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Question Number 36742 by prof Abdo imad last updated on 04/Jun/18
$${study}\:{the}\:{convergence}\:{of}\: \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {ln}\left(\mathrm{1}+\:\frac{\mathrm{1}}{{n}\left(\mathrm{1}+{x}\right)}\right). \\ $$
Commented by maxmathsup by imad last updated on 04/Aug/18
$${let}\:{f}_{{n}} \left({x}\right)\:=\left(−\mathrm{1}\right)^{{n}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}\left(\mathrm{1}+{x}\right)}\right)\:\:{for}\:{x}\neq−\mathrm{1}\:{we}\:{have}\:\:\left({f}_{{n}} \right)\:{are}\:{continues}\:{and} \\ $$$${f}_{{n}} \left({x}\right)\:\sim\left(−\mathrm{1}\right)^{{n}} \:\:\frac{\mathrm{1}}{{n}\left(\mathrm{1}+{x}\right)}\:\:{and}\:\sum_{{n}\geqslant\mathrm{1}} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}\left(\mathrm{1}+{x}\right)}\:=\frac{\mathrm{1}}{\mathrm{1}+{x}}\:\sum_{{n}\geqslant\mathrm{1}} \:\:\:\frac{\left(−\mathrm{1}\right)}{{n}}\:{is}\:{a}\:{convegent} \\ $$$${serie}\:{so}\:{the}\:{simple}\:{convergence}\:{of}\:\Sigma\:{f}_{{n}} \left({x}\right)\:{is}\:{assured} \\ $$