when is this series convergent. Let $(x,y)\in \mathbb{R^2}$ and consider the series 
$$\lim_{n\rightarrow \infty}\sum_{\ell,\,k=0}^n \frac{k^2 x^k y^\ell}{\ell!}$$
Then for what values $(x,y)$ the series converges. I have absolutely no idea how to deal with it.
 A: $$
\sum_{\ell=0}^n \left( \sum_{k=0}^n \frac{k^2 x^k y^\ell}{\ell!} \right) 
= \sum_{\ell=0}^n \left( \frac{y^\ell}{\ell!} \sum_{k=0}^n k^2 x^k \right)
$$
The step above can be done because $\dfrac{y^\ell}{\ell!}$ does not change as $k$ goes from $0$ to $n$.
This next step can be done because $\displaystyle\sum_{k=0}^n k^2 x^k$ does not change as $\ell$ goes from $0$ to $n$:
$$
\sum_{\ell=0}^n \left( \frac{y^\ell}{\ell!} \left( \sum_{k=0}^n k^2 x^k \right) \right) = \left( \sum_{\ell=0}^n \frac{y^\ell}{\ell!} \right) \left( \sum_{k=0}^n k^2 x^k \right)
$$
Now you just need to consider them separately.  A ratio test does the first one instantly if you know things like $52!/53! = 1/53$.  A ratio test also handles the second one if you know how to find things like $\displaystyle \lim_{k\to\infty} \dfrac{(k+1)^2}{k^2}$.
A: Simple calculation shows that the alternating series $$\sum_{k}^{\infty}( \sum_{l}^{\infty}\frac{k^{2}x^{k}y^{l}}{l!})$$ and $$\sum_{l}^{\infty}( \sum_{k}^{\infty}\frac{k^{2}x^{k}y^{l}}{l!})$$ are absolutely convergent for $(x,y)\in (-1,1)\times\mathbb{R}.$ So the options $1$ and $3$rd are correct. For option $2$ and $3$rd if we take $x=1$, series is not convergent as mention in question .
