Question-149378

Question Number 149378 by fotosy2k last updated on 05/Aug/21

Answered by Ar Brandon last updated on 05/Aug/21

S=ψ(4)−ψ(1)=(1/3)+(1/2)+1+ψ(1)−ψ(1)=((11)/6) S=(1−(1/4))+((1/2)−(1/5))+((1/3)−(1/6))+((1/4)−(1/7))+∙∙∙+ =1+(1/2)+(1/3)=((11)/6)

$${S}=\psi\left(\mathrm{4}\right)−\psi\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}+\psi\left(\mathrm{1}\right)−\psi\left(\mathrm{1}\right)=\frac{\mathrm{11}}{\mathrm{6}} \\ $$$${S}=\left(\mathrm{1}−\cancel{\frac{\mathrm{1}}{\mathrm{4}}}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}}−\cancel{\frac{\mathrm{1}}{\mathrm{5}}}\right)+\left(\frac{\mathrm{1}}{\mathrm{3}}−\cancel{\frac{\mathrm{1}}{\mathrm{6}}}\right)+\left(\cancel{\frac{\mathrm{1}}{\mathrm{4}}}−\cancel{\frac{\mathrm{1}}{\mathrm{7}}}\right)+\centerdot\centerdot\centerdot+ \\ $$$$\:\:\:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{11}}{\mathrm{6}} \\ $$

Answered by gsk2684 last updated on 13/Sep/21

(1−(1/4))+((1/2)−(1/5))+((1/3)−(1/6))+ ((1/4)−(1/7))+((1/5)−(1/8))+((1/6)−(1/9))+ ((1/7)−(1/(10)))+((1/8)−(1/(11)))+((1/9)−(1/(12)))+... =1+(1/2)+(1/3)

$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{5}}\right)+\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{6}}\right)+ \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{7}}\right)+\left(\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{8}}\right)+\left(\frac{\mathrm{1}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{9}}\right)+ \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{7}}−\frac{\mathrm{1}}{\mathrm{10}}\right)+\left(\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{11}}\right)+\left(\frac{\mathrm{1}}{\mathrm{9}}−\frac{\mathrm{1}}{\mathrm{12}}\right)+… \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Answered by mathmax by abdo last updated on 05/Aug/21

S_n =Σ_(k=1) ^n ((1/k)−(1/(k+3))) ⇒S_n =Σ_(k=1) ^n (1/k)−Σ_(k=1) ^n (1/(k+3)) Σ_(k=1) ^n (1/k)=H_n ,Σ_(k=1) ^n (1/(k+3))=Σ_(k=4) ^(n+1) (1/k) =Σ_(k=1) ^n (1/k)+(1/(n+1))−1−(1/2)−(1/3)=H_n +(1/(n+1))−(3/2)−(1/3)=H_n +(1/(n+1))−((11)/6) ⇒S_n =H_n −H_n −(1/(n+1))+((11)/6)=((11)/6)−(1/(n+1)) ⇒ Σ_(n=1) ^∞ ((1/n)−(1/(n+3)))=((11)/6)

$$\mathrm{S}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left(\frac{\mathrm{1}}{\mathrm{k}}−\frac{\mathrm{1}}{\mathrm{k}+\mathrm{3}}\right)\:\Rightarrow\mathrm{S}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}}−\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}+\mathrm{3}} \\ $$$$\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}}=\mathrm{H}_{\mathrm{n}} ,\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}+\mathrm{3}}=\sum_{\mathrm{k}=\mathrm{4}} ^{\mathrm{n}+\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{k}} \\ $$$$=\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}}+\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{H}_{\mathrm{n}} +\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}−\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{H}_{\mathrm{n}} +\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}−\frac{\mathrm{11}}{\mathrm{6}} \\ $$$$\Rightarrow\mathrm{S}_{\mathrm{n}} =\mathrm{H}_{\mathrm{n}} −\mathrm{H}_{\mathrm{n}} −\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}+\frac{\mathrm{11}}{\mathrm{6}}=\frac{\mathrm{11}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\:\Rightarrow \\ $$$$\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{n}}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{3}}\right)=\frac{\mathrm{11}}{\mathrm{6}} \\ $$

Commented by fotosy2k last updated on 05/Aug/21

$${thank}\:{you} \\ $$

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