Cauchy recursive sequense

Question

I am wondering if this sequence is a Cauchy one or not?

$X_{n+1}=\beta+\underbrace{k\cdot\sqrt{d}}_{\alpha}\cdot X_n$.

Here, $\beta=1$, $\alpha\in (0,+\infty)$, $X_0=1$, $k\in (0,1)$ and $d\in(0,+\infty)$.

I understand that it is defined over $\mathbb{R}$ as a Banach space. Hence, my ultimate goal is to see if the sequence is convergent.

Answer 1

You should always try writing a few terms of the sequence out: \begin{align*} X_{0} & =1\\ X_{1} & =\beta+\alpha X_{0}=\alpha^{\phantom{1}}+\beta(1)\\ X_{2} & =\beta+\alpha X_{1}=\alpha^{2}+\beta(1+\alpha)\\ X_{3} & =\beta+\alpha X_{2}=\alpha^{3}+\beta(1+\alpha+\alpha^{2})\\ \text{etc.} \end{align*} The above suggests $$ X_{n}=\alpha^{n}+\beta\sum_{k=0}^{n-1}\alpha^{k} $$ (you can prove this by induction).

Now, the summation (from $k=0$ to $n-1$) is a geometric series, which has a closed form solution. If you plug that in, then you will have an expression for $X_{n}$ whose limit you can take.

Answer 2

Let me please try to prove the raised induction part of the problem as$\colon$

Prove $X_n=\alpha^n+\beta\sum_{k=0}^{n-1}{\alpha^k}$ for all $n\ge 0$.

We start by reformulating the problem so that we have$\colon$

Prove $\sum_{k=0}^{n-1}{\alpha^k}=\frac{X_n-\alpha^n}{\beta}$

We proceed by proof as$\colon$

• Induction basis$\colon$ In case $n=1$,

$\sum_{k=0}^{n-1}{\alpha^k}=\sum_{k=0}^{0}{\alpha^k}=\alpha^0=1$.

• Induction hypothesis: we assume in case $n\ge 0$,

$\sum_{k=0}^{n-1}{\alpha^k}=\frac{X_n-\alpha^n}{\beta}$

holds true.

• Induction step: we need to show:

$\sum_{k=0}^{n}{\alpha^k}=\frac{X_n-\alpha^n}{\beta}$

holds also true. In this direction, we take the left hand side of the latter equation and expand it as:

$\sum_{k=0}^{n}{\alpha^k}=\sum_{k=0}^{n-1}{\alpha^k}+\sum_{k=n}^{n}{\alpha^k}$

or further as:

$\sum_{k=0}^{n}{\alpha^k}=\underbrace{\frac{X_n-\alpha^n}{\beta}}_{\sum_{k=0}^{n-1}{\alpha^k}}+\sum_{k=n}^{n}{\alpha^k}$

or,

$\sum_{k=0}^{n}{\alpha^k}-\sum_{k=n}^{n}{\alpha^k}=\frac{X_n-\alpha^n}{\beta}$

that could be simplified as:

$\sum_{k=0}^{n-1}{\alpha^k}=\frac{X_n-\alpha^n}{\beta}$

A final re-arrangement leads us to:

$X_n=\alpha^n+\beta\sum_{k=0}^{n-1}{\alpha^k}$

that is true based on our induction hypothesis.