@DanielFischer re :
this shouldn't it be sufficient to prove that there is a constant $c$, such that $\displaystyle \sum\limits_{k=n+1}^{\infty} \dfrac{1}{2^{k(k+1)}} \le \dfrac{c}{2^{2n(n+1)}}$ for all $n \ge N$ to prove that the number $\displaystyle \alpha = \sum\limits_{k=0}^{\infty} \dfrac{1}{2^{k(k+1)}}$ is irrational ?
Find the closed form of $$\sum_{n=1}^{\infty} \left(\frac{H_n}{n}\right)^4$$
I know the closed form for smaller powers like $2, 3$ exists, but I'm not sure if there is a
closed form for this variant. Is it possible to tackle the question in an elementary way
and find the answer, without using i...
