Determine whether the function $$g(x,y) = \sum_{k=1}^\infty \frac{(x-2y)^k \sin(kx + y)}{\sqrt{k!} (1 + x^{2k}y^{4k})}$$is continuous on $\mathbb{R}^2$.
My attempt: I tried to use Weierstrass M-test, to prove that $g$ is the uniform convergence of continuous partial sums. However, for each $k \geq 1$, I only obtained the following bound for each $(x,y) \in\mathbb{R}^2$, \begin{align*} \left|\frac{(x-2y)^k \sin(kx + y)}{\sqrt{k!} (1 + x^{2k}y^{4k})} \right| &\leq \frac{(|x| + 2 |y|)^k}{\sqrt{k!}} \leq \frac{2^{k-1} |x|^k}{\sqrt{k!}} + \frac{2^{2k-1} |y|^k}{\sqrt{k!}}\\ &\leq \frac{1}{2} \frac{|2x|^k}{\sqrt{k!}} + \frac{1}{2} \frac{|4y|^k}{\sqrt{k!}} \end{align*} However, I am unable to find a bound $M_k$ for the above expression $\frac{(x-2y)^k \sin(kx + y)}{\sqrt{k!} (1 + x^{2k}y^{4k})}$ independent of $x$ and independent of $y$ for each $k$ so that the series $\sum M_k < \infty.$
Any hints are appreciated.
Hint
To prove the continuity of $g$, it is enough to verify that the series converges uniformly on each compact. For this, you can use Weierstrass M-test and pick up the norm $\lVert (x,y) \rVert_1 = \lvert x \rvert + \lvert y \rvert$ (all norms are equivalent on $\mathbb R^2$). Then you have
$$\left\lvert \frac{(x-2y)^k \sin(kx + y)}{\sqrt{k!} (1 + x^{2k}y^{4k})} \right\rvert \le \frac{2^k A^k}{\sqrt{k!}}$$ for $\lVert (x,y) \rVert_1 \le A$. I let you continue from there.