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Question Number 172309 by mathocean1 last updated on 25/Jun/22

Determinate lim_(n→+∞) (Σ_(k=1) ^n (((−1)^(k+1) )/k))

$${Determinate}\: \\ $$$$\underset{{n}\rightarrow+\infty} {{lim}}\left(\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} }{{k}}\right) \\ $$

Commented by mr W last updated on 25/Jun/22

ln (1+x)=Σ_(k=1) ^∞ (((−1)^(k+1) x^k )/k) with x=1: ln 2=Σ_(k=1) ^∞ (((−1)^(k+1) )/k)

$$\mathrm{ln}\:\left(\mathrm{1}+{x}\right)=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} {x}^{{k}} }{{k}} \\ $$$${with}\:{x}=\mathrm{1}: \\ $$$$\mathrm{ln}\:\mathrm{2}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} }{{k}} \\ $$

Answered by Mathspace last updated on 25/Jun/22

u_n =Σ_(k=1) ^n (((−1)^(k−1) )/k) and p(x)=Σ_(k=1) ^n (((−1)^(k−1) )/k)x^k p^′ (x)=Σ_(k=1) ^n (−1)^(k−1) x^(k−1) =Σ_(k=1) ^n (−x)^(k−1) =Σ_(k=0) ^(n−1) (−x)^k =((1−(−x)^n )/(1+x)) ⇒ p(x)=∫_0 ^x ((1−(−t)^n )/(1+t)) +c c=p(0)=0 ⇒ p(x)=∫_0 ^x ((1−(−t)^n )/(1+t))dt ⇒p(1)=∫_0 ^1 (dt/(1+t)) −(−1)^n ∫_0 ^1 (t^n /(1+t))dt lim_(n→+∞) u_n =lim_(n→+∞) p(1) =ln2−lim_(n→+∞) (−1)^n ∫_0 ^1 (t^n /(1+t))dt =ln2−0 =ln2

$${u}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}} \\ $$$${and}\:{p}\left({x}\right)=\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}}{x}^{{k}} \\ $$$${p}^{'} \left({x}\right)=\sum_{{k}=\mathrm{1}} ^{{n}} \left(−\mathrm{1}\right)^{{k}−\mathrm{1}} {x}^{{k}−\mathrm{1}} \\ $$$$=\sum_{{k}=\mathrm{1}} ^{{n}} \left(−{x}\right)^{{k}−\mathrm{1}} =\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(−{x}\right)^{{k}} \\ $$$$=\frac{\mathrm{1}−\left(−{x}\right)^{{n}} }{\mathrm{1}+{x}}\:\Rightarrow \\ $$$${p}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \frac{\mathrm{1}−\left(−{t}\right)^{{n}} }{\mathrm{1}+{t}}\:+{c} \\ $$$${c}={p}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow \\ $$$${p}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \frac{\mathrm{1}−\left(−{t}\right)^{{n}} }{\mathrm{1}+{t}}{dt} \\ $$$$\Rightarrow{p}\left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\mathrm{1}+{t}}\:−\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{n}} }{\mathrm{1}+{t}}{dt} \\ $$$${lim}_{{n}\rightarrow+\infty} {u}_{{n}} ={lim}_{{n}\rightarrow+\infty} {p}\left(\mathrm{1}\right) \\ $$$$={ln}\mathrm{2}−{lim}_{{n}\rightarrow+\infty} \left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{n}} }{\mathrm{1}+{t}}{dt} \\ $$$$={ln}\mathrm{2}−\mathrm{0}\:={ln}\mathrm{2} \\ $$

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