TITLE: Show that $a_n$ is convergent.
QUESTION [5 upvotes]: 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?

REPLY [3 votes]: 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
$$
a_n^2-a_1^2=\sum_{i=1}^{n-1}\frac{1}{2^i}={1-0.5^n}
$$