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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+(1/2)+(1/3)+(1/4)−4∙(1/4))+ ((1/5)+(1/6)+(1/7)+(1/8)−4∙(1/8))+ ...+ ((1/(4n−3))+(1/(4n−2))+(1/(4n−1))+(1/(4n))−4∙(1/(4n)))= Σ_(k=1) ^(4n) (1/k)−4Σ_(k=1) ^n (1/(4k))=Σ_(k=1) ^(4n) (1/k)−Σ_(k=1) ^n (1/k)= Σ_(k=n+1) ^(4n) (1/k)

$$=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{4}\centerdot\frac{\mathrm{1}}{\mathrm{4}}\right)+ \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{8}}−\mathrm{4}\centerdot\frac{\mathrm{1}}{\mathrm{8}}\right)+ \\ $$$$...+ \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{4}{n}−\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}{n}−\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}{n}−\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{4}{n}}−\mathrm{4}\centerdot\frac{\mathrm{1}}{\mathrm{4}{n}}\right)= \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}{n}} {\sum}}\frac{\mathrm{1}}{{k}}−\mathrm{4}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{4}{k}}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}{n}} {\sum}}\frac{\mathrm{1}}{{k}}−\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}}= \\ $$$$\underset{{k}={n}+\mathrm{1}} {\overset{\mathrm{4}{n}} {\sum}}\frac{\mathrm{1}}{{k}} \\ $$

Commented by hardmath last updated on 17/Apr/25

thank you my dear professor

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

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