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# TIFR 2013 problem 15 | Problem on Convergence

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.

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