Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 1 Exercise 1.5

Solution: The $n$-th proposition is

$$

P_n: 1+\frac{1}{2}+\cdots+\frac{1}{2^n}=2-\frac{1}{2^n}.

$$ Then $P_1$ asserts $1+\dfrac{1}{2}=2-\frac{1}{2^1}$ which is clearly true and we have the induction basis.

Now we assume $P_n$ is true, that is the equation

\begin{equation}\label{eq:1-5-1}

1+\frac{1}{2}+\cdots+\frac{1}{2^n}=2-\frac{1}{2^n}.

\end{equation} We would like to show $P_{n+1}$ is true based on $P_n$. We add both sides of \eqref{eq:1-5-1} by $\dfrac{1}{2^{n+1}}$ and obtain

\begin{align*}

&\ 1+\frac{1}{2}+\cdots+\frac{1}{2^n}+\dfrac{1}{2^{n+1}}\\

=&\ 2-\frac{1}{2^n}+\dfrac{1}{2^{n+1}}=2-\frac{2}{2^{n+1}}+\dfrac{1}{2^{n+1}}\\

=&\ 2-\frac{1}{2^{n+1}}.

\end{align*} Therefore $P_{n+1}$ is true if $P_n$ is true. By the principle of mathematical induction, we make the conclusion that $P_n$ is true for all positive integers $n$.