Question-186951

Question Number 186951 by Humble last updated on 12/Feb/23

Answered by integralmagic last updated on 12/Feb/23

$$ \\ $$$$={ln}\mathrm{2} \\ $$

Answered by SEKRET last updated on 12/Feb/23

(−1)∙Σ_(k=1) ^∞ (((−1)^k )/k) = ? −1=a f(a)= −Σ_(k=1) ^∞ (a^k /k) f ′ (a)= −Σ_(k=1) ^∞ a^(k−1) = 1+a+a^2 +a^3 +...a^n +.. f ′ (a)= (1/(1−a)) c=0 f(a)=ln(1−a) f(−1)= ln(2)

$$\:\:\left(−\mathrm{1}\right)\centerdot\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{\mathrm{k}}} }{\boldsymbol{\mathrm{k}}}\:\:\:\:\:=\:\:? \\ $$$$\:\:−\mathrm{1}=\boldsymbol{\mathrm{a}} \\ $$$$\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{a}}\right)=\:−\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\boldsymbol{\mathrm{a}}^{\boldsymbol{\mathrm{k}}} }{\boldsymbol{\mathrm{k}}}\: \\ $$$$\:\boldsymbol{\mathrm{f}}\:'\:\left(\boldsymbol{\mathrm{a}}\right)=\:−\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\boldsymbol{\mathrm{a}}^{\boldsymbol{\mathrm{k}}−\mathrm{1}} =\:\mathrm{1}+\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{a}}^{\mathrm{3}} +…\boldsymbol{\mathrm{a}}^{\boldsymbol{\mathrm{n}}} +.. \\ $$$$\:\boldsymbol{\mathrm{f}}\:'\:\left(\boldsymbol{\mathrm{a}}\right)=\:\frac{\mathrm{1}}{\mathrm{1}−\boldsymbol{\mathrm{a}}}\:\:\:\:\boldsymbol{\mathrm{c}}=\mathrm{0}\:\:\:\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{a}}\right)=\boldsymbol{\mathrm{ln}}\left(\mathrm{1}−\boldsymbol{\mathrm{a}}\right) \\ $$$$\:\:\boldsymbol{\mathrm{f}}\left(−\mathrm{1}\right)=\:\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right) \\ $$$$\: \\ $$

Commented by Humble last updated on 12/Feb/23

$${thank}\:{you},\:{sir}. \\ $$

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