advanced-calculus-prove-n-1-H-n-n-3-pi-4-72-note-H-n-1-1-2-1-n-

Question Number 123456 by mnjuly1970 last updated on 25/Nov/20

\dots a d v a n c e d c a l c u l u s \dots

p r o v e ::

ϕ = \underset{n = 1}{\sum^{\infty}} (\frac{H_{n}}{n^{3}}) \overset{? ? ?}{=} \frac{π^{4}}{72}

n o t e : H_{n} = 1 + \frac{1}{2} + \dots + \frac{1}{n}