Given $y_{k+1} = 1 + \sqrt{y_k}$ for $k \geq 0$ and $y_0 = 0$, we have a limit of the form $L = \lim_{k \rightarrow \infty} y_k = 1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + ... \sqrt{1 + \sqrt{2}}}}}$.
Apparently this can be rewritten as $L = \frac{3 + \sqrt{5}}{2}$, but I am at a loss for how to approach this. Tips would be greatly appreciated.
I see that $L = \frac{3 + \sqrt{5}}{2} = 1 + \frac{1}{2} + \frac{\sqrt{5}}{2}$, but this doesn't really take me very far.
On
If the limit of $\{y_n\}$ does exist, then you should know that $\lim_{n\to \infty}y_{k+1}= \lim_{n\to \infty}y_{k}.$ So, you will have $L=1+\sqrt L$ if suppose that $\lim_{n\to \infty}y_{k}=L$.
On
To show that the limit actually exists, we can proceed as follows:
Let $$ \textstyle a_n=\overbrace{1+\sqrt{1+\sqrt{1+\sqrt{1+\dots+\sqrt{1+\sqrt2}}}}}^{n\ 1\text{s}}\tag{1} $$ so that $$ (a_n-1)^2=a_{n-1}\tag{2} $$ Subtracting $(2)$ substituting $n\mapsto n-1$ from $(2)$ and dividing by $(a_n-1)+(a_{n-1}-1)$, gives $$ a_n-a_{n-1}=\frac{a_{n-1}-a_{n-2}}{(a_n-1)+(a_{n-1}-1)}\tag{3} $$ Since $a_1=1+\sqrt2\gt2=a_0$, $(3)$ says that $a_n$ is monotonically increasing Furthermore, we have that for $n\ge1$, $(a_n-1)+(a_{n-1}-1)\ge1+\sqrt2$. Therefore, $$ 0\le a_n-a_{n-1}\le\frac{a_{n-1}-a_{n-2}}{1+\sqrt2}\tag{4} $$ $(4)$ implies that $a_n$ is Cauchy, and therefore, converges.
Letting $$ a=\lim_{n\to\infty}a_n\tag{5} $$ if we take the limit of $(2)$, we get $$ (a-1)^2=a\tag{6} $$ and since $a_n\ge2$, we must have $$ a=\frac{3+\sqrt{5}}2\tag{7} $$
Hint: $L = 1 + \sqrt{L}$. Solve for $L$.