Question Number 39378 by maxmathsup by imad last updated on 05/Jul/18
$${study}\:{the}\:{derivability}\:{of} \\ $$$${f}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{nx}\:+\mathrm{1}} \\ $$
Commented by math khazana by abdo last updated on 10/Jul/18
$${the}\:{functions}\:\:{f}_{{n}} \left({x}\right)\:=\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{nx}\:+\mathrm{1}}\:{are}\:{derivables} \\ $$$${if}\:{we}\:{consider}?{for}\:{x}\:{fixed}\:\varphi\left({t}\right)=\:\frac{\mathrm{1}}{{xt}\:+\mathrm{1}} \\ $$$$\varphi\:{is}\:{decreasing}\:{so}\:{f}\left({x}\right)\:{is}\:{a}\:{alternate}\:{serie} \\ $$$${convergent}\:\:{also}\:{the}\:{serie}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\left({nx}+\mathrm{1}\right)^{\mathrm{2}} }\:{is} \\ $$$${unif}.{convergent}\:{because}\:\mid\:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\left({nx}+\mathrm{1}\right)^{\mathrm{2}} }\mid\leqslant\frac{\mathrm{1}}{{n}^{\mathrm{2}} {x}^{\mathrm{2}} } \\ $$$${so}\:{f}\:{is}\:{derivable}\:{and} \\ $$$${f}^{'} \left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\left({nx}+\mathrm{1}\right)^{\mathrm{2}} }\:. \\ $$