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Question Number 162026 by mnjuly1970 last updated on 25/Dec/21
prove that.... ( 1+ (1/n) )^( n) < e < (1+(1/n) )^( n+1)
$$ \\ $$$$\:\:\:\:{prove}\:{that}…. \\ $$$$\: \\ $$$$\:\:\:\:\:\left(\:\mathrm{1}+\:\frac{\mathrm{1}}{{n}}\:\right)^{\:{n}} \:<\:{e}\:<\:\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\:\right)^{\:{n}+\mathrm{1}} \\ $$$$ \\ $$$$ \\ $$
Answered by mindispower last updated on 25/Dec/21
not true (1+(1/n))^(n+1) <(1+(1/n))^n ,n>0
$${not}\:{true} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}+\mathrm{1}} <\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} ,{n}>\mathrm{0} \\ $$
Commented by Ar Brandon last updated on 25/Dec/21
n=1 (2)^2 =4 (2)^1 =2 (1+(1/n))^(n+1) >(1+(1/n))^n for n=1>0
$${n}=\mathrm{1} \\ $$$$\left(\mathrm{2}\right)^{\mathrm{2}} =\mathrm{4} \\ $$$$\left(\mathrm{2}\right)^{\mathrm{1}} =\mathrm{2} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}+\mathrm{1}} >\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} \:\mathrm{for}\:{n}=\mathrm{1}>\mathrm{0} \\ $$
Answered by Ar Brandon last updated on 25/Dec/21
From e^x ≥x+1 ∀x>0 ⇒lne^x ≥ln(x+1)⇒x≥ln(x+1) ⇒(1/n)≥ln(1+(1/n))⇒1≥nln(1+(1/n)) ⇒1≥ln(1+(1/n))^n ⇒e≥(1+(1/n))^n ...(1) (1+(1/n))^(n+1) ≥(1+(1/(n+1)))^(n+1) ...(2) (1+(1/n))^(n+1) ≥(1+(1/N))^N =e , N=(1/(n+1)) (1) and (2) ⇒(1+(1/n))^(n+1) ≥e≥(1+(1/n))^n ∀n∈N^+
$$\mathrm{From}\:{e}^{{x}} \geqslant{x}+\mathrm{1}\:\forall{x}>\mathrm{0} \\ $$$$\Rightarrow\mathrm{ln}{e}^{{x}} \geqslant\mathrm{ln}\left({x}+\mathrm{1}\right)\Rightarrow{x}\geqslant\mathrm{ln}\left({x}+\mathrm{1}\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{{n}}\geqslant\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\Rightarrow\mathrm{1}\geqslant{n}\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right) \\ $$$$\Rightarrow\mathrm{1}\geqslant\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} \Rightarrow{e}\geqslant\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} …\left(\mathrm{1}\right) \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}+\mathrm{1}} \geqslant\left(\mathrm{1}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)^{{n}+\mathrm{1}} …\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}+\mathrm{1}} \geqslant\left(\mathrm{1}+\frac{\mathrm{1}}{{N}}\right)^{{N}} ={e}\:,\:{N}=\frac{\mathrm{1}}{{n}+\mathrm{1}} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{2}\right) \\ $$$$\Rightarrow\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}+\mathrm{1}} \geqslant{e}\geqslant\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} \:\:\forall{n}\in\mathbb{N}^{+} \\ $$

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