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Let $u: mathbb{N} to mathbb{R}^+$ be a positive sequence. Then it is true that
$$
int u dmu = sum_{n=1}^infty u(n).
$$
For the measure space $(mathbb{N}, mathcal{P}(mathbb{N}),mu)$, where $mu$ is the counting measure. The proof uses monotone convergence theorem, and goes as follows:

Take the measure space $(mathbb{N},mathcal{P}(mathbb{N}), underbrace{sum_{j=1}^infty delta_j}_{mu})$.
Let $u: mathbb{N} to mathbb{R}^+$ be a positive function ($u geq 0$).

We define
$$
u_n(k) =
begin{cases}
u(k) & k leq n, \
0 & k > n.
end{cases}
$$
For all $n in mathbb{N}$, $u_n$ is a normal function since we have $u_n = sum_{k=1}^n u(k)1_{{k}}$. Thus, the approximation cuts of $u$ after $n$. Moreover, we have
$$
int u_n dmu = I_mu(f) = sum_{k=1}^n u(k) mu({k}) = sum_{k=1}^n u(k).
$$
Since we have $0 leq u_n uparrow u$, then it follows by MCT that
$$
int_{mathbb{N}} u dmu stackrel{MTC}{=} lim_{n to infty} int u_n dmu = sum_{k=1}^infty u(k).
$$

Now can someone tell me under what condition this is true if $u: mathbb{N} to mathbb{R}$ is not necessarily positive?

I prove

If $u(n)$ is a sequence such that $sum_n |u(n)| <infty$ then $int u dmu = sum_n u(n)$

By the definition of integral we have $int u dmu = int u^+ dmu - int u^- dmu$

By what you just showed for $u^+,u^-$ we have $u^pm = sum_n u^pm (n)$

Since the sum is uniformly convergence it is the same to sum from $1,2,...$ or to sum first the $n$ for which $u(n)>0$ and only then the $n$ for which $u(n)<0$.