Let , $x_1 \in (0,1)$ be a real number. For $n>1$ define $x_{n+1}=x_n-x_n^{n+1}$. Then prove that $\displaystyle \lim_{n \to \infty} x_n$ exists.

We have to prove that the given sequence $\{x_n\}$ is convergent. So we have to show that $\{x_n\}$ is monotone and bounded.

I proved that the sequence is monotone decreasing. But I'm unable to show that it is bounded below. How can I show it ?

Any other way to prove that the limit exists ?


We show by induction that $x_n \in (0,1)$ for all $n$:

The case $n=1$ is clear.

Now let $n \in \mathbb N$ and $x_n \in (0,1)$

Then: $x_{n+1}=x_n(1-x_n^n)$. From $x_n \in (0,1)$ we get $x_n^n \in (0,1)$ and therefore $1-x_n^n \in (0,1)$.

Consequence: $x_{n+1} \in (0,1)$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.