Question Number 206858 by Ghisom last updated on 27/Apr/24
$$\mathrm{prove}\:\mathrm{that} \\ $$$${H}_{{n}} =\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{{t}^{{n}} −\mathrm{1}}{{t}−\mathrm{1}}{dt} \\ $$
Answered by mathzup last updated on 28/Apr/24
![I_ξ =∫_ξ ^1 ((t^n −1)/(t−1))dt we have H_n =lim_(ξ→0^+ ) I_ξ but I_ξ =∫_ξ ^1 (((t−1)(1+t+t^2 +....+t^(n−1) ))/(t−1))dt =∫_ξ ^1 (1+t+t^2 +....+t^(n−1) )dt =[t+(t^2 /2)+(t^3 /3)+....+(t^n /n)]_ξ ^1 =1+(1/2)+(1/3)+....+(1/n)−(ξ+(ξ^2 /2) +....+(ξ^n /n)) ⇒lim_(ξ→0) I_ξ =1+(1/2)+(1/3)+....+(1/n)=H_n](https://www.tinkutara.com/question/Q206860.png)
$${I}_{\xi} =\int_{\xi} ^{\mathrm{1}} \frac{{t}^{{n}} −\mathrm{1}}{{t}−\mathrm{1}}{dt}\:\:{we}\:{have}\:{H}_{{n}} ={lim}_{\xi\rightarrow\mathrm{0}^{+} } \:{I}_{\xi} \\ $$$${but}\:{I}_{\xi} =\int_{\xi} ^{\mathrm{1}} \frac{\left({t}−\mathrm{1}\right)\left(\mathrm{1}+{t}+{t}^{\mathrm{2}} +….+{t}^{{n}−\mathrm{1}} \right)}{{t}−\mathrm{1}}{dt} \\ $$$$=\int_{\xi} ^{\mathrm{1}} \left(\mathrm{1}+{t}+{t}^{\mathrm{2}} +….+{t}^{{n}−\mathrm{1}} \right){dt} \\ $$$$=\left[{t}+\frac{{t}^{\mathrm{2}} }{\mathrm{2}}+\frac{{t}^{\mathrm{3}} }{\mathrm{3}}+….+\frac{{t}^{{n}} }{{n}}\right]_{\xi} ^{\mathrm{1}} \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+….+\frac{\mathrm{1}}{{n}}−\left(\xi+\frac{\xi^{\mathrm{2}} }{\mathrm{2}}\:+….+\frac{\xi^{{n}} }{{n}}\right) \\ $$$$\Rightarrow{lim}_{\xi\rightarrow\mathrm{0}} \:\:{I}_{\xi} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+….+\frac{\mathrm{1}}{{n}}={H}_{{n}} \\ $$
Commented by Ghisom last updated on 28/Apr/24
thank you
Answered by JDamian last updated on 28/Apr/24
![((t^n −1)/(t−1))=1+t+t^2 + ∙∙∙ +t^(n−1) G.P. ∫_0 ^1 1+t+t^2 + ∙∙∙ +t^(n−1) dt= = [t+(t^2 /2)+(t^3 /3)+ ∙∙∙ +(t^n /n)]_0 ^1 = =(1+(1/2)+(1/3)+ ∙∙∙ +(1/n))−(0+0+ ∙∙∙ +0)= = 1+(1/2)+(1/3)+ ∙∙∙ +(1/n) = H_n ■](https://www.tinkutara.com/question/Q206864.png)
$$\frac{{t}^{{n}} −\mathrm{1}}{{t}−\mathrm{1}}=\mathrm{1}+{t}+{t}^{\mathrm{2}} +\:\centerdot\centerdot\centerdot\:+{t}^{{n}−\mathrm{1}} \:\:\:{G}.{P}. \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\mathrm{1}+{t}+{t}^{\mathrm{2}} +\:\centerdot\centerdot\centerdot\:+{t}^{{n}−\mathrm{1}} \:{dt}= \\ $$$$=\:\left[{t}+\frac{{t}^{\mathrm{2}} }{\mathrm{2}}+\frac{{t}^{\mathrm{3}} }{\mathrm{3}}+\:\centerdot\centerdot\centerdot\:+\frac{{t}^{{n}} }{{n}}\right]_{\mathrm{0}} ^{\mathrm{1}} = \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\:\centerdot\centerdot\centerdot\:+\frac{\mathrm{1}}{{n}}\right)−\left(\mathrm{0}+\mathrm{0}+\:\centerdot\centerdot\centerdot\:+\mathrm{0}\right)= \\ $$$$=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\:\centerdot\centerdot\centerdot\:+\frac{\mathrm{1}}{{n}}\:=\:\mathrm{H}_{{n}} \:\:\:\blacksquare \\ $$
Commented by Ghisom last updated on 28/Apr/24
thank you