Is this proof right?

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)

$$|f|_{infty} = max_{aleq x<infty}{|f(x)|}$$

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 $[0,infty)$.

For every $ngeq 1$ 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 $M_n := sup_{xgeq a} |f_n(x)|$. 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 $$ sigma_N := sup_{xgeq 0} sum_{n=N}^infty f_n(x) $$ does not converge to $0$. This fact is easily proved observing that $$ sigma_N geq f_Nleft(frac{1}{N^2}right) = e^{-1}, qquad forall Ngeq 1. $$