I would like to prove the second remark in the Wikipedia article of Abel's theorem:
Let $a_k $ be real numbers. If $\sum_{k=0}^{\infty} a_k = +\infty $ then $$\lim_{z\to 1^-} \sum_{k=0}^{\infty} a_k z^k =+\infty, $$ provided that the radius of convergence of the series is $1$.
I tried to use the same ideas as in the usual case where $\sum_{k=0}^{\infty} a_k<\infty$ but without success.
Any help would be appreciated. Thank you.
Using summation by parts with $S_n = \sum_{k=0}^n a_k$,
$$\sum_{k=0}^n a_kx^k = S_nx^n + \sum_{k=0}^{n-1}S_k(x^k - x^{k+1}) = S_nx^n + (1-x)\sum_{k=0}^{n-1}S_kx^k $$
Since $\lim_{n \to \infty} S_n = \infty$, for any $M > 0$ there exists a positive integer $N$ such that $S_n > M$ for all $n \geqslant N$.
Hence, for $n > N$ we have $$\sum_{k=0}^n a_kx^k > Mx^n + (1-x)\sum_{k=0}^{N-1}S_kx^k + (1-x)M\sum_{k=N}^{n-1}x^k \\ = Mx^n + (1-x)\sum_{k=0}^{N-1}S_kx^k + (1-x)M\frac{x^N -x^{n}}{1-x} \\= Mx^n + (1-x)\sum_{k=0}^{N-1}S_kx^k + M(x^N -x^{n}) $$
For $0 < x < 1$, the series on the LHS converges and $x^n \to 0$ as $n \to \infty$.
Thus,
$$\sum_{k=0}^\infty a_kx^k \geqslant (1-x)\sum_{k=0}^{N-1}S_kx^k + Mx^N. $$
From here you should be able to finish by showing that for any $M > 0$,
$$\lim_{x \to 1-}\sum_{k=0}^\infty a_kx^k \geqslant M $$