Is this proof right? Let $f_n(x) = n^2 x e^{- n^2 x}$, $n=1,2,3,...$.

*

*Show that $\sum_{n=1}^{\infty} f_n(x) $ convergeces uniformly on $[a,\infty]$, where $a>0$.


*Show that $\sum_{n=1}^{\infty} f_n(x) $ does not convergece uniformly on $[0,\infty]$.
Proof:
First, I found the maximum of $f_n(x)$ which occured at $x=1/2n$, from which I constructed the upper bound for $f_n(x)$ for all $n$. Namely
$$|f_n (x) | \leq M_n = f_n(1/2n),\,\,\,\, x \neq 1/2n$$
I claimed that $\sum_1^\infty M_n$ is convergent because $\int_1^\infty M_n dn = 3 e^{-1/2}$.
Therefore, since we have the requirements to apply the M-test theorem, we can say that $\sum f_n(x)$ is uniformly convergent on $[a, \infty],\,\,\, a>0$. While $\sum f_n(x)$ is not uniformly confergennt because $f_n(x)$ is not bounded at $x=0$. (there are no integer $n$ that makes $1/2n$ vanishies).
 A: For every $n\geq 1$ we have that $f_n(0) = 0$ and $f_n(x) > 0$ for every $x > 0$.
Moreover, the function $f_n$ attains its maximum when $n^2 x = 1$, i.e. at $x = 1/n^2$.
It is easily seen that the series converges pointwise in $[0,+\infty)$.
Let $a > 0$, and let $M_n := \sup_{x\geq a} |f_n(x)|$. Since
$$
M_n = f_n(a) = n^2 a \, e^{-n^2 a},
\qquad \forall n \geq 1/\sqrt{a},
$$
by the Weierstrass M-test we deduce that the series converges uniformly in $[a,+\infty)$.
Let us prove that the series is not uniformly convergent in $[0,+\infty)$.
To this end, we have to prove that
$$
\sigma_N := \sup_{x\geq 0} \sum_{n=N}^\infty f_n(x)
$$
does not converge to $0$.
This fact is easily proved observing that
$$
\sigma_N \geq f_N\left(\frac{1}{N^2}\right) = e^{-1},
\qquad \forall N\geq 1.
$$
A: The Right Solution to My Problem.
First, we check the pointwise convergence using the ratio test.
\begin{align}
L &= \lim_{n \to \infty} \bigg|  \frac{(n+1)^2 x e^{- (n+1)^2 x}}{n^2 x e^{- n^2 x}} \bigg|\\
L&= \lim_{n \to \infty} \frac{(n+1)^2}{n^2} \times \lim_{n \to \infty} e^{- x (1+2n)}\\
L&= 0 < 1
\end{align}
Thus, the series $\sum_{n=1}^{\infty} f_n (x)$ is pointwise convergent on $(0,\infty)$.
\emph{Note}: We can omit the pointwise convergent if we are aiming to prove the uniform convergence.
Since it is clear that $\lim_{n \to \infty} n^2 x e^{- n^2 x} =0$ as $n \to \infty$, then we will find $\|f_n(x) -0 \|_{\infty}$, where $\|.\|_{\infty}$ is the uniform norm, (it is also called supremum norm, the Chebyshev norm, the infinity norm and the maximum norm where the maximum value of the function is included)
$$\|f\|_{\infty} = \max_{a\leq x<\infty}{|f(x)|}$$
\begin{align}
\|f_n(x)\|_{\infty}   &= \max_{a \leq x <\infty)} |f_n(x)|\\
\|f_n(x)\|_{\infty}  &=\max_{a \leq x <\infty} \bigg| \frac{n^2 x}{e^{n^2 x}}\bigg|\\
\|f_n(x)\|_{\infty}  &=  \max_{a \leq x <\infty} \bigg\{ \frac{n^2 x}{1 +n^2 x + n^4 x^2 /2 + n^6 x^3 /6+...+ (n^2 x)^k/k!+...} \bigg\} ,\,\,\,\, m=1,2,3,...\\
\|f_n(x)|_{\infty} & < \max_{a \leq x <\infty} \bigg\{ \frac{n^2 x}{1 + \frac{1}{2} n^4 x^2} \bigg\},\\
\|f_n(x)\|_{\infty}&< \frac{n^ 2  }{\frac{1}{2} a n^4}\\
\|f_n(x)\|_{\infty} & < \frac{2}{a n^2}
\end{align}
Let $M_n = \frac{2}{a \, n^2}$, since we found that $\|f_n(x)\|_\infty < M_n$ for all $n$, and $\sum_{n=1}^{\infty} M_n$ is convergent, for
$$\sum_{n=1}^{\infty} M_n = \frac{2}{a}\sum_{n=1}^{\infty} \frac{1}{n^2}.$$
Thus, the series $\sum_{n=1}^{\infty} n^2 x e^{- n^2 x}$ converges uniformly on $[a,\infty)$ for $a>0$. To show that the series is not uniformly convergent on $[0,\infty)$.
Let's take $x=\frac{1}{m^2} \to 0$ as $m \to \infty$, and let $g(x) = \sum_{n=1}^{\infty} n^2 x e^{n^2 x}$.
$$g(\frac{1}{m^2}) \geq \sum_{n=1}^{m} \frac{n^2}{m^2} e^{\frac{-n^2}{m^2}} \ge \sum_{n=1}^{m} \frac{1}{e} $$

$$g(\frac{1}{m^2}) \ge \frac{m}{e}.$$

We can see that $g$ is unbounded on $0<x<a$ for any $a>0$. Which brakes the condition of the uniform convergence i.e.((If $\sum_{n=1}^{\infty} S_n(x)$ converges uniformly to $S$, then $S$ is bounded.)). Thus, the series $\sum_{n=1}^{\infty} n^2 x e^{-n^2 x}$ does not converge uniformly on $[0,\infty)$.
