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# Convergence (TIFR-2013 problem 15)

Question:

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.

August 4, 2017