convergence of $\sum_n\left(1+\log\left(1+\frac1n\right)\right)^{n^2}x^n$ I used the Cauchy Hadamard theorem to get the radius of convergence finding $r=\frac1e$.
Now I have to check the convergence in $|x|=r$  but I didn't manage to prove the convergence or the divergence of the numerical series.
Moreover the exercise asks for the uniform convergence of the series that depends on the previous point.
 A: For the case $x = \frac{1}{e}$ it is not difficult to show using the Taylor series expansion for $\log(1+x)$ that
$$\tag{*}\lim_{n\to \infty}\frac{\left(1+\log\left(1+\frac1n\right)\right)^{n^2}}{e^n}= \frac{1}{e} \neq 0,$$
and, hence, the term limit with $x = -\frac{1}{e}$ does not exist since
$$\limsup_{n\to \infty}\frac{\left(1+\log\left(1+\frac1n\right)\right)^{n^2}}{(-e)^n}= \frac{1}{e},\quad  \liminf_{n\to \infty}\frac{\left(1+\log\left(1+\frac1n\right)\right)^{n^2}}{(-e)^n}= -\frac{1}{e}$$
and the series does not converge at the endpoints $x = \pm\frac{1}{e}$
The series in question is a power series and, hence,  it converges on any compact subset of the open interval $\left(-\frac{1}{e}, \frac{1}{e}\right)$.
That leaves the question of uniform convergence of the series on the entire open interval  $\left(-\frac{1}{e}, \frac{1}{e}\right)$.
To show that the series  fails to converge uniformly on an open interval with $\frac1e$ as an endpoint, it is enough to show that the terms fail to converge uniformly to $0$.  This is indeed the case because with
$$f_n(x) = \left(1+\log\left(1+\frac1n\right)\right)^{n^2}x^n$$
and $x_n = \frac{\left(1-\frac1n\right)}{e} \in (0,\frac1e)$ we have
$$\lim_{n \to \infty} f_n(x_n) = \lim_{n\to \infty}\frac{\left(1+\log\left(1+\frac1n\right)\right)^{n^2}}{e^n}\left( 1 - \frac1n\right)^n= \frac{1}{e^2}\neq 0$$

To prove (*), consider
$$\log a_n =\log \frac{\left(1+\log\left(1+\frac1n\right)\right)^{n^2}}{e^n} = n^2 \log\left(1+\log\left(1+\frac1n\right)\right)-n\\ = n^2 \log \left(1+\frac{1}{n} - \frac{1}{2n^2} + \mathcal{O}\left(\frac{1}{n^3} \right)\right)- n\\= n^2 \left[\left(\frac{1}{n} - \frac{1}{2n^2} + \mathcal{O}\left(\frac{1}{n^3} \right)\right)- \frac{1}{2}\left(\frac{1}{n} - \frac{1}{2n^2} + \mathcal{O}\left(\frac{1}{n^3} \right)\right)^2 + \mathcal{O} \left(\frac{1}{n^3} \right) \right]-n\\ = -1 + \mathcal{O}\left( \frac1n\right) $$
The RHS converges to $-1$ as $n \to \infty$ and, thus, $a_n \to e^{-1}$ as $n \to \infty$.
