Question Number 40889 by abdo.msup.com last updated on 28/Jul/18
$${prove}?{that} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}−\left(\mathrm{1}−{t}\right)^{{n}} }{{t}}{dt}\:=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}} \\ $$
Answered by math khazana by abdo last updated on 30/Jul/18
![we have 1−x^n =(1−x)(1+x+x^2 +...+x^(n−1) )⇒ ((1−(1−t)^n )/t) =((t(1+(1−t)+(1−t)^2 +...+(1−t)^(n−1) )/t) ∫_0 ^1 ((1−(1−t))/t)dt =∫_0 ^1 Σ_(k=0) ^(n−1) (1−t)^k dt =Σ_(k=0) ^(n−1) ∫_0 ^1 (1−t)^k dt =Σ_(k=0) ^n [−(1/(k+1))(1−t)^(k+1) ]_0 ^1 =Σ_(k=0) ^(n−1) (1/(k+1)) =Σ_(k=1) ^n (1/k) (=H_n ).](https://www.tinkutara.com/question/Q40951.png)
$${we}\:{have}\:\mathrm{1}−{x}^{{n}} =\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \:+…+{x}^{{n}−\mathrm{1}} \right)\Rightarrow \\ $$$$\frac{\mathrm{1}−\left(\mathrm{1}−{t}\right)^{{n}} }{{t}}\:=\frac{{t}\left(\mathrm{1}+\left(\mathrm{1}−{t}\right)+\left(\mathrm{1}−{t}\right)^{\mathrm{2}} +…+\left(\mathrm{1}−{t}\right)^{{n}−\mathrm{1}} \right.}{{t}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}−\left(\mathrm{1}−{t}\right)}{{t}}{dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{{k}} {dt} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}\right)^{{k}} {dt}\:=\sum_{{k}=\mathrm{0}} ^{{n}} \left[−\frac{\mathrm{1}}{{k}+\mathrm{1}}\left(\mathrm{1}−{t}\right)^{{k}+\mathrm{1}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:\:\left(={H}_{{n}} \right). \\ $$