If $$u(x,t)=\sum_{k=0}^\infty \frac{1}{(2k)!}x^{2k}\frac{d^k}{dt^k}e^{\frac{-1}{t^2}}$$ with $x\in \mathbb R$. How do is show that $u(x,0)=0$ for $x\in \mathbb R$.
I know that the as $t\rightarrow 0$, $e^{\frac{-1}{t^2}}$ and its derivative individually goes to $0$, but I am not able to get it as the convergence of series.
Any type of help will be appreciated. Thanks in advance.
Provided that we can bound $\frac{d^k}{dt^k}e^{-\frac{1}{t^2}}$ for every $t$, we know that the series $\frac{x^{2k}}{2k!}$ converges so we can use the dominated convergence theorem, and take the limit inside of the sum. Then the series will naturally converge to zero as it will end up being the sum of all zero terms.