Try this problem based on convergence from TIFR 2013, Problem 15.
Question: TIFR 2013 problem 15
True/False?
Let \(x_1\in(0,1)\). For \(n>1\) define \(x_{n+1}=x_n-x_n^{n+1} \) Then \(lim_{n\to \infty} x_n \) exists.
Hint: A bounded monotone sequence is convergent.
Discussion:
By induction, we can prove that the sequence is between 0 and 1 always.
Suppose, \(x_n\in(0,1)\). Since we are removing a positive number from \(x_n\) to get \(x_{n+1}\), we have \(x_{n+1}<x_n<1\). This also shows that the sequence is decreasing. And since for a number in between 0 and 1, when we take positive powers it decreases, in other words \(x_n>x_n^{n+1}\) we have \(x_{n+1}>0\).
Therefore the given sequence is decreasing and bounded below by 0. Hence it is convergent.
