Question Number 30214 by abdo imad last updated on 18/Feb/18
$${study}\:{the}\:{convergence}\:{of}\:{u}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \:\:\frac{{C}_{{n}} ^{{k}} }{{k}} \\ $$$${for}\:{that}\:{use}\:{H}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:. \\ $$
Commented by prof Abdo imad last updated on 22/Feb/18
![let condider p(x)= Σ_(k=1) ^n (−1)^(k−1) (C_n ^k /k) x^k we have p^′ (x)= Σ_(k=1) ^n C_n ^k (−1)^(k−1) x^(k−1) =((−1)/x)Σ_(k=1) ^n C_n ^k (−1)^k x^k =−(1/x)(Σ_(k=0) ^n (−1)^k x^k −1) =(1/x)( 1−(1−x)^n )=((1−(1−x)^n )/x) ⇒ p(x)= ∫_0 ^x ((1−(1−t)^n )/t)dt +λ but λ=p(0)=0⇒ p(x)= ∫_0 ^x ((1−(1−t)^n )/t)dt and u_n =p(1)⇒ u_n = ∫_0 ^1 ((1−(1−t)^n )/t)dt = = ∫_0 ^1 ((1+(1−t) +(1−t)^2 +....(1−t)^(n−1) )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) [((−1)/(k+1))(1−t)^(k+1) ]_0 ^1 =Σ_(k=0) ^(n−1) (1/(k+1)) = H_n but H_n ∼ln(n) for n→∞ ⇒lim_(n→∞) u_n =+∞.](https://www.tinkutara.com/question/Q30446.png)
$${let}\:{condider}\:\:{p}\left({x}\right)=\:\sum_{{k}=\mathrm{1}} ^{{n}} \left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \frac{{C}_{{n}} ^{{k}} }{{k}}\:{x}^{{k}} \:{we}\:{have} \\ $$$${p}^{'} \left({x}\right)=\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \:{x}^{{k}−\mathrm{1}} \\ $$$$=\frac{−\mathrm{1}}{{x}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \:\left(−\mathrm{1}\right)^{{k}} \:{x}^{{k}} \:=−\frac{\mathrm{1}}{{x}}\left(\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {x}^{{k}} \:−\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{{x}}\left(\:\mathrm{1}−\left(\mathrm{1}−{x}\right)^{{n}} \right)=\frac{\mathrm{1}−\left(\mathrm{1}−{x}\right)^{{n}} }{{x}}\:\Rightarrow \\ $$$${p}\left({x}\right)=\:\int_{\mathrm{0}} ^{{x}} \:\:\:\:\frac{\mathrm{1}−\left(\mathrm{1}−{t}\right)^{{n}} }{{t}}{dt}\:+\lambda\:\:{but}\:\lambda={p}\left(\mathrm{0}\right)=\mathrm{0}\Rightarrow \\ $$$${p}\left({x}\right)=\:\int_{\mathrm{0}} ^{{x}} \:\:\frac{\mathrm{1}−\left(\mathrm{1}−{t}\right)^{{n}} }{{t}}{dt}\:\:{and}\:{u}_{{n}} ={p}\left(\mathrm{1}\right)\Rightarrow \\ $$$${u}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{1}−\left(\mathrm{1}−{t}\right)^{{n}} }{{t}}{dt}\:= \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\left(\mathrm{1}+\left(\mathrm{1}−{t}\right)\:+\left(\mathrm{1}−{t}\right)^{\mathrm{2}} \:+….\left(\mathrm{1}−{t}\right)^{{n}−\mathrm{1}} \right){dt}\right. \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{{k}} {dt}=\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\mathrm{1}−{t}\right)^{{k}} {dt} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left[\frac{−\mathrm{1}}{{k}+\mathrm{1}}\left(\mathrm{1}−{t}\right)^{{k}+\mathrm{1}} \right]_{\mathrm{0}} ^{\mathrm{1}} =\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:=\:{H}_{{n}} \:{but} \\ $$$${H}_{{n}} \:\sim{ln}\left({n}\right)\:{for}\:{n}\rightarrow\infty\:\:\Rightarrow{lim}_{{n}\rightarrow\infty} {u}_{{n}} =+\infty. \\ $$