# How to show that sequence converges and find that point $L$

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 $$L=\pm\sqrt{k}$$. I am not sure how to go about showing that it does in fact converge and to which value the limit actually is?

I have tried showing it is bounded monotonic sequence but have not been able to. Also I wonder if trying to prove it is Cauchy would be useful?

This is for an intro Real Analysis course so we are using basic techniques.

• Why duplicate? Oct 20, 2019 at 19:34
• @metamorphy because things were still unclear and used techniques that we have not yet learned. The information I have received from both posts have been helpful Oct 20, 2019 at 20:15

It is given that $$a_1 = 1 < \sqrt{k}$$.

Assume $$a_n < \sqrt{k}$$. Then $$a_n < \sqrt{k} \Rightarrow (\sqrt{k}-1)a_n < (\sqrt{k}-1)\sqrt{k}=k-\sqrt{k} \\ \Rightarrow a_n+k > \sqrt{k} + \sqrt{k}a_n = \sqrt{k}(1 + a_n) \\ \Rightarrow \sqrt{k}(a_n+k) > k(1 + a_n) \\ \Rightarrow a_{n+1} = \frac{k(1 + a_n)}{a_n+k} < \sqrt{k}$$

By induction, we have proven $$a_n < \sqrt{k}$$ for all $$n \ge 1.$$

Furthermore, $$a_n < \sqrt{k} \Rightarrow k > a_n^2 \Rightarrow k(1+a_n) > ka_n + a_n^2=a_n(k+a_n) \\ \Rightarrow a_{n+1}=\frac{k(1+a_n)}{k+a_n} > a_n.$$

Hence $$a_n$$ is a monotonically increasing sequence bounded by $$\sqrt{k}$$, and it converges to a limit $$L$$.

And you already know how to find the limit, so I stop here (note $$a_n$$ > 0).

$$a_{n+1} \leq k,\forall n \geq 1$$ and $$a_n \geq 1,\forall n \in \Bbb{N}$$

Now the function $$f(x)=\frac{k(1+x)}{k+x}$$ is increasing on $$[1,+\infty)$$

So by induction you can prove that the sequence is increasing.

Since it is also bounded,it has a limit $$L$$

• How would you find that limit $L$? Oct 20, 2019 at 18:16
• Do you really need to use induction? Isn’t it given that the sequence is increasing if the function is? Oct 20, 2019 at 18:17
• @JacFrall just plug $L$ and solve as you did.. Oct 20, 2019 at 18:26
• @JacFrall you must unse induction to assume that $a_n \leq a_{n+1}$...then by monotonicity of the function you have that : $a_{n+1}=f(a_n) \leq f(a_{n+1})=a_{n+2}$ Oct 20, 2019 at 18:37