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.

  • $\begingroup$ Why duplicate? $\endgroup$
    – metamorphy
    Oct 20, 2019 at 19:34
  • $\begingroup$ @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 $\endgroup$
    – Jac Frall
    Oct 20, 2019 at 20:15

2 Answers 2


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 $L$

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

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