Prove that 2^(n+1) > (n + 2) · sin(n) for all positive integers n

Let $P(n)$ be the proposition

Basis Step. $P(1)$ is true because

$2^{1+1}=4>3=1+2.$

Inductive Step.

Assume that $P(k)$ holds for an arbitrary positive integer $k.$ That is, we assume that

$2^{k+1}> k + 2$

Under this assumption, it must be shown that $P(k + 1)$ is true, namely, that

$2^{(k+1)+1}> (k+1) + 2$

$2^{(k+1)+1}=2(2^{k+1})$

$=2^{k+1}+2^{k+1}>k+2+k+2$

$=k+3+k+1>k+3$

This shows that $P(k + 1)$ is true under the assumption that $P(k)$ is true. This completes the inductive step.

We have completed the basis step and the inductive step, so by mathematical induction we know that $P(n)$ is true for all positive integers $n.$ That is, we have proved that $2^{n+1}> n + 2$ for all positive integers $n.$

$\sin(n)\leq1, n\in\R$

Then for all positive integers $n$

$(n+2)\cdot\sin(n)\leq n+2, n\in\R$

We have proved that $2^{n+1}> n + 2$ for all positive integers $n.$

Therefore we have proved that

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

for all positive integers $n.$