n-1-H-n-n-k-S-k-H-q-p-1-q-1-p-Is-there-a-simple-from-for-S-k-

Question Number 90038 by frc2crc last updated on 21/Apr/20

\underset{n = 1}{\sum^{\infty}} \frac{H_{n}}{n^{k}} = S_{k} H_{q} = \underset{p = 1}{\sum^{q}} \frac{1}{p}

I s t h e r e a s i m p l e f r o m f o r S_{k}