Formula for series $frac{sqrt{a}}{b}+frac{sqrt{a+sqrt{a}}}{b}+cdots+frac{sqrt{a+sqrt{a+sqrt{cdots+sqrt{a}}}}}{b}$

All variables are positive integers.

For:

$$a_1\qquad\frac{\sqrt{x}}{y}$$
$$a_2\qquad\frac{\sqrt{x\!+\!\sqrt{x}}}{y}$$
$$\cdots$$
$$a_n\qquad\frac{\sqrt{x\!+\!\sqrt{\!x+\!\sqrt{\!\cdots\!+\sqrt{x}}}}}{y}$$

Is there a formula of an unconditional form to describe series $a_n$?

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I thought of something along the lines of:

$$\sum _{k=1}^{n } \left(\sum _{j=1}^k \frac{\sqrt{x}}{y}\right)$$

but, I quickly realized that it was very incorrect; Then I thought of:

$$\sum _{k=1}^{n} \frac{\sum _{j=1}^k \sqrt{x}}{y}$$

which I also concluded as very incorrect...

I'm blank, but I would like to see an example of something along the lines of:

$$\sum _{k=1}^{n } \frac{\sqrt{x+\sqrt{x+\sqrt{\cdot\cdot\cdot+\sqrt{x}}}}}{y}$$

where each $\sqrt{x+\sqrt{\cdots}}$ addition, repeats $k$ times. (i.e $k=3 \Rightarrow \sqrt{x+\sqrt{x+\sqrt{x}}}$);
If it is possible...

Cheers!

Answer

If all you are looking for is a compact representation, let
$$s_{k}=\begin{cases} 0 & \text{if }k=0\\ \sqrt{a+s_{k-1}} & \text{if }k>0 \end{cases}.$$
Then
$$\begin{align*} S_n & =\left(\frac{\sqrt{a}}{b}\right)+\left(\frac{\sqrt{a+\sqrt{a}}}{b}\right)+\ldots+\left(\frac{\sqrt{a+\sqrt{a+\ldots+\sqrt{a}}}}{b}\right)\\ & =\frac{1}{b}\left[\sqrt{a}+\sqrt{a+\sqrt{a}}+\ldots+\sqrt{a+\sqrt{a+\ldots+\sqrt{a}}}\right]\\ & =\frac{1}{b}\sum_{k=1}^{n}s_{k}. \end{align*}$$

Assume $a\in\mathbb{R}$ (you don't have to do this). We can show that the recurrence is
stable everywhere (weakly stable at $a=-\frac{1}{4}$). Particularly,
the fixed point is given by
$$s^{2}-s-a=0,$$
which has roots
$$\frac{1\pm\sqrt{1+4a}}{2}.$$
Particularly, the locally stable fixed point is the solution with $\pm$ is $+$.
So, for large enough $k$,
$$s_k\approx\frac{1+\sqrt{1+4a}}{2}.$$

This is as good an answer as you can hope for, save for error bounds on the above expression.