Help with proving the Fatou's Lemma for discrete functions I need help to understand two steps in this proof for the Fatou's lemma in its discrete version.
Let $f_k:\mathbb{N} \rightarrow [0, \infty), k\in\mathbb{N}$, be a sequence of functions. Then
$$
\sum_{j=1}^\infty \lim_{k \rightarrow\infty} \inf f_k(j) \leq \lim_{k \rightarrow\infty} \inf \sum_{j=1}^\infty f_k(j), \forall j\in\mathbb{N},
$$
where $\lim_{k \rightarrow\infty} \inf f_k(j):= \lim_{k \rightarrow\infty}\left( \inf_{n \geq k} f_n(j) \right), \forall j\in\mathbb{N}$
The four steps for the proof are
i) For each $K\in\mathbb{N}$ there exists a number $m(K)$ such that
$$
\lim_{n \rightarrow\infty}\inf f_n(k) \leq f_m(k) + \frac{1}{K^2}, \quad k\in\{1,\dots,K \},
$$
note that $\lim_{n \rightarrow\infty}\inf f_n(k)$ might be $\infty$.
I guess I have to proceed here as in the "standard proof" for the continuous version, defining $g_n(k)= \inf f_n(k)$, and noting that $g_n(k) \leq g_{n+1}(k)$, fixing $K$ and $n$, I can always pick up a $m$ such that 
$$g_n(k) - f_m(k) \leq \frac{1}{K^2}, \quad k\in\{1,\dots,K \}, \quad m \geq m(K)$$. 
But how to proceed when $n \rightarrow \infty$?
ii) Assuming that i) is true, with $m(K)$ chosen as in i), it is easy to show that for $k\in\{1,\dots,K \}$, and all $m \geq m(K)$
$$
\sum_{k=1}^K \lim_{n \rightarrow\infty}\inf f_n(k) \leq \sum_{k=1}^Kf_m(k) + \sum_{k=1}^K\frac{1}{K^2} \leq \sum_{k=1}^\infty f_m(k) + \frac{1}{K}
$$
iii)
Conclude that for $k\in\{1,\dots,K \}$
$$
\sum_{k=1}^K \lim_{n \rightarrow\infty}\inf f_n(k) \leq \lim_{m \rightarrow\infty}\inf \sum_{k=1}^\infty f_m(k) + \frac{1}{K}
$$
I can not see this step, is this true? $$ \sum_{k=1}^\infty f_m(k) \leq 
 \lim_{m \rightarrow\infty}\inf \sum_{k=1}^\infty f_m(k)
$$
iv)
Assuming that iii) is true, we can conclude the proof by letting $K\rightarrow \infty$.
Basically, I have problems to see the steps i) and iii). Any help? Thank you very much in advance.
 A: For i) note that it is not worth considering the case when $\liminf f_n(k)=\infty$, because in this case the inequality becomes $\infty=\infty$. Note also that $m(K)$ is not unique. You can take a strictly increasing sequence of $m$'s which goes to $\infty$ and still have the desired inequality. 
(The idea behind the liminf is that if $\liminf z_n=L$ then there is a subsequence of $z_n$ which converges to $L$, and any other subsequence has a limit greater than $L$.)
As for the proof for i), it is just the definition of $\liminf$. You could argue by contradiction. Fix $K$ and then suppose that there is no $m$ such that $\liminf\limits_{n \to \infty} f_n(k) \leq f_m(k)+\frac{1}{K^2}$. That means that the opposite inequality is true. Take the liminf and you get $0 \geq 1/K^2$.
In step iii) you just take a liminf of the inequality obtained in ii) as $m \to \infty$ (in fact i) holds for an increasing sequence of $m$'s). The left side does not depend on $m$ so it is unchanged, while in the right side you get what you want.
