prove-that-1-1-n-n-lt-e-lt-1-1-n-n-1-

Question Number 162026 by mnjuly1970 last updated on 25/Dec/21

p r o v e t h a t \dots .

{(1 + \frac{1}{n})}^{n} < e < {(1 + \frac{1}{n})}^{n + 1}

Answered by mindispower last updated on 25/Dec/21

n o t t r u e

{(1 + \frac{1}{n})}^{n + 1} < {(1 + \frac{1}{n})}^{n}, n > 0

Commented by Ar Brandon last updated on 25/Dec/21

n = 1

{(2)}^{2} = 4

{(2)}^{1} = 2

{(1 + \frac{1}{n})}^{n + 1} > {(1 + \frac{1}{n})}^{n} for n = 1 > 0

Answered by Ar Brandon last updated on 25/Dec/21

From e^{x} ⩾ x + 1 \forall x > 0

\Rightarrow \ln e^{x} ⩾ \ln (x + 1) \Rightarrow x ⩾ \ln (x + 1)

\Rightarrow \frac{1}{n} ⩾ \ln (1 + \frac{1}{n}) \Rightarrow 1 ⩾ n \ln (1 + \frac{1}{n})

\Rightarrow 1 ⩾ \ln {(1 + \frac{1}{n})}^{n} \Rightarrow e ⩾ {(1 + \frac{1}{n})}^{n} \dots (1)

{(1 + \frac{1}{n})}^{n + 1} ⩾ {(1 + \frac{1}{n + 1})}^{n + 1} \dots (2)

{(1 + \frac{1}{n})}^{n + 1} ⩾ {(1 + \frac{1}{N})}^{N} = e, N = \frac{1}{n + 1}

(1) and (2)

\Rightarrow {(1 + \frac{1}{n})}^{n + 1} ⩾ e ⩾ {(1 + \frac{1}{n})}^{n} \forall n \in N^{+}