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?


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}
$.
