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This is a very beautiful sample problem from ISI MStat PSB 2009 Problem 2 based on Convergence of a sequence. Let’s give it a try !!

## Problem– ISI MStat PSB 2009 Problem 2

Let $\{x_{n}: n \geq 0\}$ be a sequence of real numbers such that
$x_{n+1}=\lambda x_{n}+(1-\lambda) x_{n-1}, n \geq 1,$ for some $0<\lambda<1$
(a) Show that $x_{n}=x_{0}+(x_{1}-x_{0}) \sum_{k=0}^{n-1}(\lambda-1)^{k}$
(b) Hence, or otherwise, show that $x_{n}$ converges and find the limit.

### Prerequisites

Limit

Sequence

Linear Difference Equation

## Solution :

(a) We are given that $x_{n+1}=\lambda x_{n}+(1-\lambda) x_{n-1}, n \geq 1,$ for some $0<\lambda<1$

So, $x_{n+1} – x_{n} = -(1- \lambda)( x_n-x_{n-1})$ —- (1)

Again using (1) we have $( x_n-x_{n-1})= -(1- \lambda)( x_{n-1}-x_{n-2})$ .

Now putting this in (1) we have , $x_{n+1} – x_{n} = {(-(1- \lambda))}^2 ( x_{n-1}-x_{n-2})$ .

So, proceeding like this we have $x_{n+1} – x_{n} = {(-(1- \lambda))}^n ( x_{1}-x_{0})$ for all $n \geq 1$ and for some $0<\lambda<1$—- (2)

So, from (2) we have $x_{n} – x_{n-1} = {(-(1- \lambda))}^{n-1} ( x_{1}-x_{0})$ , $\cdots , (x_2-x_1)=-(\lambda-1)(x_1-x_{0})$ and $x_1-x_{0}=x_{1}-x_{0}$

Adding all the above n equation we have $x_{n}-x_{0}=(x_{1}-x_{0}) \sum_{k=0}^{n-1} {(\lambda-1)}^{k}$

Hence , $x_{n}=x_{0}+(x_{1}-x_{0}) \sum_{k=0}^{n-1}(\lambda-1)^{k}$ (proved ) .

(b) As we now have an explicit form of $x_{n}=x_{0}+(x_{1}-x_{0}) \times \frac{1-{( \lambda -1)}^n}{1-(\lambda -1)}$ —-(3)

Hence from (3) we can say $x_{n}$ is bounded and monotonic ( verify ) so , it’s convergent .

Now let’s take $\lim_{n\to\infty}$ both side of (3) we get , $\lim_{x\to\infty} x_{n} = x_{0}+(x_{1}-x_{0}) \times \frac{1}{2 – \lambda}$ .

Since , $\lim_{x\to\infty} {( \lambda – 1)}^{n} = 0$ as , $-1 < \lambda -1 < 0$ .

## Food For Thought

$\mathrm{m}$ and $\mathrm{k}$ are two natural number and $a_{1}, a_{2}, \ldots, a_{m}$ and $b_{1}, b_{2}, \ldots, b_{k}$ are two sets of positive real numbers such that $a_{1}^{\frac{1}{n}}+a_{2}^{\frac{1}{n}}+\cdots+a_{m}^{\frac{1}{n}}$ = $b_{1}^{\frac{1}{n}}+\cdots+b_{k}^{\frac{1}{n}}$

for all natural number $\mathrm{n} .$ Then prove that $\mathrm{m}=\mathrm{k}$ and $a_{1} a_{2} \ldots a_{m}=b_{1} b_{2} . . b_{k}$ .