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Question Number 174439 by mnjuly1970 last updated on 01/Aug/22

       x∈ ( 0 , 1 ) , k ∈ N        prove  that :    kx^( k)  < (x/(1−x))  (math  analysis)

$$ \\ $$ $$\:\:\:\:\:{x}\in\:\left(\:\mathrm{0}\:,\:\mathrm{1}\:\right)\:,\:{k}\:\in\:\mathbb{N} \\ $$ $$\:\:\:\:\:\:{prove}\:\:{that}\:: \\ $$ $$\:\:{kx}^{\:{k}} \:<\:\frac{{x}}{\mathrm{1}−{x}}\:\:\left({math}\:\:{analysis}\right) \\ $$ $$ \\ $$

Answered by behi834171 last updated on 01/Aug/22

for:  k=1:  1×x^1 <^? (x/(1−x))⇒x<^? (x/(1−x))  0<x<1⇒−1<−x<0⇒0<1−x<1  so: 1−x<1⇒(x/(1−x))>x ,  i.e:it is true for : k=1  let assume that is true for k=n  we should prove that is true for k=n+1  (n+1)x^(n+1) <^? (x/(1−x))  (n+1)x^(n+1) (1−x)=(n+1)x^(n+1) −(n+1)x^(n+2) <  <nx^n .x+x^(n+1) −nx^n .x^2 −x^(n+2)   <(x/(1−x)).x−(x/(1−x)).x^2 +x^(n+1) −x^(n+2) =  =((x^2 −x^3 )/(1−x))+x^(n+1) −x^(n+2) =x^2 +x^(n+1) −x^(n+2) <  <x+x−x=x  .hence proved. ■  [0<x<1⇒   x^(p∈N) <x]

$${for}:\:\:{k}=\mathrm{1}:\:\:\mathrm{1}×{x}^{\mathrm{1}} \overset{?} {<}\frac{{x}}{\mathrm{1}−{x}}\Rightarrow{x}\overset{?} {<}\frac{{x}}{\mathrm{1}−{x}} \\ $$ $$\mathrm{0}<{x}<\mathrm{1}\Rightarrow−\mathrm{1}<−{x}<\mathrm{0}\Rightarrow\mathrm{0}<\mathrm{1}−{x}<\mathrm{1} \\ $$ $${so}:\:\mathrm{1}−{x}<\mathrm{1}\Rightarrow\frac{{x}}{\mathrm{1}−{x}}>{x}\:, \\ $$ $${i}.{e}:{it}\:{is}\:{true}\:{for}\::\:{k}=\mathrm{1} \\ $$ $${let}\:{assume}\:{that}\:{is}\:{true}\:{for}\:{k}={n} \\ $$ $${we}\:{should}\:{prove}\:{that}\:{is}\:{true}\:{for}\:{k}={n}+\mathrm{1} \\ $$ $$\left({n}+\mathrm{1}\right){x}^{{n}+\mathrm{1}} \overset{?} {<}\frac{{x}}{\mathrm{1}−{x}} \\ $$ $$\left({n}+\mathrm{1}\right){x}^{{n}+\mathrm{1}} \left(\mathrm{1}−{x}\right)=\left({n}+\mathrm{1}\right){x}^{{n}+\mathrm{1}} −\left({n}+\mathrm{1}\right){x}^{{n}+\mathrm{2}} < \\ $$ $$<{nx}^{{n}} .{x}+{x}^{{n}+\mathrm{1}} −{nx}^{{n}} .{x}^{\mathrm{2}} −{x}^{{n}+\mathrm{2}} \\ $$ $$<\frac{{x}}{\mathrm{1}−{x}}.{x}−\frac{{x}}{\mathrm{1}−{x}}.{x}^{\mathrm{2}} +{x}^{{n}+\mathrm{1}} −{x}^{{n}+\mathrm{2}} = \\ $$ $$=\frac{{x}^{\mathrm{2}} −{x}^{\mathrm{3}} }{\mathrm{1}−{x}}+{x}^{{n}+\mathrm{1}} −{x}^{{n}+\mathrm{2}} ={x}^{\mathrm{2}} +{x}^{{n}+\mathrm{1}} −{x}^{{n}+\mathrm{2}} < \\ $$ $$<{x}+{x}−{x}=\boldsymbol{{x}}\:\:.\boldsymbol{{hence}}\:\boldsymbol{{proved}}.\:\blacksquare \\ $$ $$\left[\mathrm{0}<\boldsymbol{{x}}<\mathrm{1}\Rightarrow\:\:\:\boldsymbol{{x}}^{\boldsymbol{{p}}\in\boldsymbol{{N}}} <\boldsymbol{{x}}\right] \\ $$

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