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Define a sequence of real numbers $(a_n)$ by setting $a_1 = 1$ and $$a_{n+1}=\sqrt{(a_n)^{2}+\frac{1}{2^{n}}}$$

Prove that $a_n$ converges.

My idea: "We will show that $(a_n)$ is a Cauchy sequence so that it converges."

First, note that for any $n$, $a_n>0$. This is because $(a_n)^2 \geq 0$, and $\frac{1}{2^{n}}>0$ for any $n$.

And note that since $(a_{n+1})^2-(a_n)^2=\frac{1}{2^{n}}>0$, for any natural number $x,y$ that satisfies $x>y$, we have $a_x>a_y>a_1=1$.

Let $c_n=(a_{n+1})^2-(a_n)^2=\frac{1}{2^{n}}$.

Since $(c_n)$ converges to $0$, it is cauchy sequence.

So, $\forall \varepsilon >0, \; \; \exists N$ such that

$$|c_n-c_m|<\varepsilon \; \; \; \; \forall n>m>N$$

$$\Rightarrow \left | (a_{n+1})^{2}-(a_n)^2-(a_{m+1})^2+(a_m)^2 \right |<\varepsilon $$

$$\Rightarrow \left | (a_{n+1}+a_{m+1})(a_{n+1}-a_{m+1})-(a_n+a_m)(a_n-a_m) \right |< \varepsilon $$

I want to leave the left-hand side only with $(a_n-a_m)$ so that I could get a Cauchy criterion, but I'm stuck here.

Is this idea totally wrong? If so, may I have some suggestions to prove the statement?

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  • $\begingroup$ Hint : $$a_{n+1}^2-a_{n}^2 =\frac{1}{2^n}$$ $\endgroup$ May 6, 2021 at 8:31
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    $\begingroup$ I think that recursive sequences do not need the power of proof of Cauchy sequences. It is enough that you see that it is monotonous and bounded $\endgroup$
    – bravoralph
    May 6, 2021 at 8:31
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    $\begingroup$ Sticking to commenting on your proof : you want to show that $a_n$ is a Cauchy sequence. So you want something regarding $a_m - a_n$ for different $m$ and $n$. However, the problem with what follows the line : "$c_n$ is Cauchy" is that this statement is not strong enough to actually give you want you want : you'll only be able to get differences in increments of $a_n$, The truth is that $c_n$ has a far stronger property : it decays so fast to zero that $\sum_{n=1}^\infty c_n$ converges. This particular property should be used to show that $a_n$ is Cauchy... $\endgroup$ May 6, 2021 at 8:35
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    $\begingroup$ ... everything that includes and precedes "$c_n$ is Cauchy" is correct, useful for the problem and would be part of an ideal attempt to solve the problem. Good job, great presentation of answer, happy to see the correctness of computation as well. The fact that you proved monotonicity would mean that boundedness is enough as well, but I'll stick to critiquing your proof first, other answers have enough alternate suggestions. (+1) $\endgroup$ May 6, 2021 at 8:36

2 Answers 2

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You could use the Cauchy criterion to prove convergence, but it is not needed here. Define $b_n=a_n^2$, then from $b_{n+1}=b_n+2^{-n}$ we get $\lim_{n\to\infty}b_n=\sum_{k=0}^\infty2^{-k}=2$, so $a_n$ converges to $\sqrt2$.

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If you are not restricted to use the Cauchy sequences and the completeness property of $\mathbb R$, you can write $$ a^2_{n+1}-a_n^2=\frac{1}{2^n}, $$ where telescopic series gives $$ a_n^2-a_1^2=\sum_{i=1}^{n-1}\frac{1}{2^i}={1-0.5^n} $$

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