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Question Number 39378 by maxmathsup by imad last updated on 05/Jul/18

study the derivability of  f(x)=Σ_(n=0) ^∞    (((−1)^n )/(nx +1))

$${study}\:{the}\:{derivability}\:{of} \\ $$$${f}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{nx}\:+\mathrm{1}} \\ $$

Commented by math khazana by abdo last updated on 10/Jul/18

the functions  f_n (x) = (((−1)^n )/(nx +1)) are derivables  if we consider?for x fixed ϕ(t)= (1/(xt +1))  ϕ is decreasing so f(x) is a alternate serie  convergent  also the serie Σ_(n=0) ^∞  (((−1)^(n+1) )/((nx+1)^2 )) is  unif.convergent because ∣ (((−1)^(n+1) )/((nx+1)^2 ))∣≤(1/(n^2 x^2 ))  so f is derivable and  f^′ (x)=Σ_(n=0) ^∞  (((−1)^(n+1) )/((nx+1)^2 )) .

$${the}\:{functions}\:\:{f}_{{n}} \left({x}\right)\:=\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{nx}\:+\mathrm{1}}\:{are}\:{derivables} \\ $$$${if}\:{we}\:{consider}?{for}\:{x}\:{fixed}\:\varphi\left({t}\right)=\:\frac{\mathrm{1}}{{xt}\:+\mathrm{1}} \\ $$$$\varphi\:{is}\:{decreasing}\:{so}\:{f}\left({x}\right)\:{is}\:{a}\:{alternate}\:{serie} \\ $$$${convergent}\:\:{also}\:{the}\:{serie}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\left({nx}+\mathrm{1}\right)^{\mathrm{2}} }\:{is} \\ $$$${unif}.{convergent}\:{because}\:\mid\:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\left({nx}+\mathrm{1}\right)^{\mathrm{2}} }\mid\leqslant\frac{\mathrm{1}}{{n}^{\mathrm{2}} {x}^{\mathrm{2}} } \\ $$$${so}\:{f}\:{is}\:{derivable}\:{and} \\ $$$${f}^{'} \left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\left({nx}+\mathrm{1}\right)^{\mathrm{2}} }\:. \\ $$

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