Chapter 9.4, Problem 20E

To determine

To prove: The formula $\left(1+\frac{1}{1}\right)+\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)\mathrm{.......}\left(1+\frac{1}{n}\right)=n+1$ for every integer $n$ . Using mathematical induction.

Explanation of Solution

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

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

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

Put $n=1$

LHS: $P\left(1\right)=2$

RHS: $\begin{array}{l}n+1\\ =1+1=2\end{array}$

  $LHS=RHS=2$

  $P\left(n\right)$ is true for $n=2$

So, $P\left(k\right)$ is true for $n=k$ where, $k\in Z^{+}$ .

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

Now, prove for $P\left(k+1\right)$ .

Put $n=k+1$

For LHS,

  $\begin{array}{c}P\left(k+1\right)=\left(1+\frac{1}{1}\right)+\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)\mathrm{.......}\left(1+\frac{1}{k+1}\right)\\ =\left(1+\frac{1}{1}\right)+\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)\mathrm{.......}\left(1+\frac{1}{k}\right)\left(1+\frac{1}{k+1}\right)\\ =k+1\left(1+\frac{1}{k+1}\right)\left[\text{Using}\left(i\right)\right]\\ =(k+1)\left(\frac{k+1+1}{k+1}\right)\\ =k+2\end{array}$

For RHS,

  $k+2$

  $LHS=RHS=k+2$

Hence, $P\left(k+1\right)$ is true.