Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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!
$f(x) = 1/x^4$ 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 $$ 2 \sum_{k=n+1}^\infty \left(\frac{1}{(2n)^4} - \frac{1}{(2n+1)^4} \right) < \sum_{k=n+1}^\infty \left(\frac{1}{(2n-1)^4} - \frac{1}{(2n+1)^4} \right) = \frac{1}{(2k+1)^4} \, . $$
$$\Delta_k=\frac{1}{(2k+1)^4} - 2 \sum_{n=k+1}^\infty \left(\frac{1}{(2n)^4} - \frac{1}{(2n+1)^4}\right)$$
$$\sum_{n=k+1}^\infty \frac{1}{(2n)^4}=\frac{1}{96} \psi ^{(3)}(k+1)\qquad \qquad\sum_{n=k+1}^\infty \frac{1}{(2n+1)^4}=\frac{1}{96} \psi ^{(3)}\left(k+\frac{3}{2}\right)$$ Using asymptotics $$\Delta_k=\frac 1 {16k^5} \left(1-\frac{5}{2 k}+\frac{25}{8 k^2}+O\left(\frac{1}{k^3}\right)\right)$$
