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Question Number 145756 by puissant last updated on 07/Jul/21

Answered by phanphuoc last updated on 07/Jul/21

$$u_n=ln(n+1)$$

Commented by puissant last updated on 07/Jul/21

$$\mathrm{How}?$$

Commented by puissant last updated on 07/Jul/21

$$\mathrm{comment}?$$

Answered by mathmax by abdo last updated on 09/Jul/21

$$\mathrm{let}{\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{k}=2}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}-1}=1-\frac{1}{2}+\frac{1}{3}-....+\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}-1}\Rightarrow$$$${\mathrm{S}}_{\mathrm{n}}{=}_{\mathrm{k}-1=\mathrm{p}}\sum _{\mathrm{p}=1}^{\mathrm{n}-1}\frac{{(-1)}^{\mathrm{p}+1}}{\mathrm{p}}=-\sum _{\mathrm{p}=1}^{\mathrm{n}-1}\frac{{(-1)}^{\mathrm{p}}}{\mathrm{p}}$$$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\sum _{\mathrm{p}=1}^{\mathrm{n}-1}\frac{{(-1)}^{\mathrm{p}}}{\mathrm{p}}{\mathrm{x}}^{\mathrm{p}}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=\sum _{\mathrm{p}=1}^{\mathrm{n}-1}{(-1)}^{\mathrm{p}}{\mathrm{x}}^{\mathrm{p}-1}$$$$=-\sum _{\mathrm{p}=1}^{\mathrm{n}-1}{(-\mathrm{x})}^{\mathrm{p}-1}=-\sum _{\mathrm{k}=0}^{\mathrm{n}-2}{(-\mathrm{x})}^{\mathrm{k}}=-\frac{1-{(-\mathrm{x})}^{\mathrm{n}-1}}{1+\mathrm{x}}$$$$=-\frac{1}{1+\mathrm{x}}+\frac{{(-\mathrm{x})}^{\mathrm{n}-1}}{1+\mathrm{x}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=-\log \mid 1+\mathrm{x}\mid +\int \frac{{(-\mathrm{x})}^{\mathrm{n}-1}}{1+\mathrm{x}}\mathrm{dx}\to -\log \mid 1+\mathrm{x}\mid$$$$\mathrm{if}\mid \mathrm{x}\mid <1\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{S}}_{\mathrm{n}}=-{\lim }_{\mathrm{x}\to 1}\mathrm{f}\left(\mathrm{x}\right)=\log 2$$

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