proof-that-2-n-1-gt-n-2-sin-n-

Question Number 162305 by mathlove last updated on 28/Dec/21

p r o o f t h a t

2^{n + 1} > (n + 2) \sin n

Answered by aleks041103 last updated on 28/Dec/21

s i n (n) ⩽ 1

\Rightarrow (n + 2) s i n (n) ⩽ n + 2

f o r n = 1 :

2^{n + 1} = 4 > 3 = n + 2

f o r n = k :

2^{k + 1} > k + 2

f o r n = k + 1 :

2^{(k + 1) + 1} = 22^{k + 1} > 2 (k + 2) = 2 k + 4

k ⩾ 1 \Rightarrow 2 k + 4 ⩾ k + 1 + 4 = k + 5 > (k + 1) + 2

\Rightarrow 2^{(k + 1) + 1} > (k + 1) + 2

\Rightarrow f o r \forall n \in N, 2^{n + 1} > n + 2 ⩾ (n + 2) s i n (n)

\Rightarrow n = 1, 2, \dots, 2^{n + 1} > (n + 2) s i n (n)

f o r n = 0 :

2^{n + 1} = 2 > 0 = (n + 2) s i n (n)

\Rightarrow 2^{n + 1} > (n + 2) s i n (n), \forall n \in N^{0}

Answered by mr W last updated on 28/Dec/21

r e c a l l :

1 ⩾ \sin n f o r a n y n \in R

{(1 + a)}^{n} > 1 + n a f o r a > 0

2^{n + 1} = {(1 + 1)}^{n + 1}

> 1 + (n + 1) \times 1 = n + 2 = (n + 2) \times 1

⩾ (n + 2) \sin n

\Rightarrow 2^{n + 1} > (n + 2) \sin n ✓

Commented by aleks041103 last updated on 28/Dec/21

I t i s w o r t h n o t i n g t h a t

(n + 2) ⩾ (n + 2) s i n (n)

i s t r u e f o r n ⩾ - 2

Commented by mr W last updated on 29/Dec/21

i a s s u m e d n \in N w h a t t h e q u e s t i o n e r

a l s o m e a n t, i t h i n k .

i f n \in Z, t h e n 2^{n + 1} > (n + 2) \sin n i s

t r u e f o r n ⩾ - 6 .

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