advanced-calculus-evaluate-n-1-H-n-2-n-2-where-H-n-k-1-n-1-k-harmonic-number-

Question Number 121774 by mnjuly1970 last updated on 11/Nov/20

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

e v a l u a t e :

\underset{n = 1}{\sum^{\infty}} \frac{{(H_{n})}^{2}}{n^{2}} = ?

w h e r e H_{n} = \underset{k = 1}{\sum^{n}} \frac{1}{k} (h a r m o n i c n u m b e r)