Radius of convergence of $\lim_{n \to +∞} \sum_{k , l = 0}^n \frac{k^2 x^k y^l}{l!}$. Radius of convergence of $$\lim\limits_{n \to +∞} \sum_{k , l = 0}^n \frac{k^2 x^k y^l}{l!}.$$       
The domain of $(x,y)$ in $\mathbb{R}^2$ is to be found where the power series converges.  I have the knowledge of one variable power series. But I can not solve this problem. 
Can you please help me by letting me know what to read to be able to solve thi problem?
 A: The sum decouples to
$$ \sum_{k,l=0}^{n} \frac{k^2 x^k y^l}{l!} = \left( \sum_{k=0}^{n} k^2 x^k \right)\left( \sum_{l=0}^{n} \frac{y^l}{l!} \right). $$
Since the second sum converges to $e^y$ for any $y\in\mathbb{R}$ and $e^y$ is non-zero, the limit converges if and only if the first sum converges, which happens exactly when $|x| < 1$. So the set of pairs $(x, y) \in \mathbb{R}^2$ for which the limit converges is $(-1, 1)\times \mathbb{R}$.
A: It might be helpful to start by writing the single sum as an iterated summation:
$$
\lim_{n\to\infty} \sum_{k , \ell = 0}^n \frac{k^2 x^k y^\ell}{\ell!}
= \lim_{n\to\infty} \sum_{k=0}^{n} \sum_{\ell=0}^{n}  \left( \frac{k^2 x^k y^\ell}{\ell!} \right). $$
The inner sum depends only on $\ell$ (and not on $k$), hence we can factor out all of the terms that don't involve $\ell$.  This gives us
$$
\lim_{n\to\infty} \sum_{k=0}^{n} \sum_{\ell=0}^{n}  \left( \frac{k^2 x^k y^\ell}{\ell!} \right)
= \lim_{n\to\infty} \sum_{k=0}^{n} \left[ k^2x^k \sum_{\ell=0}^{n} \frac{y^{\ell}}{\ell!} \right].
$$
The outer sum depends only on $k$ (and not $\ell$), thus we can again factor in order to get
$$
\lim_{n\to\infty} \sum_{k=0}^{n} \left[ k^2x^k \sum_{\ell=0}^{n} \frac{y^{\ell}}{\ell!} \right]
=\lim_{n\to\infty} \left[ \sum_{k=0}^{n} k^2x^k \right] \left[ \sum_{\ell=0}^{n} \frac{y^{\ell}}{\ell!} \right].
$$
Now, recall the elementary result that if $a_n \to a$ and $b_n \to b$, then $a_n b_n \to ab$.  If each of the two series above converges, then the original series will converge to the product of those two series.  Note that this is not an "if and only if" statement, and that is it (in principle) possible for the product series to converge even if both of the series involved diverge, or if one of the series diverges, but the other converges to zero fast enough.
The second series is the Taylor series for the exponential function, which converges for all $y\in\mathbb{R}$.  Thus we have
$$
\lim_{n\to\infty} \sum_{\ell=0}^{n} \frac{y^{\ell}}{\ell!} = \mathrm{e}^{y},
\qquad\forall y\in\mathbb{R}. $$
Indeed, for $y\in\mathbb{R}$, this series will always converge to a positive real number.  Thus the only circumstance in which the original series will diverge is if the series depending on $x$ and $k$ diverges, thus the original series converges if and only if $\sum k^2 x^k$ converges.  To deal with this series, recall that the radius of convergence of a power series $\sum a_k x^k$ is given by
$$ \frac{1}{R} = \limsup |a_k^{1/k}|.  $$
Here, the coefficients are given by $a_k = k^2$, thus we have
$$ \frac{1}{R}
= \limsup |(k^2)^{1/k}|
= \limsup |k^{2/k}|
=1
\quad \implies \quad
R = 1. $$
Rather than giving a proof that this $\limsup$ is 1, let me point you in the direction of several arguments which justify the relatively elementary fact.  In any event, we have that the series
$$ \sum_{k=0}^{\infty} k^2 x^k $$
converges whenever $|x| < 1$.  If $|x|=1$, it is not too difficult to see that the series diverges (the modulus of each term will be equal to $k^2$, which does not tend to zero, hence the series does not converge).
Summarizing all of the above, if $x,y\in\mathbb{R}$, we conclude that
$$ \lim_{n\to\infty} \sum_{k , \ell = 0}^n \frac{k^2 x^k y^\ell}{\ell!} $$
converges if and only if $x \in (-1,1)$ and $y \in \mathbb{R}$.
