Show that $a_n$ is convergent.

Question

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?

Answer 1

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 $\sqrt2$.

Answer 2

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 $$ a_n^2-a_1^2=\sum_{i=1}^{n-1}\frac{1}{2^i}={1-0.5^n} $$