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Question Number 34697 by abdo imad last updated on 10/May/18

let f(x)=((ln(1+x))/(1+x))  developp f at integr serie

$${let}\:{f}\left({x}\right)=\frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}} \\ $$$${developp}\:{f}\:{at}\:{integr}\:{serie} \\ $$

Commented by math khazana by abdo last updated on 11/May/18

we have ln^′ (1+x)= (1/(1+x))=Σ_(n=0) ^∞ (−1)^n x^n   with∣x∣<1  ⇒ln(1+x)= Σ_(n=0) ^∞  (((−1)^n )/(n+1))x^(n+1)  = Σ_(n=1) ^∞  (((−1)^(n−1) )/n)x^n   f(x) =(Σ_(n=0) ^∞  (−1)^n x^n ) (Σ_(n=1) ^∞  (((−1)^n )/n) x^n )  =Σ_(n=1) ^∞   (((−1)^n )/n) x^n   +(Σ_(n=1) ^∞ (−1)^n x^n )( Σ_(n=1) ^∞  (((−1)^n )/n)x^n )  =Σ_(n=1) ^∞  (((−1)^n )/n) x^n     +Σ_(n=1) ^∞   c_n  x^n    with  c_n = Σ_(i+j=n) a_i  b_j  = Σ_(i=1) ^n  a_i  b_(n−i)   =Σ_(i=1) ^(n−1)   (−1)^i   .(((−1)^(n−i) )/(n−i)) = Σ_(i=1) ^(n−1)    (((−1)^n )/(n−i)) so  f(x)= Σ_(n=1) ^∞    (((−1)^n )/n) x^n   + Σ_(n=1) ^∞ ( Σ_(i=1) ^(n−1)  (((−1)^n )/(n−i)))x^n   .

$${we}\:{have}\:{ln}^{'} \left(\mathrm{1}+{x}\right)=\:\frac{\mathrm{1}}{\mathrm{1}+{x}}=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {x}^{{n}} \:\:{with}\mid{x}\mid<\mathrm{1} \\ $$$$\Rightarrow{ln}\left(\mathrm{1}+{x}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} \:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}{x}^{{n}} \\ $$$${f}\left({x}\right)\:=\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} {x}^{{n}} \right)\:\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:{x}^{{n}} \right) \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:{x}^{{n}} \:\:+\left(\sum_{{n}=\mathrm{1}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {x}^{{n}} \right)\left(\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}{x}^{{n}} \right) \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:{x}^{{n}} \:\:\:\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:{c}_{{n}} \:{x}^{{n}} \:\:\:{with} \\ $$$${c}_{{n}} =\:\sum_{{i}+{j}={n}} {a}_{{i}} \:{b}_{{j}} \:=\:\sum_{{i}=\mathrm{1}} ^{{n}} \:{a}_{{i}} \:{b}_{{n}−{i}} \\ $$$$=\sum_{{i}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\left(−\mathrm{1}\right)^{{i}} \:\:.\frac{\left(−\mathrm{1}\right)^{{n}−{i}} }{{n}−{i}}\:=\:\sum_{{i}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}−{i}}\:{so} \\ $$$${f}\left({x}\right)=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:{x}^{{n}} \:\:+\:\sum_{{n}=\mathrm{1}} ^{\infty} \left(\:\sum_{{i}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}−{i}}\right){x}^{{n}} \:\:. \\ $$$$ \\ $$

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