When reviewing my notes on series' convergence, I thought of applying a workaround on why $\sum_{n=0}^{\infty}(-1)^nn$ should or shouldn't be $-1/4$ (I recalled this page).
I started by considering the following function series:
$$ f(x)=\sum_{n=0}^{\infty}\frac{n}{x^n} $$
by the ratio test, it can converge if $x\in\mathcal{D}:(-\infty;-1)\cup(1;+\infty)$, so it can be proved by taking the finite sum that it's pointwise convergent on $\mathcal{D}$: $$ f_j(x)=\sum_{n=0}^{j}\frac{n}{x^n}=\frac{1}{x-1}\sum_{n=0}^{j-1}\frac{1}{x^n}-\frac{j}{x^j(x-1)}=\frac{x(x^{j}-1)}{x^j(x-1)^2}-\frac{j}{x^j(x-1)} $$ from the above calculation it follows that $\lim_{j\to+\infty}f_j(x)=f(x)$ iff $x\in\mathcal{D}$. However, since: $$ f(x)=x/(1-x)^2 $$ actually is defined on $x=-1$, therefore it would seem that $f(-1)=\sum_{n=0}^{\infty}(-1)^nn=-1/4$.
To prove the falsity of this possibilty I thought of searching the general convergence on $x=-1$: at that exact point if the distance from the finite sum and the function tends to zero as $j\to+\infty$, then the previous statement is true; if false, then the series would deviate at the limit from the function even after magnifying infinite times, and therefore will never reach $-1/4$.
It can be shown that the function is in fact uniformly convergent on entire $\mathcal{D}$: \begin{align*} 0&=\limsup_{\substack{j\to+\infty\\x\in\mathcal{D}}}\left\lvert\frac{x}{(x-1)^2}-\frac{x(x^{j}-1)}{x^j(x-1)^2}+\frac{j}{x^j(x-1)}\right\rvert\\ &=\limsup_{\substack{j\to+\infty\\x\in\mathcal{D}}}\left\lvert\frac{x}{(x-1)^2}-\frac{x}{(x-1)^2}+\frac{o(x^j)}{x^j(x-1)}\right\rvert\\ &=\limsup_{\substack{j\to+\infty\\x\in\mathcal{D}}}\frac{o(x^j)}{|x|^j|x-1|}=0\hskip6cm\square \end{align*} but now, when the point is fixed on $x=-1$ this result is obtained (it shows neither pointwise nor uniform since UC$\implies$PC and $\neg$PC$\implies$$\neg$UC), at first PC: $$ \lim_{\substack{j\to+\infty\\x=-1}}\frac{x(x^{j}-1)}{x^j(x-1)^2}-\frac{j}{x^j(x-1)}=\lim_{\substack{j\to+\infty\\x=-1}}\frac{1-(-1)^{j}+2j}{4(-1)^j}=\lim_{\substack{j\to+\infty\\x=-1}}\frac{j}{2(-1)^j}\to\nexists $$ then by testing UC, as predicted above: \begin{align*} 0&=\lim_{\substack{j\to+\infty\\k=-1}}\left\lvert-\frac{1}{4}+\frac{(-1)^{j}-1}{(-1)^j4}-\frac{j}{(-1)^j2}\right\rvert\\ &=\lim_{\substack{j\to+\infty\\k=-1}}\left\lvert\frac{-(-1)^{j}+(-1)^{j}-1}{(-1)^j4}-\frac{j}{(-1)^j2}\right\rvert\\ &=\lim_{\substack{j\to+\infty\\k=-1}}\left\lvert\frac{-1-2j}{(-1)^j4}\right\rvert\\ &=\lim_{\substack{j\to+\infty\\k=-1}}\frac{|-2j|}{4}=+\infty \end{align*} (edited to show both types of divergence)
This sums up my question: is this method sufficient to prove that $\sum_{n=0}^{\infty}(-1)^nn\neq-1/4$ because the distance from the function $x/(x-1)^2$ at $x=-1$ will deviate as $\infty$ from $-1/4$?