Chapter 9.4, Problem 20E

To determine

To proof: The expression $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\cdots \left(1+\frac{1}{n}\right)=n+1$ with help of mathematical induction for integers $n\ge 1$ .

Explanation of Solution

Given information:

The expression $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\cdots \left(1+\frac{1}{n}\right)=n+1$ .

Formula used:

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

Step 1. Show that the statement is true for $n=1$ .

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 expression $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\cdots \left(1+\frac{1}{n}\right)=n+1$ .

Denote the expression by $S_n$ .

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

Step 1. Show that the statement is true for $n=1$ .

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 $n=1$ in the expression,

  $\begin{array}{c}\left(1+\frac{1}{1}\right)=1+1\\ 1+1=1+1\\ 2=2\end{array}$

The above equation holds true.

Therefore, the statement is true for $n=1$ so $S_1$ is true.

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

  $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\cdots \left(1+\frac{1}{k}\right)=k+1$  ...(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 $n=k+1$ . Group the terms that involve $S_k$ .

  $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\cdots \left(1+\frac{1}{k}\right)\left(1+\frac{1}{k+1}\right)=S_k\left(1+\frac{1}{k+1}\right)$

Substitute the value of $S_k$ from equation (1),

  $\begin{array}{c}\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\cdots \left(1+\frac{1}{k}\right)\left(1+\frac{1}{k+1}\right)=S_k\left(1+\frac{1}{k+1}\right)\\ =\left(k+1\right)\left(1+\frac{1}{k+1}\right)\\ =\left(k+1\right)\left(\frac{k+1+1}{k+1}\right)\\ =\left(k+1\right)+1\end{array}$

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

  $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\cdots \left(1+\frac{1}{k}\right)\left(1+\frac{1}{k+1}\right)=\left(k+1\right)+1$

Hence, by the principle of mathematical induction the statement $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\cdots \left(1+\frac{1}{n}\right)=n+1$ is true for all integers $n\ge 1$ .