Just wanted to verify my proof. Please take a look:
Let $\{f_n\}_{n=1}^\infty$ be a sequence with $f_n:\mathbb{N} \rightarrow \mathbb{R}^+$. Suppose $\sum\limits_{n = 1}^\infty \Big (\sum\limits_{k=1}^\infty f_n(k)\Big ) = \infty$. We need to show that $\sum\limits_{k = 1}^\infty \Big (\sum\limits_{n=1}^\infty f_n(k)\Big ) = \infty$.
Assume for the sake of contradiction that $\sum\limits_{k = 1}^\infty \Big (\sum\limits_{n=1}^\infty f_n(k)\Big )$. Since for all $n \in \mathbb{N}$, $f: \mathbb{N} \rightarrow \mathbb{R}^+$, every term is positive, and since this is a series, it is a monotonically increasing sequence bounded below by the first term. And since it does not diverge to $\infty$, it is bounded above. Thus, $\sum\limits_{k = 1}^\infty \Big (\sum\limits_{n=1}^\infty f_n(k)\Big )$ converges.
Let $\varepsilon > 0$. So by the Cauchy criterion there exists a $N_0 \in \mathbb{N}$ such that for all $m_0 \ge n_0 \ge N_0$, we have $\Big |\sum\limits_{k = n_0}^{m_0} \Big (\sum\limits_{n=1}^\infty f_n(k)\Big )\Big | < \varepsilon$. So, $$\Big |\sum\limits_{n = 1}^\infty f_n(n_0) + \sum\limits_{n = 1}^\infty f_n(n_0 + 1) + \dots + \sum\limits_{n = 1}^\infty f_n(m_0) \Big | < \varepsilon.$$ If any term were to diverge to infinity, then we would get an immediate contradiction, so every term must be bounded. Thus, the same as above, every term converges. Again, by the Cauchy criterion, there exists $N_j \in \mathbb{N}$ such that for all $m_j \ge n_j \ge N_j$, we have $$\Big |\sum\limits_{n = n_j}^{m_j} f_n(n_0 + i - 1) \Big| < \frac{\varepsilon}{m_0 - n_0 + 1}$$ for $j \in \Omega = \{1,2, \dots, m_0 - n_0 + 1 \}$.Thus, it follows (by the triangle inequality) that $$ \Big | \sum\limits_{n = n_1}^{m_1} f_n(n_0) + \sum\limits_{n = n_2}^{m_2} f_n(n_0) + \dots + \sum\limits_{n = n_{m_0 - n_0 + 1}}^{m_{m_0 - n_0 + 1}} f_n(n_0) \Big | < \varepsilon.$$ Now take $N' = \max\limits_{j \in \Omega \cup \{0\}} \{N_j\}$ so that for all $M \ge N \ge N'$, we have $$\Big | \sum\limits_{k = N}^M \Big (\sum\limits_{n=N}^M f_n(k)\Big ) \Big | < \varepsilon.$$ Since this is a finite sum, we can rearrange the terms so that for all $M \ge N \ge N'$, we have $$\Big | \sum\limits_{n = N}^M \Big (\sum\limits_{k=N}^M f_n(k)\Big ) \Big | < \varepsilon.$$ Thus, $\sum\limits_{n = 1}^\infty \Big (\sum\limits_{k=1}^\infty f_n(k)\Big )$ converges, which is a contradiction. Hence, $\sum\limits_{k = 1}^\infty \Big (\sum\limits_{n=1}^\infty f_n(k) \Big ) = \infty$.