Let $k>1$ and define a sequence $\left\{a_{n}\right\}$ by $a_{1}=1$ and $$a_{n+1}=\frac{k\left(1+a_{n}\right) }{\left(k+a_{n}\right)}$$
(a) Show that $\left\{a_{n}\right\}$ is monotonic increasing.
Assume $a_n \geq a_{n-1}$. Then,
$$a_{n+1} = \frac{k(1+a_n)}{k+a_n} \geq \frac{k(1+a_{n-1})}{k+a_n}….$$
But I get hung up on the $a_n$ in the denominator. I cannot replace it with $a_{n-1}$ since $a_n \geq a_{n-1}$. Is there a trick to get around this?
Best Answer
Turn the question of whether $(a_n)$ is monotone increasing into an inequality purely in terms of a single term $a_n$. In particular, $$a_n \le a_{n+1} = \frac{k(1 + a_n)}{k + a_n}.$$ Simplifying, making the temporary assumption that $k + a_n > 0$, $$a_n(k + a_n) \le k(1 + a_n) \iff a_n^2 - k \le 0 \iff a_n \in [-\sqrt{k}, \sqrt{k}].$$ Addressing this assumption, it is extremely easy to show $a_n > 0$ for all $n$ by induction, hence $k + a_n > k > 1 > 0$. Thus, you really just need to show $a_n \le \sqrt{k}$.
You can now show this by induction. In fact, you can bundle the positivity proof in as well. That is, you can show that $0 \le a_n \le \sqrt{k}$ for all $n$ by induction.
Let's begin. Since $k \ge 1$, note that $0 \le 1 \le \sqrt{k}$, hence $0 \le a_1 \le \sqrt{k}$.
Now, assume $0 \le a_n \le \sqrt{k}$. Then, \begin{align*} a_{n+1} &= \frac{k(1 + a_n)}{k + a_n} \\ &= \frac{k + k^2 + ka_n - k^2}{k + a_n} \\ &= \frac{k(k + a_n)}{k + a_n} + \frac{k - k^2}{k + a_n} \\ &= k - \frac{k^2 - k}{k + a_n}. \end{align*} Note that $k^2 - k > 0$, hence $x \mapsto k - \frac{k^2 - k}{k + x} = \frac{k(1 + x)}{k + x}$ is increasing for $x > -k$. So, since $-k < 0 \le a_n \le \sqrt{k}$, we have $$\frac{k(1 + 0)}{k + 0} \le \frac{k(1 + a_n)}{k + a_n} \le \frac{k(1 + \sqrt{k})}{k + \sqrt{k}} \implies 1 \le a_{n+1} \le \sqrt{k},$$ as required.