It's the sequence of approximations obtained when you approximate the perimeter of the circle of diameter $1$ with inscribed regular $n$-gons for $n$ a power of $2$.
As I happen to have this TeXed' up, I'll offer:
Suppose regular $2^n$-gons are inscribed in a circle of radius $r$.
Suppose the side length(the length of one "face") $a_n$ of a the inscribed $2^{n}$-gon is known (so, $a_2$ is the side length of the square).
To find the side length of the $2^{n+1}$-gon, one may apply the Pythagorean Theorem twice to obtain
$$
\tag{1}a_{n+1} = r\sqrt{2-\sqrt{4-{a_n^2\over r^2}}}
$$
Now, starting with a square, $$a_2=\sqrt 2 r.$$
Using the recursion formula (1) repeatedly gives:
$$
a_3%= r\sqrt{2-\sqrt{4-{2r^2\over r^2}}}
= r\sqrt{2-\sqrt2},
$$
$$
a_4%= r\sqrt{2-\sqrt{4-{ ( r\sqrt{2-\sqrt2})^2 \over r^2}}} = r\sqrt{2-\sqrt{4-{ ({2-\sqrt2} ) }}}
= r\sqrt{2-\sqrt{{ {2+\sqrt2} }}},
$$
and
$$
a_5%= r\sqrt{2-\sqrt{4-{ ( r\sqrt{2-\sqrt{{ {2+\sqrt2} }}} )^2\over r^2}}}
=
r\sqrt{2-\sqrt{ 2+\sqrt{{ {2+\sqrt2} }}} }.
$$
$$\vdots$$
Let $b_n=2^n a_n$. Let $P_n=r\cdot b_n$ be the perimeter of the $2^n$-gon. Let $P$ be the perimeter of the circle.
Then $$
\lim_{ n\rightarrow \infty} P_n = P.
$$
Note that from the above identity, it follows that the ratio of the perimeter of a circle to its diameter must be a constant, namely $\lim\limits_{n \rightarrow \infty} b_n$. We call this number $\pi$.
Below are some particular calculations when the radius of the circle is
$1/2$:
$$\eqalign{
P_2&=2^1\cdot\sqrt 2 \approx 2.82842712\cr
P_3&=2^2\cdot\sqrt{2-\sqrt2}\approx 3.06146746\cr
P_4&=2^3\cdot\sqrt{2-\sqrt{2+\sqrt2}}\approx3.12144515 \cr
P_5&=2^4\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt2}}}\approx 3.13654849\cr
P_6&=2^5\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt2}}}}\approx 3.14033116\cr
P_7&=2^6\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt2}}}}}\approx 3.14127725\cr
P_8&=2^7\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt2}}}}}}\approx 3.1415138 \cr
P_9&=2^8\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt2}}}}}}}\approx 3.14157294 \cr
}
$$
For completedness:
Remark 1:
Here is the proof that the recursion formula (1) holds:
Let $a_n$ be the side length of the $2^n$-gon.
To obtain the $2^{n+1}$-gon: take the "outer end point" of the radii of the circle that bisect the faces of the $2^n$-gon to form the new vertices of the $2^{n+1}$-gon.
We then have, for $a_{n+1}$, the scenario shown in the following diagram (not to scale):
Now $$
b^2=r^2-{a_n^2\over4};
$$
whence
$$\eqalign{
a_{n+1}^2={a_n^2\over4} + \Biggl((r-\sqrt{ r^2-{a_n^2\over4}}\ \Biggr)^2
&={a_n^2\over4}+ r^2-2r\sqrt{r^2-{a_n^2\over4}}+r^2 -{a_n^2\over4}\cr
&= 2r^2-2r\sqrt{r^2-{a_n^2\over4}}\cr
&= 2r^2-r^2\sqrt{4-{a_n^2\over r^2}}\cr
&= r^2 \Biggl(2-\sqrt{4-{a_n^2\over r^2}}\ \Biggr).}$$
And, thus
$$
a_{n+1}= r \sqrt{2-\sqrt{4-{a_n^2\over r^2}}}.
$$
Remark 2: To explain why limit $\lim\limits_{n\rightarrow\infty} P_n=P\ $ holds, I can do no better than refer you to Eric Naslund's comment in his answer.
See also, here.
Set $y = \sqrt{x + \sqrt{x + \dots}}$.
$y^2 = x+\sqrt{x+\dots} = x + y$
So $y^2 - y = x$.
But $\displaystyle y = \frac{1 + \sqrt{53}}{2}$
Now just substitute.
Addendum for people who want to see something a bit more rigorous.
We will take the infinite square root expression to be defined as the limit of the sequence defined recursively:
$$s(1) = \sqrt{x}$$
$$s(n+1) = \sqrt{x + s(n)}$$
We want to take $\lim_{n \to \infty} s(n)$.
For $x > 0, s(n) \le s(n+1)$ by induction on $n$:
$s(1) \le s(2)$ because $s(2) = \sqrt{x + s(1)} > \sqrt{x}$ since $s(1) > 0$.
If $s(n) \le s(n+1)$, then $s(n+2) = \sqrt{x + s(n+1)} \ge \sqrt{x+s(n)} = s(n+1)$.
So $s$ is monotonic.
Moreover, $s$ is bounded above by $\max(x,2)$. This will also be proven by induction.
$s(1) = \sqrt{x} \le \max(x,1) \le \max(x,2)$
If $s(n) \le 2$, then $s(n+1) = \sqrt{x+2} \le \sqrt{4} \le 2$.
Similarly, if $2 < s(n) \le x$, then $s(n+1) = \sqrt{x+s(n)} \le \sqrt{2x} \le \sqrt{x^2} = x$
Since $s$ is bounded and monotonic, it has a limit as $n \to \infty$.
Since $f(y) = \sqrt{ x + y }$ is continuous, $$\lim_{n \to \infty} s(n) = \lim_{n \to \infty} s(n+1) = \lim_{n \to \infty} \sqrt{ x + s(n) } = \sqrt{ x + \lim_{n \to \infty} s(n)}$$
Justifying the reasoning above.
Best Answer
Your edit suggests using induction. Define $f(0)=\sqrt 6, f(n+1)=\sqrt{6+f(n)}$ We would like to prove $f(n)\gt 3-\frac 1{5^n}$. Certainly this is true for $n=0$, as it asserts $\sqrt 6 \gt 2$, $n=1$ where it asserts $\sqrt{6+\sqrt 6} \approx 2.906 \gt 3-\frac 15$ and $n=2$ where $f(2)\approx 2.9844 \gt 3-\frac 1{5^2}$. Now assume it is true for $k$, then $f(k+1)=\sqrt {6+f(k)}\gt \sqrt{9-\frac 1{5^k}}=3\sqrt{1-\frac1{9\cdot 5^k}}\gt 3(1-\frac1{16\cdot 5^k})\gt3-\frac1{ 5^{k+1}}$
The next to last inequality comes from $\sqrt{1-\frac1{9\cdot 5^k}} \approx 1-\frac 1{2\cdot 9 \cdot 5^k} - \frac 1{8\cdot 9^2 \cdot 5^{2k}} \gt 1-\frac 1{16\cdot 5^k}$ because $\frac 1{16}-\frac 1{18}=\frac 1{144} $ which dominates all the remaining terms in the expansion.