Prove $\tan x > \sum\limits_{ n=1 }^{ \infty } \frac{x^{2n+1}}{4^n -1}$ for $ 0

Question

Prove $\tan x > \sum\limits_{ n=1 }^{ \infty } \frac{x^{2n+1}}{4^n -1}$ for $ 0

Answer 1

Oh, that is very nice. We may notice that $\tan x$ is related with the logarithmic derivative of the cosine function. Since: $$ \cos(x) = \prod_{n\geq 0}\left(1-\frac{4x^2}{(2n+1)^2\pi^2}\right)\tag{1}$$ by the Weierstrass product, by applying $\frac{d}{dx}\log(\cdot)$ to both sides we get:

$$ \tan x = \sum_{n\geq 0}\frac{8x}{\pi^2(2n+1)^2-4x^2}\tag{2} $$ and by expanding the terms in the RHS of $(2)$ as geometric series, $$ \frac{\tan x}{2x} = \sum_{n\geq 0}\left(\frac{4}{\pi^2(2n+1)^2}+\frac{4^2}{\pi^4(2n+1)^4}x^2+\ldots\right)=\sum_{m\geq 1}(4^m-1)\frac{\zeta(2m)}{\pi^{2m}}x^{2m-2}\tag{3} $$ so we may prove the given inequality by simply exploiting $\zeta(2m)\geq 1$.