Answer to Question #225718 in Calculus for Seif El-Den Ashra

Question

prove that 2ⁿ⁺¹ > (n+2) sin(n) for all positive integers n

Answer 1

Consider the function $f(x)=2^{x+1}-(x+2).$

$$f'(x)=2^{x+1}\ln2-1\geq2^{1+1}\ln2-1=\ln(\dfrac{4}{e})>0, x\geq1$$

Then the function $f(x)$ increases for $x\geq1.$

$$f(1)=2^{1+1}-(1+2)=1>0$$

Hence

$$2^{n+1}>(n+2), n\geq 1$$

$$1\geq\sin n, n>1$$

Then

$$2^{n+1}>n+2\geq(n+2)\cdot\sin n, n\geq1$$

Therefore $2^{n+1}>(n+2)\cdot\sin n,$ for all positive integers $n.$