For $(x,y)\in R^2$, consider the series $\lim_{n \to \infty} \sum_{l,k=0}^n \frac{k^2x^ky^l}{l!}$. Then the series converges for $(x,y)$ in $ \\ 1.(-1,1)\times(0,\infty)\\ 2. R\times(-1,1) \\ 3. (-1,1)\times(-1,1) \\ 4. R\times R $
I considered the series separately. The series $\sum_{k=0}^n {k^2x^k}$ converges for $x<1$ and the series $\sum_{l=0}^n \frac{y^l}{l!}$ converges in (0,$\infty$) and in $[-1,1]$. I don't how to solve further, can we apply Able's test here?
As $n \to \infty$, the series $\sum_{k=0}^n {k^2x^k} $ converges for $|x| < 1$ and $\sum_{l=0}^n \frac{y^l}{l!} $ converges for all finite $y$ so I would say that the result is $(-1, 1) \times R $.