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 !!
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.
Limit
Sequence
Linear Difference Equation
(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 , \( -1 < \lambda -1 < 0 \) .
\(\mathrm{m}\) 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 \( a_{1} a_{2} \ldots a_{m}=b_{1} b_{2} . . b_{k} \) .
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