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Question Number 5900 by 314159 last updated on 04/Jun/16
Prove that log n! >((3n)/(10))((1/2)+(1/3)+(1/4)+...+(1/n)−1).
$${Prove}\:{that} \\ $$$${log}\:{n}!\:>\frac{\mathrm{3}{n}}{\mathrm{10}}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+…+\frac{\mathrm{1}}{{n}}−\mathrm{1}\right). \\ $$
Answered by Yozzii last updated on 05/Jun/16
Let for n∈N,n≥2, P(n): logn!>((3n)/(10))((1/2)+(1/3)+(1/4)+...+(1/n)−1). For n=2⇒log2>((3×2)/(10))((1/2)−1)=−0.3 and log2>0>−0.3. So, P(n) is true when n=2. Assume when n=k, P(k) is true ⇒logk!>((3k)/(10))((1/2)+(1/3)+(1/4)+...+(1/k)−1) ⇒logk!−((3k)/(10))((1/2)+(1/3)+(1/4)+...+(1/k)−1)>0. Consider when n=k+1. Let φ=log(k+1)!−((3(k+1))/(10))((1/2)+(1/3)+(1/4)+...+(1/k)+(1/(k+1))−1). Let u=(1/2)+(1/3)+(1/4)+...+(1/k)−1. ∴φ=log(k+1)!−((3(k+1))/(10))(u+(1/(k+1))) φ=logk!+log(k+1)−((3ku)/(10))−((3k)/(10(k+1)))−((3u)/(10))−(3/(10(k+1))) φ=(logk!−((3ku)/(10)))+logk(1+(1/k))−(3/(10))(1+u) φ=(logk!−((3ku)/(10)))+logk+log(1+(1/k))−(3/(10))((1/2)+(1/3)+(1/4)+...+(1/k)) Now, Σ_(i=1) ^k (1/i)=H(k)≤1+logk. ⇒−(H(k)−1)≥−logk ⇒((−3)/(10))(H(k)−1)≥−(3/(10))logk ⇒logk+log(1+k^(−1) )−(3/(10))(H(k)−1)≥(7/(10))logk+log(1+k^(−1) )>0 (k≥2) ⇒(logk!−((3ku)/(10)))+log(k+1)−(3/(10))(H(k)−1)>0 since P(k)⇒(logk!−((3uk)/(10)))>0. So, P(k)⇒P(k+1). ∴ Since P(2) is true⇒P(n) is true by P.M.I for all n≥2.
$${Let}\:{for}\:{n}\in\mathbb{N},{n}\geqslant\mathrm{2},\: \\ $$$${P}\left({n}\right):\:{logn}!>\frac{\mathrm{3}{n}}{\mathrm{10}}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+…+\frac{\mathrm{1}}{{n}}−\mathrm{1}\right). \\ $$$${For}\:{n}=\mathrm{2}\Rightarrow{log}\mathrm{2}>\frac{\mathrm{3}×\mathrm{2}}{\mathrm{10}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)=−\mathrm{0}.\mathrm{3} \\ $$$${and}\:{log}\mathrm{2}>\mathrm{0}>−\mathrm{0}.\mathrm{3}.\:{So},\:{P}\left({n}\right)\:{is} \\ $$$${true}\:{when}\:{n}=\mathrm{2}. \\ $$$$ \\ $$$${Assume}\:{when}\:{n}={k},\:{P}\left({k}\right)\:{is}\:{true} \\ $$$$\Rightarrow{logk}!>\frac{\mathrm{3}{k}}{\mathrm{10}}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+…+\frac{\mathrm{1}}{{k}}−\mathrm{1}\right) \\ $$$$\Rightarrow{logk}!−\frac{\mathrm{3}{k}}{\mathrm{10}}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+…+\frac{\mathrm{1}}{{k}}−\mathrm{1}\right)>\mathrm{0}. \\ $$$${Consider}\:{when}\:{n}={k}+\mathrm{1}. \\ $$$${Let}\:\phi={log}\left({k}+\mathrm{1}\right)!−\frac{\mathrm{3}\left({k}+\mathrm{1}\right)}{\mathrm{10}}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+…+\frac{\mathrm{1}}{{k}}+\frac{\mathrm{1}}{{k}+\mathrm{1}}−\mathrm{1}\right). \\ $$$${Let}\:{u}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+…+\frac{\mathrm{1}}{{k}}−\mathrm{1}. \\ $$$$\therefore\phi={log}\left({k}+\mathrm{1}\right)!−\frac{\mathrm{3}\left({k}+\mathrm{1}\right)}{\mathrm{10}}\left({u}+\frac{\mathrm{1}}{{k}+\mathrm{1}}\right) \\ $$$$\phi={logk}!+{log}\left({k}+\mathrm{1}\right)−\frac{\mathrm{3}{ku}}{\mathrm{10}}−\frac{\mathrm{3}{k}}{\mathrm{10}\left({k}+\mathrm{1}\right)}−\frac{\mathrm{3}{u}}{\mathrm{10}}−\frac{\mathrm{3}}{\mathrm{10}\left({k}+\mathrm{1}\right)} \\ $$$$\phi=\left({logk}!−\frac{\mathrm{3}{ku}}{\mathrm{10}}\right)+{logk}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}}\right)−\frac{\mathrm{3}}{\mathrm{10}}\left(\mathrm{1}+{u}\right) \\ $$$$\phi=\left({logk}!−\frac{\mathrm{3}{ku}}{\mathrm{10}}\right)+{logk}+{log}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}}\right)−\frac{\mathrm{3}}{\mathrm{10}}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+…+\frac{\mathrm{1}}{{k}}\right) \\ $$$$ \\ $$$${Now},\:\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}\frac{\mathrm{1}}{{i}}={H}\left({k}\right)\leqslant\mathrm{1}+{logk}. \\ $$$$\Rightarrow−\left({H}\left({k}\right)−\mathrm{1}\right)\geqslant−{logk} \\ $$$$\Rightarrow\frac{−\mathrm{3}}{\mathrm{10}}\left({H}\left({k}\right)−\mathrm{1}\right)\geqslant−\frac{\mathrm{3}}{\mathrm{10}}{logk} \\ $$$$\Rightarrow{logk}+{log}\left(\mathrm{1}+{k}^{−\mathrm{1}} \right)−\frac{\mathrm{3}}{\mathrm{10}}\left({H}\left({k}\right)−\mathrm{1}\right)\geqslant\frac{\mathrm{7}}{\mathrm{10}}{logk}+{log}\left(\mathrm{1}+{k}^{−\mathrm{1}} \right)>\mathrm{0}\:\:\left({k}\geqslant\mathrm{2}\right) \\ $$$$\Rightarrow\left({logk}!−\frac{\mathrm{3}{ku}}{\mathrm{10}}\right)+{log}\left({k}+\mathrm{1}\right)−\frac{\mathrm{3}}{\mathrm{10}}\left({H}\left({k}\right)−\mathrm{1}\right)>\mathrm{0}\:{since}\:{P}\left({k}\right)\Rightarrow\left({logk}!−\frac{\mathrm{3}{uk}}{\mathrm{10}}\right)>\mathrm{0}. \\ $$$${So},\:{P}\left({k}\right)\Rightarrow{P}\left({k}+\mathrm{1}\right). \\ $$$$\therefore\:{Since}\:{P}\left(\mathrm{2}\right)\:{is}\:{true}\Rightarrow{P}\left({n}\right)\:{is}\:{true}\:{by}\:{P}.{M}.{I} \\ $$$${for}\:{all}\:{n}\geqslant\mathrm{2}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by 314159 last updated on 05/Jun/16
Thanks a lot!
$${Thanks}\:{a}\:{lot}! \\ $$

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