Prove-that-log-n-gt-3n-10-1-2-1-3-1-4-1-n-1-

Question Number 5900 by 314159 last updated on 04/Jun/16

P r o v e t h a t

l o g n! > \frac{3 n}{10} (\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{n} - 1) .

Answered by Yozzii last updated on 05/Jun/16

L e t f o r n \in N, n ⩾ 2,

P (n) : l o g n! > \frac{3 n}{10} (\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{n} - 1) .

F o r n = 2 \Rightarrow l o g 2 > \frac{3 \times 2}{10} (\frac{1}{2} - 1) = - 0.3

a n d l o g 2 > 0 > - 0.3 . S o, P (n) i s

t r u e w h e n n = 2 .

A s s u m e w h e n n = k, P (k) i s t r u e

\Rightarrow l o g k! > \frac{3 k}{10} (\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{k} - 1)

\Rightarrow l o g k! - \frac{3 k}{10} (\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{k} - 1) > 0 .

C o n s i d e r w h e n n = k + 1 .

L e t ϕ = l o g (k + 1)! - \frac{3 (k + 1)}{10} (\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{k} + \frac{1}{k + 1} - 1) .

L e t u = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{k} - 1 .

∴ ϕ = l o g (k + 1)! - \frac{3 (k + 1)}{10} (u + \frac{1}{k + 1})

ϕ = l o g k! + l o g (k + 1) - \frac{3 k u}{10} - \frac{3 k}{10 (k + 1)} - \frac{3 u}{10} - \frac{3}{10 (k + 1)}

ϕ = (l o g k! - \frac{3 k u}{10}) + l o g k (1 + \frac{1}{k}) - \frac{3}{10} (1 + u)

ϕ = (l o g k! - \frac{3 k u}{10}) + l o g k + l o g (1 + \frac{1}{k}) - \frac{3}{10} (\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{k})

N o w, \underset{i = 1}{\sum^{k}} \frac{1}{i} = H (k) ⩽ 1 + l o g k .

\Rightarrow - (H (k) - 1) ⩾ - l o g k

\Rightarrow \frac{- 3}{10} (H (k) - 1) ⩾ - \frac{3}{10} l o g k

\Rightarrow l o g k + l o g (1 + k^{- 1}) - \frac{3}{10} (H (k) - 1) ⩾ \frac{7}{10} l o g k + l o g (1 + k^{- 1}) > 0 (k ⩾ 2)

\Rightarrow (l o g k! - \frac{3 k u}{10}) + l o g (k + 1) - \frac{3}{10} (H (k) - 1) > 0 s i n c e P (k) \Rightarrow (l o g k! - \frac{3 u k}{10}) > 0 .

S o, P (k) \Rightarrow P (k + 1) .

∴ S i n c e P (2) i s t r u e \Rightarrow P (n) i s t r u e b y P . M . I

f o r a l l n ⩾ 2 .

Commented by 314159 last updated on 05/Jun/16

T h a n k s a l o t!

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