Multivariable Taylor series convergence Let us have a Taylor series expansion for multuvariable function
$$
f(x_1,x_2,\ldots,x_N)=\sum_{n=0}^{\infty}{\sum_{j_1+j_2+\ldots +j_N=n}{\dfrac{1}{j_1!j_2!\ldots j_N!}\dfrac{\partial^{n}f}{\partial x_1^{j_1}\partial x_2^{j_2}\ldots \partial x_N^{j_N}}\Bigg|_{x=0}x_1^{j_1}x_2^{j_2}\ldots x_N^{j_N}}},
$$
where in the second sum summation goes over all possible combinations of non-negative integer solutions of the equation $j_1+j_2+\ldots + j_N=n$. In particular I'm interested in the expansion of the function
$$
f(x_1,x_2,\ldots,x_N)=\exp\left\{\dfrac{1}{2}D_{st}{x_s}{x_t}\right\},
$$
where $D_{st}x_sx_t$ means sum $\sum_{s=1}^{N}{\sum_{t=1}^{N}{D_{st}x_sx_t}}$, because that is a moment-generating function for multivariable normal distribution with a covariance matrix $D$.
I need to prove that in the case of such a function it's Taylor expansion converges to this function for every possible vector $(x_1,x_2,\ldots,x_N)\in \mathbb{R}^{N}$.
I know that for a function of one variable its Taylor series
$$
\sum_{i=0}^{n}{\dfrac{f^{(i)}(0)}{i!}x^{i}}+R_{n}(x)
$$
converges to it if and only if its remainder term $R_n(x)$ has a limit
$$
\lim_{n\rightarrow \infty}{R_n(x)}=0,
$$
and for any given interval $x\in (-R,R)$ this is true if, for example,
$$
\left|f^{(n)}(\xi)\right|\leq M
$$
for any $n$ and $\xi\in(-R,R)$.
I'm interested if there some way to prove that for multivariable function in question its series converges to it on all $\mathbb{R}^N$ so that its series unambiguously defines it.  
This question has stemmed out of that The theorem inverse to Isserlis' theorem
 A: I hit this problem myself, having just read this, which lends itself nicely towards this problem:
Let $\hat{\Delta\mathbf{r}}=\mathbf{a}$ be a unit vector, and $|\Delta\mathbf{r}|=t$:
$$\phi(\mathbf{r}+\mathbf{a}t)=\sum_{n=0}^\infty{\frac{t^n}{n!}(\mathbf{a}\cdot\nabla)^n\phi(\mathbf{r})}$$
Then we then get a boundary within which we have convergence, of:
$$\lim_{n\to\infty}\left[\frac{\frac{t^{n+1}}{(n+1)!}(\mathbf{a}\cdot\nabla)^{n+1}\phi(\mathbf{r})}{\frac{t^n}{n!}(\mathbf{a}\cdot\nabla)^n\phi(\mathbf{r})}\right]<1$$
Hence:
$$t<\lim_{n\to\infty}\left[(n+1) \frac{(\mathbf{a}\cdot\nabla)^n\phi(\mathbf{r})}{(\mathbf{a}\cdot\nabla)^{n+1}\phi(\mathbf{r})}\right]$$
Or, substituting in $\Delta\mathbf{r}$:
$$\phi(\mathbf{r}+\Delta\mathbf{r})=\sum_{n=0}^\infty{\frac{1}{n!}(\Delta\mathbf{r}\cdot\nabla)^n\phi(\mathbf{r})}$$
$$|\Delta\mathbf{r}|<\lim_{n\to\infty}\left[(n+1) \frac{(\hat{\Delta\mathbf{r}}\cdot\nabla)^n\phi(\mathbf{r})}{(\hat{\Delta\mathbf{r}}\cdot\nabla)^{n+1}\phi(\mathbf{r})}\right]$$
Interestingly, the magnitude of the radius of convergence is now a function of the angle, and one which I suspect could be pretty much any reasonable continuous curve in n-spherical polar coordinates.
