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Question Number 95903 by 675480065 last updated on 28/May/20

Commented by PRITHWISH SEN 2 last updated on 28/May/20

∵ x>0 ∴ x^2 >0 1+x>1 1−x^2 <1 ∴ (1/(1+x))<1 1−x<(1/(1+x)) ∴ 1−x<(1/(1+x)) < 1 as we know that if f(x)<>g(x) then ∫_a ^b f(x)dx<>∫_a ^b g(x)dx ∴ ∫_0 ^1 (1−x)dx<∫_0 ^1 (dx/(1+x)) < ∫_0 ^1 dx = x−(x^2 /2) < ln∣1+x∣ < x proved

$$\because\:\mathrm{x}>\mathrm{0} \\ $$$$\therefore\:\mathrm{x}^{\mathrm{2}} >\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}+\mathrm{x}>\mathrm{1} \\ $$$$\mathrm{1}−\mathrm{x}^{\mathrm{2}} <\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\therefore\:\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}}<\mathrm{1} \\ $$$$\mathrm{1}−\mathrm{x}<\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}} \\ $$$$\therefore\:\mathrm{1}−\mathrm{x}<\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}}\:<\:\mathrm{1} \\ $$$$\boldsymbol{\mathrm{as}}\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{know}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)<>\boldsymbol{\mathrm{g}}\left(\boldsymbol{\mathrm{x}}\right)\:\boldsymbol{\mathrm{then}}\:\int_{\boldsymbol{\mathrm{a}}} ^{\boldsymbol{\mathrm{b}}} \boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{dx}}<>\int_{\boldsymbol{\mathrm{a}}} ^{\boldsymbol{\mathrm{b}}} \boldsymbol{\mathrm{g}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{dx}} \\ $$$$\therefore\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\mathrm{x}\right)\mathrm{dx}<\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}}\:<\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{dx} \\ $$$$=\:\boldsymbol{\mathrm{x}}−\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }{\mathrm{2}}\:\:<\:\boldsymbol{\mathrm{ln}}\mid\mathrm{1}+\boldsymbol{\mathrm{x}}\mid\:<\:\boldsymbol{\mathrm{x}}\:\:\:\boldsymbol{\mathrm{proved}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by 675480065 last updated on 28/May/20

thanks . what about the second part

$$\mathrm{thanks}\:. \\ $$$$\mathrm{what}\:\mathrm{about}\:\mathrm{the}\:\mathrm{second}\:\mathrm{part} \\ $$

Commented by PRITHWISH SEN 2 last updated on 28/May/20

lim_(n→∞) lnU_n =lim_(n→∞) { ln(1+(1/n^2 ))+........+ln(1+(n/n^2 ))} = lim_(t→∞) (1/t)Σ_1 ^(√t) tln(1+(k/t)) let n^2 =t n→∞⇒t→∞ = lim_(t→∞) (1/t)Σ_1 ^(√t) ln(1+(k/t))^t =lim_(t→∞) (1/t) Σ_1 ^(√t) k {∵ lim_(t→∞) (1+(k/t))^t =e^k } =lim_(t→∞) (1/t).(((√t)((√t)+1))/2) = lim_(t→∞) (((1+(1/(√t))))/2) = (1/2) ∴ U_n = e^(1/2) = (√e)

$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}lnU}_{\mathrm{n}} =\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left\{\:\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\right)+........+\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{n}}{\mathrm{n}^{\mathrm{2}} }\right)\right\} \\ $$$$\:=\:\underset{\mathrm{t}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{t}}\underset{\mathrm{1}} {\overset{\sqrt{\mathrm{t}}} {\sum}}\mathrm{tln}\left(\mathrm{1}+\frac{\mathrm{k}}{\mathrm{t}}\right)\:\:\:\:\mathrm{let}\:\mathrm{n}^{\mathrm{2}} =\mathrm{t}\:\:\mathrm{n}\rightarrow\infty\Rightarrow\mathrm{t}\rightarrow\infty \\ $$$$\:=\:\underset{\mathrm{t}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{t}}\underset{\mathrm{1}} {\overset{\sqrt{\mathrm{t}}} {\sum}}\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{k}}{\mathrm{t}}\right)^{\mathrm{t}} =\underset{\mathrm{t}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{t}}\:\underset{\mathrm{1}} {\overset{\sqrt{\mathrm{t}}} {\sum}}\:\mathrm{k}\:\:\:\left\{\because\:\underset{\mathrm{t}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}+\frac{\mathrm{k}}{\mathrm{t}}\right)^{\mathrm{t}} =\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{k}}} \right\} \\ $$$$=\underset{\mathrm{t}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{t}}.\frac{\sqrt{\mathrm{t}}\left(\sqrt{\mathrm{t}}+\mathrm{1}\right)}{\mathrm{2}}\:=\:\underset{\mathrm{t}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\sqrt{\mathrm{t}}}\right)}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\therefore\:\boldsymbol{\mathrm{U}}_{\boldsymbol{\mathrm{n}}} \:=\:\boldsymbol{\mathrm{e}}^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\sqrt{\boldsymbol{\mathrm{e}}} \\ $$$$\:\:\:\: \\ $$

Answered by Rio Michael last updated on 28/May/20

Alternatively, for x > 0 , x−(x^2 /2) is decreasing, hence if f(x) = x−(x^2 /2) ⇒ f ′(x) < 0 for x > 0, ln(1 + x) is increasing , hence if g(x) = ln(1 + x) ⇒ g′(x) > 0 for x > 0, x is increasing strictly, hence if h(x) = x ⇒ h′(x) >>0 thus x−(x^2 /2) < ln(1 + x) but ln(1 +x) < x for x > 0 ⇒ x−(x^2 /2) < ln( 1 + x) < x for x > 0 proved!

$$\mathrm{Alternatively},\: \\ $$$$\:\mathrm{for}\:{x}\:>\:\mathrm{0}\:,\:\:{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\mathrm{is}\:\mathrm{decreasing},\:\mathrm{hence}\:\mathrm{if}\:{f}\left({x}\right)\:=\:{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\:{f}\:'\left({x}\right)\:<\:\mathrm{0} \\ $$$$\:\:\mathrm{for}\:{x}\:>\:\mathrm{0},\:\mathrm{ln}\left(\mathrm{1}\:+\:{x}\right)\:\mathrm{is}\:\mathrm{increasing}\:,\:\mathrm{hence}\:\mathrm{if}\:\mathrm{g}\left({x}\right)\:=\:\mathrm{ln}\left(\mathrm{1}\:+\:{x}\right)\:\Rightarrow\:\mathrm{g}'\left({x}\right)\:>\:\mathrm{0} \\ $$$$\:\mathrm{for}\:{x}\:>\:\mathrm{0},\:{x}\:\mathrm{is}\:\mathrm{increasing}\:\mathrm{strictly},\:\mathrm{hence}\:\mathrm{if}\:{h}\left({x}\right)\:=\:{x}\:\Rightarrow\:{h}'\left({x}\right)\:>>\mathrm{0} \\ $$$$\mathrm{thus}\:\:{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:<\:\mathrm{ln}\left(\mathrm{1}\:+\:{x}\right) \\ $$$$\mathrm{but}\:\mathrm{ln}\left(\mathrm{1}\:+{x}\right)\:<\:{x}\:\mathrm{for}\:{x}\:>\:\mathrm{0} \\ $$$$\Rightarrow\:\:{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:<\:\mathrm{ln}\left(\:\mathrm{1}\:+\:{x}\right)\:<\:{x}\:\mathrm{for}\:{x}\:>\:\mathrm{0}\:\mathrm{proved}! \\ $$

Answered by mathmax by abdo last updated on 28/May/20

1) for x>0 let f(x)=x−ln(1+x)⇒f^′ (x)=1−(1/(1+x)) =(x/(1+x))>0 f is increasing and f(0) =0 ⇒f(x)>0 ⇒ln(1+x)<x let g(x)=ln(1+x)−x+(x^2 /2) g^′ (x) =(1/(1+x))−1+x =((1+x^2 −1)/(1+x)) =(x^2 /(1+x))>0 ⇒g is increazing and g(0)=0 ⇒ g(x)>0 ⇒x−(x^2 /2)<ln(1+x) ⇒x−(x^2 /2)<ln(1+x)<x 2)U_n >0 ⇒ln(U_n ) =Σ_(k=1) ^n ln(1+(k/n^2 )) we have (k/n^2 )−(k^2 /(2n^4 ))<ln(1+(k/n^2 ))<(k/n^2 ) ⇒ Σ_(k=1) ^n ((k/n^2 ))−Σ_(k=1) ^n (k^2 /(2n^4 ))<Σ_(k=1) ^n ln(1+(k/n^2 ))<Σ_(k=1) ^n (k/n^2 ) ⇒ (1/n^2 )×((n(n+1))/2) −(1/(2n^4 ))×((n(n+1)(2n+1))/6)<ln(U_n )<((n(n+1))/(2n^2 )) ⇒ lim_(n→+∞) ln(U_n ) =(1/2) ⇒lim_(n→+∞) U_n =e^(1/2) =(√e)

$$\left.\mathrm{1}\right)\:\mathrm{for}\:\mathrm{x}>\mathrm{0}\:\:\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}−\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)\Rightarrow\mathrm{f}^{'} \left(\mathrm{x}\right)=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}}\:=\frac{\mathrm{x}}{\mathrm{1}+\mathrm{x}}>\mathrm{0}\:\mathrm{f}\:\mathrm{is}\:\mathrm{increasing} \\ $$$$\mathrm{and}\:\mathrm{f}\left(\mathrm{0}\right)\:=\mathrm{0}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)>\mathrm{0}\:\Rightarrow\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)<\mathrm{x}\:\:\mathrm{let}\:\mathrm{g}\left(\mathrm{x}\right)=\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)−\mathrm{x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{g}^{'} \left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}}−\mathrm{1}+\mathrm{x}\:\:=\frac{\mathrm{1}+\mathrm{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}+\mathrm{x}}\:=\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}}>\mathrm{0}\:\Rightarrow\mathrm{g}\:\mathrm{is}\:\mathrm{increazing}\:\mathrm{and}\:\mathrm{g}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{g}\left(\mathrm{x}\right)>\mathrm{0}\:\Rightarrow\mathrm{x}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}<\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)\:\Rightarrow\mathrm{x}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}<\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)<\mathrm{x} \\ $$$$\left.\mathrm{2}\right)\mathrm{U}_{\mathrm{n}} >\mathrm{0}\:\Rightarrow\mathrm{ln}\left(\mathrm{U}_{\mathrm{n}} \right)\:=\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{k}}{\mathrm{n}^{\mathrm{2}} }\right)\:\:\mathrm{we}\:\mathrm{have}\:\frac{\mathrm{k}}{\mathrm{n}^{\mathrm{2}} }−\frac{\mathrm{k}^{\mathrm{2}} }{\mathrm{2n}^{\mathrm{4}} }<\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{k}}{\mathrm{n}^{\mathrm{2}} }\right)<\frac{\mathrm{k}}{\mathrm{n}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left(\frac{\mathrm{k}}{\mathrm{n}^{\mathrm{2}} }\right)−\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{k}^{\mathrm{2}} }{\mathrm{2n}^{\mathrm{4}} }<\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{k}}{\mathrm{n}^{\mathrm{2}} }\right)<\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{k}}{\mathrm{n}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }×\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2n}^{\mathrm{4}} }×\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)}{\mathrm{6}}<\mathrm{ln}\left(\mathrm{U}_{\mathrm{n}} \right)<\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2n}^{\mathrm{2}} }\:\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{ln}\left(\mathrm{U}_{\mathrm{n}} \right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\mathrm{U}_{\mathrm{n}} =\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\sqrt{\mathrm{e}} \\ $$

Commented by PRITHWISH SEN 2 last updated on 28/May/20

excellent sir. By using squeeze theorem.

$$\mathrm{excellent}\:\mathrm{sir}.\:\mathrm{By}\:\mathrm{using}\:\mathrm{squeeze}\:\mathrm{theorem}. \\ $$

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