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Question Number 69374 by mathmax by abdo last updated on 22/Sep/19

let S_n =Σ_(k=1) ^n (((−1)^k )/k) 1) calculate S_n interms of n 2) find lim_(n→+∞) S_n

$${let}\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{S}_{{n}} \:{interms}\:{of}\:{n} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} \\ $$

Commented by mathmax by abdo last updated on 15/Oct/19

1) S_n =W(−1) with w(x)=Σ_(k=1) ^n (x^k /k) w^′ (x) =Σ_(k=1) ^n x^(k−1) =Σ_(k=0) ^(n−1) x^k =((x^n −1)/(x−1)) (we suppose x≠1)⇒ w(x)=∫_0 ^x ((t^n −1)/(t−1))dt +c with c=w(0)=0 ⇒w(x)=∫_0 ^x (t^n /(t−1))dt−∫_0 ^x (dt/(t−1)) =∫_0 ^x (t^n /(t−1))dt −ln∣x−1∣⇒S_n =w(−1)=∫_0 ^(−1) (t^n /(t−1))dt−ln(2) t=−u give ∫_0 ^(−1) (t^n /(t−1))dt =∫_0 ^1 (((−u)^n )/(−u−1))(−du) =(−1)^n ∫_0 ^1 (u^n /(1+u))du ⇒ S_n =(−1)^n ∫_0 ^1 (u^n /(1+u))du−ln(2) 2)we have ∣ S_n +ln(2)∣ =∫_0 ^1 (u^n /(1+u))du≤∫_0 ^1 u^n du =(1/(n+1)) ⇒ lim_(n→+∞) ∣S_n +ln(2)∣=0 ⇒lim_(n→+∞) S_n =−ln(2).

$$\left.\mathrm{1}\right)\:{S}_{{n}} ={W}\left(−\mathrm{1}\right)\:{with}\:{w}\left({x}\right)=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{{x}^{{k}} }{{k}} \\ $$$${w}^{'} \left({x}\right)\:=\sum_{{k}=\mathrm{1}} ^{{n}} \:{x}^{{k}−\mathrm{1}} =\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{x}^{{k}} \:=\frac{{x}^{{n}} −\mathrm{1}}{{x}−\mathrm{1}}\:\:\left({we}\:{suppose}\:{x}\neq\mathrm{1}\right)\Rightarrow \\ $$$${w}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \:\frac{{t}^{{n}} −\mathrm{1}}{{t}−\mathrm{1}}{dt}\:+{c}\:\:{with}\:{c}={w}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow{w}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \:\frac{{t}^{{n}} }{{t}−\mathrm{1}}{dt}−\int_{\mathrm{0}} ^{{x}} \:\frac{{dt}}{{t}−\mathrm{1}} \\ $$$$=\int_{\mathrm{0}} ^{{x}} \:\frac{{t}^{{n}} }{{t}−\mathrm{1}}{dt}\:−{ln}\mid{x}−\mathrm{1}\mid\Rightarrow{S}_{{n}} ={w}\left(−\mathrm{1}\right)=\int_{\mathrm{0}} ^{−\mathrm{1}} \:\frac{{t}^{{n}} }{{t}−\mathrm{1}}{dt}−{ln}\left(\mathrm{2}\right) \\ $$$${t}=−{u}\:{give}\:\int_{\mathrm{0}} ^{−\mathrm{1}} \:\frac{{t}^{{n}} }{{t}−\mathrm{1}}{dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\left(−{u}\right)^{{n}} }{−{u}−\mathrm{1}}\left(−{du}\right)\:=\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{u}^{{n}} }{\mathrm{1}+{u}}{du}\:\Rightarrow \\ $$$${S}_{{n}} =\left(−\mathrm{1}\right)^{{n}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{u}^{{n}} }{\mathrm{1}+{u}}{du}−{ln}\left(\mathrm{2}\right) \\ $$$$\left.\mathrm{2}\right){we}\:{have}\:\mid\:{S}_{{n}} +{ln}\left(\mathrm{2}\right)\mid\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{u}^{{n}} }{\mathrm{1}+{u}}{du}\leqslant\int_{\mathrm{0}} ^{\mathrm{1}} \:{u}^{{n}} \:{du}\:=\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \mid{S}_{{n}} +{ln}\left(\mathrm{2}\right)\mid=\mathrm{0}\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =−{ln}\left(\mathrm{2}\right). \\ $$$$ \\ $$

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