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Question Number 218651 by hardmath last updated on 13/Apr/25

Answered by vnm last updated on 14/Apr/25

$$=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-4\cdot \frac{1}{4})+$$$$(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}-4\cdot \frac{1}{8})+$$$$...+$$$$(\frac{1}{4n-3}+\frac{1}{4n-2}+\frac{1}{4n-1}+\frac{1}{4n}-4\cdot \frac{1}{4n})=$$$$\underset{k=1}{\sum ^{4n}}\frac{1}{k}-4\underset{k=1}{\sum ^n}\frac{1}{4k}=\underset{k=1}{\sum ^{4n}}\frac{1}{k}-\underset{k=1}{\sum ^n}\frac{1}{k}=$$$$\underset{k=n+1}{\sum ^{4n}}\frac{1}{k}$$

Commented by hardmath last updated on 17/Apr/25

$$\mathrm{thank}\mathrm{you}\mathrm{my}\mathrm{dear}\mathrm{professor}$$

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