Q:

in which subset of $R^2$ the series is convergent?

For $(x,y) \in \Bbb R^2 $ ,consider the series $\lim_{n \to \infty } \sum_{l,k=o}^n \frac{k^2x^ky^l}{l !} $ .Then the series converges for $ (x,y)$ in 
1.$(-1,1)\times (0, \infty )$                     2.$\Bbb R \times (-1,1)$ 
3.$(-1,1)\times(-1,1)$                            4.$\Bbb R \times \Bbb R $
I don't know how to find the convergence of series in $\Bbb R^2$ . 

A:

Let $P_n(y) = \sum_{l=0}^{n} y^l\,/l!.$ We can write the sum as
$$\sum_{k=0}^{n }\sum_{l=0}^{n }k^2x^k\frac{y^l}{l!} = P_n(y)\sum_{k=0}^{n}k^2x^k.$$
As $n\to \infty,$ $P_n(y) \to e^y$ for all $y\in \mathbb R.$ If $x \in (-1,1),$ then $\sum_{k=0}^{n }k^2x^k$ converges (absolutely). Thus the domain of convergence as $n\to \infty$ is at least $(-1,1)\times \mathbb R.$ That implies none of the answers on offer describes the domain of convergence exactly (although it is true that the sets in 1. and 3. both lead to convergence).

