Infinite sum of analytic function still analytic Consider $$ f_n(x)  = n e^{-n^6(x-n)^2} : \mathbb R \rightarrow \mathbb R$$
and the series $$ f(x) = \sum_{n=1}^{\infty} f_n(x). $$
Is $f$ analytic on $\mathbb R$?

A function is analytic if for every compact set $K$ we can find $C$ s.t. the $k$-th derivative can be bounded as $$|f^{(k)}(x) | \leq C^{k+1} k! \quad \forall x \in K.$$
From the form of $f_n$, we know that $f^{(k)}$ will be a series of a polynomial of degree $k$ times $f_n$. However I find it hard to actually estimate $|f^{(k)}(x) |$ as required...

Another try: since $f_n$ is analytic we can write $f_n(x) = \sum_k a_{k,n} x^k $ so 
$$ f(x) = \sum_n \sum_k a_{k,n} x^k.$$
Now if we switch the sums we get
$$ f(x) = \sum_k \left( \sum_n a_{k,n} \right) x^k. $$
But can we switch the sum, and it is true that $\forall k, \sum_n a_{k,n} < \infty$?

Any suggestion or faster way to prove that $f$ is analytic?
 A: Very often, the fastest or most convenient way to show that a function $h \colon \mathbb{R}\to \mathbb{R}$ is (real-)analytic is to extend the problem to an open subset of $\mathbb{C}$, using the fact that a function is real-analytic on some interval if and only if it is the restriction of a holomorphic function defined in an open neighbourhood of that interval in $\mathbb{C}$.
Here, we have entire functions $f_n \colon z \mapsto n e^{-n^6(z-n)^2}$, and once we have shown that the series
$$\sum_{n=1}^\infty f_n(z)\tag{$\ast$}$$
converges locally uniformly, it follows that
$$f(z) = \sum_{n=1}^\infty f_n(z)$$
is an entire function, and hence its restriction to $\mathbb{R}$ is real-analytic (and moreover, that the power series expansions about any $r\in\mathbb{R}$ converge on all of $\mathbb{R}$).
The crucial point is that the locally uniform limit of holomorphic functions is again holomorphic. The Weierstraß approximation theorem shows that we don't have a corresponding property for real-analytic functions.
So let us see whether we can show that the series $(\ast)$ is locally uniformly convergent.
For any compact $K \subset \mathbb{C}$, there is an $N\in \mathbb{N}$ such that $\lvert \operatorname{Re} z\rvert \leqslant N$ and $\lvert\operatorname{Im} z\rvert \leqslant N$ for all $z\in K$. Then, for $n \geqslant 2N+1$ and $z\in K$, we have
$$\operatorname{Re}\bigl( (z-n)^2\bigr) = (\operatorname{Re} z - n)^2 - (\operatorname{Im} z)^2 \geqslant (n-N)^2 - N^2 \geqslant (N+1)^2 - N^2 = 2N+1 \geqslant 1,$$
and therefore
$$\lvert f_n(z)\rvert = n \exp\Bigl(-n^6\operatorname{Re} \bigl((z-n)^2\bigr)\Bigr) \leqslant n e^{-n^6} \leqslant n e^{-n}$$
on $K$ for $n \geqslant 2N+1$. Since
$$\sum_{n=1}^\infty n e^{-n} < \infty,$$
it follows that $(\ast)$ converges uniformly on $K$. Since $K$ was an arbitrary compact subset of $\mathbb{C}$, the series converges locally uniformly, and hence
$$f(z) = \sum_{n=1}^\infty f_n(z)$$
is an entire function, and its restriction to $\mathbb{R}$ is real-analytic.
