Chapter 12, Problem 14CT

To determine

To prove: The statement $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\mathrm{...}\left(1+\frac{1}{n}\right)=n+1$ is true for all natural numbers by using the Principle of Mathematical Induction.

Explanation of Solution

Given information:

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

Formula used:

The principle of mathematical induction:

The following two conditions are satisfied by given statement about natural numbers.

Condition I: The statement is true for the natural number $k=1$.

Condition II: The statement is true for some natural number $k$ and it can be shown to be true for the next natural number $k+1$.

Then the statement is true for all natural numbers.

Proof:

Let us first prove that the statement holds for $n=1$.

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

The equality is true for $n=1$.

Next, assume that $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\mathrm{...}\left(1+\frac{1}{n}\right)=n+1$ is true for some k and check whether the formula holds for $k+1$.

Assume,

$\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\mathrm{...}\left(1+\frac{1}{k}\right)=k+1$ … Induction assumption

To show,

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

Consider,

$\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\mathrm{...}\left(1+\frac{1}{k}\right)\left(1+\frac{1}{k+1}\right)$

$=\left[\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\mathrm{...}\left(1+\frac{1}{k}\right)\right]\left(1+\frac{1}{k+1}\right)$

$=\left(k+1\right)\left(1+\frac{1}{k+1}\right)$….using the induction assumption

$=\left(k+1\right)\cdot 1+\left(k+1\right)\cdot \frac{1}{k+1}$

$=k+1+1$

$=k+2$

Thus the statement $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\mathrm{...}\left(1+\frac{1}{n}\right)=n+1$ is true for all natural numbers.