The definition is $\forall \epsilon>0 \ \exists n_0=n_0(\epsilon)\in \mathbb{N} \ \forall m\geq n_0 \ \forall x\in D : |s_m(x)-f(x)|<\epsilon$.
We have that $s_m=\frac{1}{x-1}+\sum_{n=2}^{m}\frac{1}{n‌​x-n^2}$ and $f(x)=\frac{1}{x-1}+\sum_{n=2}^{\infty}\frac{1}{n‌​x-n^2}$, so $|s_m(x)-f(x)|=|\sum_{n=2}^{m}\frac{1}{n‌​x-n^2}-\sum_{n=2}^{\infty}\frac{1}{n‌​x-n^2}|$, which is the definition that $\sum_{n=2}^{\infty}\frac{1}{nx-n^2}$ converges uniformly, right?

Since we cannot compute the limit function, the only way to check the pointwise convergence is to say that since the series converges …

Let $V$ be a $\mathbb{K}$-vector space and $\psi:V\rightarrow V$ be a linear mapping, with $\psi^k\neq 0$ and $\psi^{k+1}=0$ for some $k>0$. Show that there is an element $x\in V$ such that the set $\{x, \psi (x), \ldots , \psi^k(x)\}$ is linearly independent.

Let $x\in V$ such that $\psi^k(x) \ne 0$. We wil show that the set $\{x, \psi (x), \ldots , \psi^k\}$ ist linearly independent, i.e., that it holds it when $c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0$ then $c_0=\ldots c_k=0$.