Nested Radical Strikes Again!

Assuming that the next fractions in the sequence are $\frac{7}{256}, \frac{7}{65536}, ....$ , we can rewrite the expression as

$X = \displaystyle\sqrt{7 - \dfrac{A}{\sqrt{2}}}$

where $A = \displaystyle\sqrt{7 + \sqrt{7 - \sqrt{7 + \sqrt{7 - ....}}}}$ .

Now $(A^{2} - 7)^{2} = 7 - A$ , which by observation has $A = 3$ as a solution. (There is another positive root, namely $\frac{1}{2} \sqrt{-1 + \sqrt{29}}$ , but since this root is less than $\sqrt{7}$ it can be discarded.)

So this leaves us with $X = \displaystyle\sqrt{7 - \dfrac{3}{\sqrt{2}}} = \boxed{2.208773}$ to $6$ decimal places.

Let $S=\sqrt{7-\frac{1}{\sqrt{2}}\sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7+\ldots}}}}}$

Let $\sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7+\ldots}}}}=x$

also, let $\sqrt{7-\sqrt{7+\sqrt{7+\ldots}}}=y$

then, we see that

$\sqrt{7+y}=x~~~~~(1)$

and

$\sqrt{7-x}=y~~~~~(2)$

squaring both $(1)$ and $(2)$ and subtract to get:

$y+x=x^2-y^2=(x+y)(x-y)\\1=x-y\implies{y=x-1}$

Substitution $y=x-1$ to $(1)$ , we have $x^2-x-6=0$ . So, we have $x=3$ and $x=-2$ . But, since $S$ is positive, then $x$ should be positive, so $x=-2$ is discarded.

Finally, $S=\sqrt{7-\frac{1}{\sqrt{2}}\times{3}}=\boxed{2.20877}$