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

Answered by integralmagic last updated on 12/Feb/23

$$$$$$=ln2$$

Answered by SEKRET last updated on 12/Feb/23

$$(-1)\cdot \underset{\mathbf{k}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathbf{k}}}{\mathbf{k}}=?$$$$-1=\mathbf{a}$$$$\mathbf{f}\left(\mathbf{a}\right)=-\underset{\mathbf{k}=1}{\sum ^{\mathrm{\infty }}}\frac{{\mathbf{a}}^{\mathbf{k}}}{\mathbf{k}}$$$$\mathbf{f}{}^{\prime }\left(\mathbf{a}\right)=-\underset{\mathbf{k}=1}{\sum ^{\mathrm{\infty }}}{\mathbf{a}}^{\mathbf{k}-1}=1+\mathbf{a}+{\mathbf{a}}^2+{\mathbf{a}}^3+...{\mathbf{a}}^{\mathbf{n}}+..$$$$\mathbf{f}{}^{\prime }\left(\mathbf{a}\right)=\frac{1}{1-\mathbf{a}}\mathbf{c}=0\mathbf{f}\left(\mathbf{a}\right)=\mathbf{ln}(1-\mathbf{a})$$$$\mathbf{f}(-1)=\mathbf{ln}\left(2\right)$$$$$$

Commented by Humble last updated on 12/Feb/23

$$thankyou,sir.$$

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