real analysis - Find $sum_{n=1}^{infty} x^{leftlfloor {n over 2}rightrfloor} y^{leftlfloor {n + 1 over 2}rightrfloor}$

Let $x,y > 0, xy <1$. Find the sum

$$\sum_{n=1}^{\infty} x^{\left\lfloor {n \over 2}\right\rfloor} y^{\left\lfloor {n + 1 \over 2}\right\rfloor}$$

While I have some ideas how to test convergence, I don't quite know how to get started on the actual sum.

Edit:
If we break the sum down to a sum of two infinite series:

$$(y + xy^2 + x^2y^3 + \dots) + (xy + x^2y^2 + \dots)$$

and use the formula for the sum of the first n terms of a geometric series twice, we get:

$$y {1 - (xy)^n \over1 - xy} + xy {1 - (xy)^n \over1 - xy} \to {y \over1 - xy} + {xy \over1 - xy} = {y(1+x) \over 1 - xy}$$
Is this reasoning correct?

Answer

Note that for $n \in \mathbf N$ we have
$$\def\fl#1{\left\lfloor#1\right\rfloor}\fl{\frac n2} + 1 = \fl{\frac{n+2}2}$$
and
$$\fl{\frac{n+1}2}+1 = \fl{\frac{n+3}2}$$
So, if we call the sum $s := \sum_{n=1}^\infty x^{\fl{n/2}}y^{\fl{(n+1)/2}}$ and assume it converges we have,

$$\begin{align*} xys &= \sum_{n=1}^\infty x^{\fl{(n+2)/2}}y^{\fl{(n+3)/2}}\\ &= \sum_{k=3}^\infty x^{\fl{k/2}}y^{\fl{(k+1)/2}}\\ &= s - x^0y^1 - x^1y^1\\ &= s - y(1+x) \end{align*}$$
Now solve for $s$, we have

$$xys = s - y(1+x) \iff s(1-xy) = y(1+x) \iff s = \frac{y(1+x)}{1-xy}$$

If you do not want to assume convergence, you can do the following: Note that
$$\begin{align*} \sum_{n=1}^\infty x^{\fl{n/2}}y^{\fl{(n+1)/2}} &= \sum_{k=1}^\infty x^{\fl{(2k-1)/2}}y^{\fl{2k/2}} + \sum_{k=1}^\infty x^{\fl{2k/2}}y^{\fl{(2k+1)/2}}\\ &= \sum_{k=1}^\infty x^{k-1}y^k + \sum_{k=1}^\infty x^k y^k\\ &= y\sum_{k=0}^\infty (xy)^k + xy \sum_{k=0}^\infty (xy)^k\\ &= (y+xy) \sum_{k=0}^\infty (xy)^k\\ &= \frac{y(1+x)}{1-xy} \end{align*}$$