Prove the sequence $C_n$ is decreasing. My question is: Prove that the following sequence is decreasing.
\begin{align}
‎‎C_n = ‎\frac{2(2^{2n}-1)}{3‎\times‎2^{2n}} ‎\zeta (2n)
\end{align}
where $\zeta$ is the Riemann zeta function.
I know that, I should show $\frac{C_n}{C_{n+1}}‎‎>‎1$ for every $n$. To check this, I did as follows
\begin{align}
\frac{‎\frac{2(2^{2n}-1)}{3‎\times‎2^{2n}} ‎\zeta (2n)}{‎\frac{2(2^{2(n+1)}-1)}{3‎\times‎2^{2(n+1)}} ‎\zeta (2(n+1))},
\end{align}
But I could not achieve the result. Also, I know that $\lim_{n\to\infty}C_n=\frac{2}{3}$. Anyone can help me. Thanks in advance.
 A: Hint: Use the definition $$\sum_{k = 1}^{\infty}\dfrac{1}{k^{2n}} = \zeta(2n)$$ to show that $$\sum_{k = 0}^{\infty}\dfrac{1}{(2k+1)^{2n}} = \dfrac{2^{2n}-1}{2^{2n}}\zeta(2n) =  \dfrac{3}{2}C_n. $$
Once you do this, it is easy to see that $C_n$ is decreasing since each term of that summation is decreasing w.r.t. $n$ and the sum is convergent for all integers $n \ge 1$.
A: Try to use
$$\zeta(2n)=(-1)^{n+1}\frac{ 2^{2 n-1} \pi ^{2 n}}{(2 n)!} B_{2 n}$$ makes
$$\frac{C_n}{C_{n+1}}=-\frac{2 \left(4^n-1\right) (n+1) (2 n+1) B_{2 n}}{\pi ^2 \left(4^{n+1}-1\right)
   B_{2 n+2}}$$
$$B_{2 n} \sim (-1)^{n+1}4 \sqrt{\pi n }   \left(\frac n {\pi e} \right)^{2 n}
$$
A: Can you use the fact that $\lim_{n \to \infty} \zeta (n) = 1$ when restricting to $(1, \infty)$? If so,
\begin{align}
‎‎\lim_{n \to \infty} C_n = \lim_{n \to \infty} ‎\frac{2(2^{2n}-1)}{3‎\times‎2^{2n}} ‎\zeta (2n) = \lim_{n \to \infty} ‎\frac{2(2^{2n}-1)}{3‎\times‎2^{2n}} \times \lim_{m \to \infty} ‎\zeta (2m) = \lim_{n \to \infty} ‎\frac{2(2^{2n}-1)}{3‎\times‎2^{2n}} = \frac{2}{3}
\end{align}
In this case, looking at $\lim_{n \to \infty} \frac{C_n}{C_{n+1}} = 1$, which won't help you. The steps and the last claim should be on the level of Calculus 1, so I skipped them.
