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

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} }{{n}^{{k}} }={S}_{{k}} \:\:\:\:\:\:\:{H}_{{q}} =\underset{{p}=\mathrm{1}} {\overset{{q}} {\sum}}\frac{\mathrm{1}}{{p}} \\ $$$${Is}\:{there}\:{a}\:{simple}\:{from}\:{for}\:{S}_{{k}} \\ $$

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