Question

Show that n!<n^n, n>1, by induction.

Answer 1

Answer on Question #83832 – Math – Real Analysis

Question

Show that n!<n^n, n="">1, by induction.

Solution

We shall prove it by the method of mathematical induction.

1. Take $n = 2$ and substitute the inequality, we get: $2! = 1 \times 2 = 2 < 2^2 = 2 \times 2 = 4$, i.e. $2 < 4$, therefore the inequality holds.

2. Let it be true for $n = k$, where $k \geq 2$, i.e. $k! < k^k$.

3. Let us prove the fulfillment of the inequality for $n = k + 1$, where $k \geq 2$.

We need to show $(k + 1)! < (k + 1)^{k + 1}$, or $k! (k + 1) < (k + 1)^{k + 1}$, or $k! < (k + 1)^k$. Transform the right hand side of the inequality as follows: $(k(1 + \frac{1}{k}))^k = k^k (1 + \frac{1}{k})^k$. We know that $\lim_{k \to \infty} \left(1 + \frac{1}{k}\right)^k = e$, i.e. always $1 < (1 + \frac{1}{k})^k < e$. And, at the same time, we have assumed that $k! < k^k$, then $k!$ also will be less than $k^k (1 + \frac{1}{k})^k$, and then $(k + 1)! < (k + 1)^{k + 1}$.

4. By the method of mathematical induction, the inequality $n! < n^n$ holds for all $n > 1$.

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