How to show an infinite series is $C^\infty$ - smooth Consider 
$$
u(x,t) = \sqrt{\frac{2}{\pi}}\sum_{k=1}^\infty a_k e^{-k^2 t}\sin(kx),
$$
where
$$a_k = \sqrt{\frac{2}{\pi}} \int_0^\pi u_0(x) \sin(kx)dx.$$
Here, $u_0(x)$ is assumed to be in $L^2(0, \pi)$. This $u$ is the solution that arises from performing separation of variables on the PDE $u_{t} - u_{xx} = 0$ with initial data $u(x,0) = u_0(x)$ and Dirichlet boundary conditions $u(0,t) = u(\pi,t) = 0$.
I am trying to understand the proof for why $u(x,t)$ is $C^\infty$ smooth. To do this, my textbook begins by noting that formally, for any nonnegative integers $i$ and $j$, 
$$
\partial_x^i \partial_t^j u(x,t) = \sqrt{\frac{2}{\pi}}\sum_{k=1}^\infty a_k \frac{d^j}{dt^j}(e^{-k^2 t})\frac{d^i}{dx^i}(\sin{kx})
$$
for any $x\in [0,\pi]$ and $t \in (0,\infty)$. 
The author writes he wants to prove that this series is convergent uniformly and absolutely for any $(x,t) \in [0,\pi]\times[t_0, \infty)$, for fixed $t_0 > 0$. The author uses the Weierstrass M-test to prove that the series is convergent uniformly and absolutely. I can follow the application of the Weierstass M-test, but why does proving the series is convergent uniformly and absolutely show that it is $C^\infty$ smooth in $[0, \pi]\times[t_0, \infty)$?
 A: This uses the relationship between uniform convergence and differentiability. If a sequence of differentiable functions $f_n:(a,b)\to\mathbb{R}$ has the property that $\lim_{n\to\infty}f_n(x_0)$ exists at some $x_0\in(a,b)$ and $(f_n')$ converges uniformly on $(a,b)$, then $(f_n)$ converges uniformly to a limit function $f$ and $f' = \lim_{n\to\infty} f_n'$ on $(a,b)$.
So here the author verifies that for each fixed $i,j$ the series
$$
\sqrt{\frac{2}{\pi}}\sum_{k=1}^\infty a_k\frac{d^j}{dt^j}(e^{-k^2t})\frac{d^i}{dx^i}(\sin kx)
$$
converges uniformly. From here you proceed by induction on $i$ and $j$. Basically, for $(i,j) = (1,0)$, what is proved is that the sequence of finite sums
$$
\sum_{k=1}^N a_k e^{-k^2t}\sin(kx)
$$
has derivatives in $x$ which converge uniformly. This verifies the conditions required to use the aforementioned relationship between differentiation and uniform convergence. Therefore the limit $u$ of this sequence exists (which you knew, because you already knew the sequence of $0$-th derivatives converges uniformly), but also the limit is once differentiable in $x$, and $u_x$ is obtained via term-by-term differentiation. You then show also that $u_t$ is obtained via term-by-term differentiation in the same way, then proceed to $u_{xx}$, $u_{tx}$, $u_{tt}$, and so on; by induction, you can show that you can do this for any order derivative $(i,j)$, which shows that $u$ is smooth.
