Show that the sequence converges to a limit $L$ 
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 $\left\{a_{n}\right\}$ converges.
  (b) Find $\lim a_{n}$

I have no problem finding the limit by taking the limit of both sides and then solving for $L$. I am not sure how to go about showing that it does in fact converge?
I tried induction on the recurrence but am unable to find a better form. 
Edit:
Assume $a_n \geq a_{n-1}$ then $a_{n+1} = \frac{k(1+a_n)}{k+a_n} \geq \frac{k(1+a_{n-1})}{k+a_n}....$
 A: Manipulating the recurrence relation, it is easy to check that
$$ a_{n+1}-a_n=\frac{k-a_n^2}{k+a_n}\qquad\text{and}\qquad a_{n+1}-\sqrt{k}=\frac{k-\sqrt{k}}{k+a_n}(a_{n}-\sqrt{k}). \tag{*}$$
Now we inductively prove the following claim.

Claim. both $0 \leq a_n \leq \sqrt{k}$ and $[(n = 1) \text{ or } (a_{n-1} \leq a_n) ]$ hold for all $n \geq 1$.

(Remark. The weird logical connective $[(n = 1) \text{ or } (a_{n-1} \leq a_n) ]$ is introduced simply because it makes the proof shorter. Indeed, this allows us to ignore the inequality $a_{n-1} \leq a_n$ when $n = 1$. One can as well work with a more natural claim $0 \leq a_n \leq a_{n+1} \leq \sqrt{k}$, but this may result in repeating the same argument in both the base step and the induction step.)


*

*Base case. When $n = 1$, we only need to check whether $0 \leq a_1 \leq \sqrt{k}$ holds or not, which is obvious from the assumption on $a_1$ and $k$.

*Induction step. Assume that the claim holds for $n \geq 1$. Then we know that $0 \leq a_n \leq \sqrt{k}$. Now the first identity of $\text{(*)}$ shows
$$a_{n+1} - a_{n} = \frac{k-a_n^2}{k+a_n^2} \geq 0$$
and hence $a_{n+1} \geq a_n \geq 0$. Also, the second identity tells that
$$a_{n+1}-\sqrt{k}=\frac{k-\sqrt{k}}{k+a_n}(a_n-\sqrt{k}) \leq 0,$$
and so, $a_{n+1} \leq \sqrt{k}$. Combining altogether, we have proved that $0 \leq a_{n+1} \leq \sqrt{k}$ and $ a_n \leq a_{n+1}$. So the induction step is established.
Therefore by the principle of mathematical induction, the claim holds for all $n$ and hence $(a_n)$ is non-decreasing and bounded.
