proof: the sequence $\{x_n\}_{n=1}^{\infty}$ converges, and find $\lim x_n$. (check).

Question: Suppose $x_1= \frac 12$ and $x_{n+1}=x_n^2$. Show that $\{x_n\}$ converges and find $\lim x_n$. Hint: you cannot divide by zero!

From the question, I know that $\{x_n\}=\{\frac 12, \frac 14, \frac 1{16}, ....\} = \{\frac 1{2^{2^n}}: n\in N\}$.

Then, $\lim x_n = 0$. So, to test convergence, let $\varepsilon > \frac 1{2^{2^M}}$, and $M \in N$. Then, for $\varepsilon >0$, there exists $M$ such that$|\frac 1{2^{2^n}}-0|<\varepsilon $ for $n\ge M$ because $2^{2^M}\le2^{2^n} \rightarrow \frac 1{2^{2^M}}\ge \frac 1{2^{2^n}} \rightarrow \frac 1{2^{2^n}}<\varepsilon$.

Could you check this proof is fine? I am not sure about my answer since I don't know when I make use of hint.

Thank you in advance.

I edited. Thanks for pointing out. Could you check anything wrong??

  • $\begingroup$ You might want to check the third term of your sequence... it certainly isn't $1/2^3$! $\endgroup$
    – user296602
    Mar 20, 2018 at 1:30
  • $\begingroup$ You made a mistake in the calculation of your sequence it's $\{x_n\} = \{ \frac 12, \frac 14, \frac 1{16}, \frac 1{16^2} ..... \}$ $\endgroup$
    – 3nondatur
    Mar 20, 2018 at 1:32
  • $\begingroup$ Further, your statement "$\lim x_n = 0$" is what you are trying to prove, so you can't use it in your next steps, but you are pretty close. Your aim is to find an $N$ for every $\varepsilon$ such that $|x_n - 0| < \varepsilon$. Try to do that. $\endgroup$
    – 3nondatur
    Mar 20, 2018 at 1:39
  • $\begingroup$ I was also thinking that it is wrong to use $\lim x_n =0$ before I prove the convergence. But, without assuming $\lim x_n =0$, is it allowed to say $|x_n-0|< \varepsilon$?? $\endgroup$
    – shk910
    Mar 20, 2018 at 1:44
  • $\begingroup$ yeah, because you suppose it, then the whole estimation thing with the $|x_n-0|< \varepsilon$ term is the proof to your assumption. $\endgroup$
    – 3nondatur
    Mar 20, 2018 at 1:52

2 Answers 2


Careful! $$ x_{n+1}= x_n^2 $$

This $$\{x_n\}=\{\frac 12, \frac 14, \frac 18, ....\} = \{\frac 1{2^n}:n>0, n\in N\}$$

is not correct.

You do not have a $1/8$ after $1/4$

you get $ (1/4)^2= 1/16 $

Also your sequence is not

$$\{\frac 1{2^n}:n>0, n\in N\}$$


You can use the well known Bolzano's lemma: "A monotonic decreasing sequence that is bounded below is convergent ". The boundedness is clear since $x_n > 0, \forall n \ge 1$. The monotone part is also clear since $x_{n+1} = x_n^2 < x_n, \forall n \ge 1$ since you can check by induction that $x_n \le 1/2, \forall n \ge 1$. From this the lemma is invoked and it says: $L = L^2 \implies L = 0$ or $L = 1$. Since $L \le 1/2, L = 0$ .


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