Chapter 9.4, Problem 25E

To determine

To proof: The inequality $n!>2^n$ with help of mathematical induction for integers $n\ge 4$ .

Explanation of Solution

Given information:

The inequality $n!>2^n$ .

Formula used:

The steps to prove a statement by mathematical induction are as follows:

Step 1. Show that the statement is true for given starting value of n .

Step 2. Assume that the statement is true for $n=k$ .

Step 3. Prove that the statement is true for $n=k+1$ with help of step 2.

Proof:

Consider inequality $n!>2^n$ .

Recall that the steps to prove a statement by mathematical induction are as follows:

Step 1. Show that the statement is true for given starting value of n .

Step 2. Assume that the statement is true for $n=k$ .

Step 3. Prove that the statement is true for $n=k+1$ with help of step 2.

Substitute $n=4$ in the expression,

  $\begin{array}{c}4!>2^4\\ 24>16\end{array}$

The above inequality holds true.

Therefore, the statement is true for $n=4$ so $S_4$ is true.

Now, assume that the statement if true for $n=k$ .

  $k!>2^k$  ...(1)

Therefore, the statement holds true for $n=k$ so $S_k$ is true.

Now, prove it for $n=k+1$ . Write the expression for $n=k+1$ .

  $\left(k+1\right)!>2^{k+1}$

Rewrite the left hand side of the above inequality as,

  $\left(k+1\right)!=\left(k+1\right)k!$

From equation (1), it is clear that $k!>2^k$ . Therefore,

  $\begin{array}{c}\left(k+1\right)!=\left(k+1\right)k!\\ >\left(k+1\right)2^k\end{array}$

Now, it is obvious that $\left(k+1\right)<\left(k+2\right)$ , so,

  $\begin{array}{c}\left(k+1\right)!>\left(k+1\right)2^k\\ >\left(k+2\right)2^k\\ >2^{k}k+2^{k+1}\\ >2^{k+1}\end{array}$

Because if $\left(k+1\right)!>2^{k}k+2^{k+1}$ then $\left(k+1\right)!>2^{k+1}$ .

Result obtained is that $S_k$ implies $S_{k+1}$ that is,

  $\left(k+1\right)!>2^{k+1}$

Hence, by the principle of mathematical induction the inequality $n!>2^n$ is true for all integers $n\ge 4$ .