \approx \frac1{2^k}+\frac1{6^k}$ so, since $\sqrt{1+x} \approx 1+\frac{x}{2}-\frac{x^2}{8}$,

$\begin{array}\\ 1+\sqrt{1+4\left(\sum_{n=1}^{\infty}\dfrac{1}{n^k(n+1)^k}\right)^2} &\approx 1+\sqrt{1+4\left(\frac1{2^k}+\frac1{6^k}\right)^2}\\ &\approx 1+\left(1+2\left(\frac1{2^k}+\frac1{6^k}\right)^2 -2\left(\frac1{2^k}+\frac1{6^k}\right)^4\right)\\ &\approx 2+\left(2\left(\frac1{4^k}+\frac{2}{12^k}\right) -2\frac1{16^k}\left(1+\frac1{3^k}\right)^4\right)\\ &\approx 2+2\left(\frac1{4^k}+\frac{1}{12^k}-\frac1{16^k}\right)\\ \end{array}$

so the inequality becomes, approximately,

$1+\frac1{4^k}+\frac1{9^k} > 1+\frac1{4^k}+\frac1{12^k}-\frac1{16^k}$ so the difference of the two sides is approximately $\frac1{9^k}-\frac1{12^k}+\frac1{16^k}$, and this is decreasing.

Computationally, using just these first few terms is not to accurate for small $k$. This difference is, for $k=1,2,3$,

${0.0902778, 0.00930748, 0.00103718}$.