Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 658 by 123456 last updated on 22/Feb/15

proof that n!>((n/3))^n ,n∈N^∗

$${proof}\:{that}\:{n}!>\left(\frac{{n}}{\mathrm{3}}\right)^{{n}} ,{n}\in\mathbb{N}^{\ast} \\ $$

Commented by123456 last updated on 20/Feb/15

n=1⇒1!=1>(1/3)=((1/3))^1   n=1⇒0!=1>(1/3)≈0.33  n=2⇒2!=2>(4/9)=((2/3))^2   n=2⇒1!=1>(4/9)≈0.44  n=3⇒3!=6>1=((3/3))^3   n=3⇒2!=2>1  n=4⇒4!=24>((256)/(81))=((4/3))^4   n=4⇒3!=6>((256)/(81))≈3.16  n=5⇒5!=120>((3125)/(243))=((5/3))^5   n=5⇒4!=24>((3125)/(243))≈13.23  n=6⇒6!=720>64=((6/3))^6   n=6⇒5!=120>64  n=7⇒7!=5040>((823543)/(2187))=((7/3))^7   n=7⇒6!=720>((823543)/(2187))≈376.52

$${n}=\mathrm{1}\Rightarrow\mathrm{1}!=\mathrm{1}>\frac{\mathrm{1}}{\mathrm{3}}=\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{1}} \\ $$ $${n}=\mathrm{1}\Rightarrow\mathrm{0}!=\mathrm{1}>\frac{\mathrm{1}}{\mathrm{3}}\approx\mathrm{0}.\mathrm{33} \\ $$ $${n}=\mathrm{2}\Rightarrow\mathrm{2}!=\mathrm{2}>\frac{\mathrm{4}}{\mathrm{9}}=\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$ $${n}=\mathrm{2}\Rightarrow\mathrm{1}!=\mathrm{1}>\frac{\mathrm{4}}{\mathrm{9}}\approx\mathrm{0}.\mathrm{44} \\ $$ $${n}=\mathrm{3}\Rightarrow\mathrm{3}!=\mathrm{6}>\mathrm{1}=\left(\frac{\mathrm{3}}{\mathrm{3}}\right)^{\mathrm{3}} \\ $$ $${n}=\mathrm{3}\Rightarrow\mathrm{2}!=\mathrm{2}>\mathrm{1} \\ $$ $${n}=\mathrm{4}\Rightarrow\mathrm{4}!=\mathrm{24}>\frac{\mathrm{256}}{\mathrm{81}}=\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{4}} \\ $$ $${n}=\mathrm{4}\Rightarrow\mathrm{3}!=\mathrm{6}>\frac{\mathrm{256}}{\mathrm{81}}\approx\mathrm{3}.\mathrm{16} \\ $$ $${n}=\mathrm{5}\Rightarrow\mathrm{5}!=\mathrm{120}>\frac{\mathrm{3125}}{\mathrm{243}}=\left(\frac{\mathrm{5}}{\mathrm{3}}\right)^{\mathrm{5}} \\ $$ $${n}=\mathrm{5}\Rightarrow\mathrm{4}!=\mathrm{24}>\frac{\mathrm{3125}}{\mathrm{243}}\approx\mathrm{13}.\mathrm{23} \\ $$ $${n}=\mathrm{6}\Rightarrow\mathrm{6}!=\mathrm{720}>\mathrm{64}=\left(\frac{\mathrm{6}}{\mathrm{3}}\right)^{\mathrm{6}} \\ $$ $${n}=\mathrm{6}\Rightarrow\mathrm{5}!=\mathrm{120}>\mathrm{64} \\ $$ $${n}=\mathrm{7}\Rightarrow\mathrm{7}!=\mathrm{5040}>\frac{\mathrm{823543}}{\mathrm{2187}}=\left(\frac{\mathrm{7}}{\mathrm{3}}\right)^{\mathrm{7}} \\ $$ $${n}=\mathrm{7}\Rightarrow\mathrm{6}!=\mathrm{720}>\frac{\mathrm{823543}}{\mathrm{2187}}\approx\mathrm{376}.\mathrm{52} \\ $$

Commented by123456 last updated on 20/Feb/15

k!>(k^k /3^k )⇒(k+1)!>((k^k (k+1))/3^k )  k!>(k^k /3^k )⇒((k!(k+1)^(k+1) )/(3k^k ))>(((k+1)^(k+1) )/3^(k+1) )

$${k}!>\frac{{k}^{{k}} }{\mathrm{3}^{{k}} }\Rightarrow\left({k}+\mathrm{1}\right)!>\frac{{k}^{{k}} \left({k}+\mathrm{1}\right)}{\mathrm{3}^{{k}} } \\ $$ $${k}!>\frac{{k}^{{k}} }{\mathrm{3}^{{k}} }\Rightarrow\frac{{k}!\left({k}+\mathrm{1}\right)^{{k}+\mathrm{1}} }{\mathrm{3}{k}^{{k}} }>\frac{\left({k}+\mathrm{1}\right)^{{k}+\mathrm{1}} }{\mathrm{3}^{{k}+\mathrm{1}} } \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com