Convergence (TIFR-2013 problem 15)



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.


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.

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