Proving that $\frac{1}{(2k+1)^4} - 2 \sum_{n=k+1}^\infty \left(\frac{1}{(2n)^4} - \frac{1}{(2n+1)^4}\right)$ is non negative

Prove that the sequence

$$s_k = \frac{1}{(2k+1)^4} - 2 \sum_{n=k+1}^\infty \left(\frac{1}{(2n)^4} - \frac{1}{(2n+1)^4}\right)$$ is non-negative.

I would appreciate an elementary proof. I tried using series / integral comparison without success.

This question is a follow up of this one. There is an answer to that question that uses the integral representation of Hurwitz zeta function, that I'd like to avoid... if possible!

• Just at first glance, have you tried factoring the $2n$ from the denominator of the right-hand term and using derivatives of geometric series to obtain some bounds? I haven’t worked through it but would first attempt it this way. (I’m guessing you have tried it since you mention attempts involving series, but I don’t know what exactly you’ve tried so leaving a comment) Sep 20 '20 at 17:30

$$f(x) = 1/x^4$$ is strictly convex, so that $$\frac{1}{(2n)^4} < \frac 12 \left(\frac{1}{(2n-1)^4} + \frac{1}{(2n+1)^4} \right)$$ or $$\frac{1}{(2n)^4} - \frac{1}{(2n+1)^4} < \frac 12 \left(\frac{1}{(2n-1)^4} - \frac{1}{(2n+1)^4} \right)$$ It follows that $$2 \sum_{k=n+1}^\infty \left(\frac{1}{(2n)^4} - \frac{1}{(2n+1)^4} \right) < \sum_{k=n+1}^\infty \left(\frac{1}{(2n-1)^4} - \frac{1}{(2n+1)^4} \right) = \frac{1}{(2k+1)^4} \, .$$
• I think you made a typo - $\frac{1}{(2n)^2}$ should be $\frac{1}{(2n)^4}$. Sep 20 '20 at 20:04
$$\Delta_k=\frac{1}{(2k+1)^4} - 2 \sum_{n=k+1}^\infty \left(\frac{1}{(2n)^4} - \frac{1}{(2n+1)^4}\right)$$
$$\sum_{n=k+1}^\infty \frac{1}{(2n)^4}=\frac{1}{96} \psi ^{(3)}(k+1)\qquad \qquad\sum_{n=k+1}^\infty \frac{1}{(2n+1)^4}=\frac{1}{96} \psi ^{(3)}\left(k+\frac{3}{2}\right)$$ Using asymptotics $$\Delta_k=\frac 1 {16k^5} \left(1-\frac{5}{2 k}+\frac{25}{8 k^2}+O\left(\frac{1}{k^3}\right)\right)$$