Problem source

Suppose that
$$3=\frac{2}{ x_1}=x_1+\frac{2}{ x_2}
=x_2+\frac{2}{ x_3}=x_3+\frac{2}{ x_4}=\cdots.$$
Guess an expression, in terms of $n$, for $x_n$.
Then, by induction or otherwise,
prove the correctness of your guess.

Solution source

\begin{align*}
x_1 &= \frac{2}{3} \\
x_n &= \frac{2}{3-x_{n-1}} \\
x_2 &= \frac{2}{3 - \frac23} \\
&= \frac{6}7 \\
x_3 &= \frac{2}{3-\frac67} \\
&= \frac{14}{15} \\
x_4 &= \frac{2}{3 - \frac{14}{15}} \\
&= \frac{30}{31}
\end{align*}

Guess: $x_n = \frac{2^{n+1}-2}{2^{n+1}-1}$.

Proof: (By induction)

(Base case): We have checked several initial cases.

(Inductive step): Suppose our formula is true for $n = k$, then consider:

\begin{align*}
x_{k+1} &= \frac{2}{3 - x_{k}} \\
&= \frac{2}{3 - \frac{2^{k+1}-2}{2^{k+1}-1}}\tag{assumption} \\
&= \frac{2\cdot(2^{k+1}-1)}{3 \cdot(2^{k+1}-1) - (2^{k+1}-2) } \\
&= \frac{2^{k+2}-2}{2\cdot 2^{k+1} - 3 + 2 } \\
&= \frac{2^{k+2}-2}{ 2^{k+2} - 1 } \\
\end{align*}

Therefore, if our formula is true for $n = k$ it is true for $n = k+1$. Therefore by the principle of mathematical induction it is true for $n \geq 1, n \in \mathbb{Z}$