Convergence of a series and its derivative I have to study punctual, uniform and total  convergence of the following sums
$$u(x)=\sum_{n=1}^{+\infty}u_n(x)=\sum_{n=1}^{+\infty}\dfrac{1}{n^2}\int_0^1\exp(-nxt^4)dt$$
$$v(x)=\sum_{n=1}^{+\infty}u_n'(x)$$
I'm looking for exercises of sequences of functions which involve the study of sequences of functions in integral form but I haven't study Lebesgue integrability yet. I'm trying to solve the exercises with a quite standard theory about convergence of function sequences and function series.
I initially found a little bit confusing the fact that the integrand has both the variables $x$ and $t$.
What I've done is considering $t\in[0,1]$ and $x\in \mathbb R$. For the punctual convergence I discarded the case $x<0$ because in this case the integrand would be, for $n\to \infty$, $\dfrac{\exp(-nxt^4)}{n^2}=\dfrac{1}{e^{nxt^4}n^2}\sim\dfrac{1}{e^{nxt^4}}\to\infty,\forall x\in\mathbb R^-$. At this point I think I can say that since the integral from zero to one can be divergent or can assume a real value, the sum would have a non zero general term so it can't be punctual convergent in $\mathbb R^-$.
For $x\ge 0$ I have $u_n(0)=\dfrac{1}{n^2}\forall n\in\mathbb N$ and I can apply
$$\dfrac{1}{e^{nxt^4}}\le \dfrac{1}{n}\implies u_n\le\dfrac{1}{n^3}\implies \sum u_n\le\sum\dfrac{1}{n^3}$$
so we have punctual, uniform and total convergence in $[0,+\infty)$.
I really have problems with $u'_n(x)$... Shouldn't it be zero?
 A: Your conclusions and general approach for $\sum u_n(x)$ are correct but the arguments have a few errors.
For $x < 0$:
You claim that
$$\displaystyle \frac{\exp(-nxt^4)}{n^2}=\frac{1}{e^{nxt^4}n^2}\sim_{n \to \infty}\frac{1}{e^{nxt^4}},$$
but the symbolism $f(n)\sim g(n)$ means $\lim_{n \to \infty}f(n)/g(n) = 1$, which is not true in this case.
The correct argument for $x < 0$ is
$$\frac{\exp(-nxt^4)}{n^2}=\frac{\exp(n|x|t^4)}{n^2} > \frac{1}{n^2}\frac{(n|x|t^4)^3}{3!} = \frac{n|x|^3t^{12}}{6},$$
and the series diverges, since,
$$u_n(x) = \frac{1}{n^2} \int_0^1\exp(-nxt^4)\, dt > \frac{n |x|^3}{6}\int_0^1 t^{12} \, dt \underset{n \to \infty} \longrightarrow +\infty $$
For $x \geqslant 0$:
The inequality $\frac{1}{e^{nxt^4}}\le \frac{1}{n}$ holds eventually for fixed $x$ and $t$, but this can't be used for uniform convergence.
Uniform convergence holds by the Weierstrass M-test simply because $\exp(-nxt^4) \leqslant 1$ and
$$u_n(x) = \frac{1}{n^2} \int_0^1\exp(-nxt^4)\, dt \leqslant  \frac{1}{n^2}$$

Regarding $v(x)$:
We can switch the order of the derivative and integral  to obtain
$$\frac{d}{dx}\int_0^1\exp(-nxt^4) \, dt = \int_0^1(-nt^4)\exp(-nxt^4) \, dt,$$
and, thus,
$$v(x) = \sum_{n=1}^\infty u'_n(x) = -\sum_{n=1}^\infty \frac{1}{n}\int_0^1t^4\exp(-nxt^4) \, dt $$
Using a similar argument as before, the series diverges for $x < 0$. We also have divergence for $x = 0$, since
$$u'_n(0) = - \frac{1}{n} \int_0^1 t^4 \, dt = - \frac{1}{5n}$$
For $x > 0$, we have
$$u'_n(x) = \frac{1}{n}\int_0^1t^4\exp(-nxt^4) \, dt < \frac{1}{n}\int_0^1 \frac{t^4}{nxt^4} \, dt = \frac{1}{n^2x},$$
and it follows that the series converges pointwise and absolutely, as well as uniformly for $x \in [a,\infty)$ for any $a > 0$.
There is one  remaining question that I will leave to you. Does   $\sum u'_n(x)$ converge uniformly on $(0,\infty)$?
