For all real numbers $x, x < 2^x$.
Use this fact to show that for any positive integer $n$ :
$x^n < (n^n)(2^x)$, for all real numbers $x > 0$.
Let f(x,n) be x^n < (n^n)(2^x) then I proved that f(1,1) = 1 < 2 is True
Assume that f(x,1) is true too, x < 2^x. Prove that f(x+1,1) is true, x+1 < 2^(x+1).
x + 1 - 2*2^x < 2^x - 2^x - 2^x + 1
x + 1 - 2*2^x < 1 - 2^x (Since x > 0)
x + 1 - 2*2^x < 0
x+1 < 2^(x+1) (Proven)
Let f(x,n) be x^n < (n^n)(2^x). Prove that f(x,n+1) is true, x^(n+1) < [(n+1)^(n+1)] 2^x
That's the problem, no matter how hard i tried, I can't prove that f(x,n+1) is true...
Can anyone out there please help me with it asap. Thank you in advance! (:
We are told that $$t\lt 2^t\tag{1}$$ for all $t$. The inequality we want to establish is equivalent to $\frac{x^n}{n^n}\lt 2^x$. This is equivalent to $\left(\frac{x}{n}\right)^n \lt 2^x$, which is equivalent, for positive $x$, to $\frac{x}{n}\lt 2^{x/n}$. Now put $t=\frac{x}{n}$ in (1).
Remark: One could alternately write the same argument as follows. We are told to assume that the inequality $t\lt 2^t$ holds for all real $t$. Let $t=x/n$, where $x$ is positive. Then $$\frac{x}{n}\lt 2^{x/n}.$$ Since $x$ is positive, taking the $n$-th power of both sides we obtain $$\frac{x^n}{n^n}\lt 2^x,$$ and therefore $x^n\lt n^n2^x$.