Prove that fuction series $\sum_{n=1}^{+\infty} \frac{2(n-x)}{e^{(n-x)^2}}$ is convergent uniformly to function $G$ and find this $G$ 
Prove that fuction series $$\sum_{n=1}^{+\infty} \frac{2(n-x)}{e^{(n-x)^2}}$$ is  convergent uniformly to function $G$ and find this $G$.

I tried to calculate $$||f_n||=\sup _{x \in \mathbb R} |\frac{2(n-x)}{e^{(n-x)^2}}|$$ but I could not reach any meaningful conclusion because I know only that $|\frac{2(n-x)}{e^{(n-x)^2}}| \le 1$ but it is not helpfull for me because $\sum 1$ is divergent. Moreover I need formula for $ G $.Have you get any hints how to do it?
 A: Let us denote $$f_n(x) = 2(n-x)e^{-(n-x)^2}$$
You can check the point-wise convergence by noticing that for any fixed $x$ the series $S(x) = \sum_{n=0}^\infty f_n(x)$ is convergent; that follows from the fact that $$\frac{f_{n+1}}{f_n} = \frac{n+1-x}{n-x}e^{-(n+1-x)^2+(n-x)^2} = \frac{1 + \frac{1-x}{n}}{1-\frac{x}{n}}e^{-1-2n+2x} \rightarrow^{n\rightarrow\infty} 0 $$
Let us check the uniform convergence on $\mathbb R$.
Let us consider $$S_n(x) = \sum_{k=1}^n f_k(x)$$ To prove the uniform convergence  on $\mathbb R$, we'd need to show that $$ \lim_{n\rightarrow \infty} \sup_{x\in\mathbb R} |S(x)-S_n(x)| = 0$$ that is
$$ \lim_{n\rightarrow \infty} \sup_{x\in\mathbb R} \left|\sum_{k=n+1}^\infty f_k(x)\right| = 0$$
However it's not true. We have 
$$ \forall x \le k: f_k(x) \ge 0$$
so for a given $n$, if we take $x<n$ and $k>n$ we have
$$ \forall x \le n \,\forall k \ge n:f_k(x) \ge 0 $$
and from this it follows that
 $$ \forall x \le n : \sum_{k=n+1}^\infty f_k(x) \ge f_{n+1}(x) \ge 0$$
We have then $$ \sup_{x\in\mathbb R} \left|\sum_{k=n+1}^\infty f_k(x)\right| \ge \sup_{x \le n} \left|\sum_{k=n+1}^\infty f_k(x)\right| \ge \sup_{x \le n} f_{n+1}(x) \ge f_{n+1}(n-1) = 2e^{-1}$$
Therefore it's impossible that $$ \lim_{n\rightarrow \infty} \sup_{x\in\mathbb R} \left|\sum_{k=n+1}^\infty f_k(x)\right| = 0$$
so the series isn't uniformly convergent on $\mathbb R$.
It is however almost uniformly convergent on $\mathbb R$. It's easy to check that $f_n(x)$ is increasing for $x \le n+1$ That means that if $x\in[a,b]$ and $n\ge b+1$ then $$ f_n(x) \le f_n(b)$$
Since $\sum_n f_n(b)$ is convergent, it means that $\sum_n f_n(x)$ is convergent uniformly for $x\in[a,b]$, and that means that it is convergent almost uniformly for $x\in\mathbb R$.
