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Question Number 83214 by mathmax by abdo last updated on 28/Feb/20

calculate U_n =Σ_(k=0) ^n (1/(3k+1)) interms of H_n =Σ_(k=1) ^n (1/k)

$${calculate}\:\:{U}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{3}{k}+\mathrm{1}}\:{interms}\:{of}\:{H}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}} \\ $$

Answered by ~blr237~ last updated on 29/Feb/20

let n≥1 H_(3n) =Σ_(k=1) ^(3n) (1/k) = Σ_(k=1) ^n (1/(3k)) +Σ_(k=0) ^(n−1) (1/(3k+1))+Σ_(k=0) ^(n−1) (1/(3k+2)) =(1/3)H_n +U_(n−1) +V_(n−1) (1) where V_n =Σ_(k=0) ^n (1/(3k+2)) H_(3n+1) =Σ_(k=1) ^n (1/(3k)) +Σ_(k=0) ^n (1/(3k+1))+Σ_(k=0) ^(n−1) (1/(3k+2)) = (1/3)H_n +U_n +V_(n−1) (2) (2)−(1) ⇒ U_n −U_(n−1) = H_(3n+1) −H_(3n) so Σ_(k=1) ^n (U_k −U_(k−1) )=Σ_(k=1) ^n (H_(3k+1) −H_(3k) ) Then U_n =U_0 +Σ_(k=1) ^n (H_(3k+1) −H_(3k) )

$${let}\:\:{n}\geqslant\mathrm{1}\:\: \\ $$$${H}_{\mathrm{3}{n}} =\underset{{k}=\mathrm{1}} {\overset{\mathrm{3}{n}} {\sum}}\frac{\mathrm{1}}{{k}}\:\:=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{3}{k}}\:+\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{3}{k}+\mathrm{1}}+\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{3}{k}+\mathrm{2}}\: \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{3}}{H}_{{n}} +{U}_{{n}−\mathrm{1}} +{V}_{{n}−\mathrm{1}} \:\:\:\:\left(\mathrm{1}\right)\:\:\:\:{where}\:\:{V}_{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{3}{k}+\mathrm{2}} \\ $$$${H}_{\mathrm{3}{n}+\mathrm{1}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{3}{k}}\:+\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{3}{k}+\mathrm{1}}+\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{3}{k}+\mathrm{2}} \\ $$$$\:\:\:=\:\frac{\mathrm{1}}{\mathrm{3}}{H}_{{n}} +{U}_{{n}} +{V}_{{n}−\mathrm{1}} \:\:\:\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{2}\right)−\left(\mathrm{1}\right)\:\Rightarrow\:{U}_{{n}} −{U}_{{n}−\mathrm{1}} =\:{H}_{\mathrm{3}{n}+\mathrm{1}} −{H}_{\mathrm{3}{n}} \\ $$$$\:\:{so}\:\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({U}_{{k}} −{U}_{{k}−\mathrm{1}} \right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({H}_{\mathrm{3}{k}+\mathrm{1}} −{H}_{\mathrm{3}{k}} \right) \\ $$$${Then}\:\:\:\:{U}_{{n}} ={U}_{\mathrm{0}} +\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({H}_{\mathrm{3}{k}+\mathrm{1}} −{H}_{\mathrm{3}{k}} \right) \\ $$

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