How to show that this sequence is monotonic increasing? 
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?

A possible way is to investigate the derivative of the function underlying the recursion and then use MVT together with induction:


*

*$\boxed{f(x)} = k\frac{1+x}{k+x}= \frac{k+x + (k-1)x}{k+x}= 1 + (k-1)\frac{x}{k+x} = \boxed{1+ (k-1)\left(1 - \frac{k}{k+x} \right)}$

*Note that $f(x) \geq 1$ for $x \geq 0$.

*$f'(x)= \frac{k(k-1)}{(k+x)^2} >0$ for $k>1$

*The induction start gives
$$a_2 = \frac{2k}{k+1} = 1+\frac{k-1}{k+1}>1 = a_1$$

*The induction step uses the MVT and the positivity of $f'$:
$$a_{n+1}-a_n \stackrel{a_{n-1}< \xi_n < a_n}{=} \underbrace{f'(\xi_n)}_{>0}(a_n - a_{n-1}) > 0$$
So, the sequence is strictly increasing.
