Q:

Nekrasov Partition Function: $F^{inst}(\epsilon_1,\epsilon_2,\mathbf{a},\mathbf{q})$ analytic at $\epsilon_1 = \epsilon_2 = 0$?

Nakajima & Yoshioka [1] showed that 
\begin{equation}
F^{inst}(\epsilon_1,\epsilon_2,\mathbf{a},\mathbf{q}) = \sum_{n = 1}^\infty \mathbf{q}^nF^{inst}_n(\epsilon_1,\epsilon_2,\mathbf{a}) := \epsilon_1\epsilon_2\log Z^{inst}(\epsilon_1,\epsilon_2,\mathbf{a},\mathbf{q})
\end{equation}
is regular at $\epsilon_1,\epsilon_2 = 0$. If I'm not mistaken, `regular' in this case means that $F^{inst}(\epsilon_1,\epsilon_2,\mathbf{a},\mathbf{q})$ is analytic as a function of $\epsilon_1,\epsilon_2$ near $\epsilon_1 = \epsilon_2 = 0$ ([2] said the same thing in Theorem 5.14). However I struggle to understand why this result is true. 
The proof of this result is given in Proposition 7.3 [1]. My understanding of the argument is that they comparing the coefficients of $\mathbf{q}^n$ in the blow-up formula for each $n$ to recursively relates $F^{inst}_n(\epsilon_1,\epsilon_2,\mathbf{a})$ as a regular function of $F^{inst}_1(\epsilon_1,\epsilon_2,\mathbf{a}),...,F^{inst}_{n-1}(\epsilon_1,\epsilon_2,\mathbf{a})$. By induction we can determine $F^{inst}_n(\epsilon_1,\epsilon_2,\mathbf{a})$ and argue that it must be regular at $\epsilon_1 =\epsilon_2 = 0$ for all $n$. What I fail to see is why this is sufficient to conclude that $F^{inst}(\epsilon_1,\epsilon_2,\mathbf{a},\mathbf{q})$ must be regular at $\epsilon_1 = \epsilon_2 = 0$ as well since I have no idea about the nature of convergence of the sum $\sum_{n = 1}^\infty \mathbf{q}^nF^{inst}_n(\epsilon_1,\epsilon_2,\mathbf{a})$. 
I also cannot convince myself intuitively why this should be true. The partition function $Z^{inst}(\epsilon_1,\epsilon_2,\mathbf{a},\mathbf{q})$ is a sum over a vector of Young diagrams $\mathbf{k} = (k_1,...,k_N)$:
\begin{align}
Z^{inst}(\epsilon_1,\epsilon_2,\mathbf{a},\mathbf{q}) = \sum_{\mathbf{k}}\mathbf{q}^{|\mathbf{k}|}\prod_{l,m = 1}^N\Big[&\prod_{(i,j)\in k_l}((i - k'_{mj})\epsilon_1 + (k_{li} - j + 1)\epsilon_2 + a_l - a_m)^{-1}\\
\times&\prod_{(i,j)\in k_m}((k'_{lj} - i + 1)\epsilon_1 + (j - k_{mi})\epsilon_2 + a_l - a_m)^{-1}\Big].
\end{align}
For this to converge we must require that $\mathbf{a}$ satisfies $a_l - a_m \notin \mathbb{Z}\epsilon_1 + \mathbb{Z}\epsilon_2$ for all $l \neq m$. When $\epsilon_1,\epsilon_2$ are small, the terms involving $\mathbf{k}$ with small $|\mathbf{k}|$ becomes analytic since $|(...)\epsilon_1 + (...)\epsilon_2| << a_l - a_m$. When $l = m$ we get $1/(\epsilon_1\epsilon_2)^n$ factors but this should be suppressed after taking $\epsilon_1\epsilon_2\log(...)$ so it makes sense to me why $F^{inst}_n(\epsilon_1,\epsilon_2,\mathbf{q})$ are analytic at $\epsilon_1,\epsilon_2 = 0$ for all $n$. But we can always choose a sequence $\epsilon_1 = -\epsilon_2 = h_n \rightarrow 0, k = 1,2,3,...$ such that there exists a Young diagram vector $\mathbf{k}_n$ (ofcourse $|\mathbf{k}_n|\rightarrow \infty$) such that the denominator of the corresponding term goes to zero as fast as we wanted $|(...)\epsilon_1 + (...)\epsilon_2 + a_l - a_m| = |(...)h_n - (a_l - a_m)| \rightarrow 0$. I cannot see why $\lim_{n\rightarrow \infty}-h^2_n\log Z^{inst}(h_n,-h_n,\mathbf{a},\mathbf{q})$ should converge which means $\lim_{\epsilon_1,\epsilon_2\rightarrow \infty}\epsilon_1\epsilon_2\log Z^{inst}(\epsilon_1,\epsilon_2,\mathbf{a},\Lambda)$ shouldn't even exists (I see people write this all the time, so I need to understand this).
What did I misunderstand? Could someone clarify this for me please, thank you!
References:
[1] https://arxiv.org/abs/math/0306198
[2] https://arxiv.org/pdf/0808.0884.pdf

A:

We meant that each coefficient of $\mathfrak q^n$ in $\varepsilon_1 \varepsilon_2 \log Z^{\mathrm{inst}}(\varepsilon_1,\varepsilon_2, \vec{a}, \mathfrak q)$ is regular at $\varepsilon_1 = \varepsilon_2 = 0$.

