Find $\sum_{n=1}^{\infty} x^{\left\lfloor {n \over 2}\right\rfloor} y^{\left\lfloor {n + 1 \over 2}\right\rfloor}$ 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?
 A: 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*}
