Interchanging limit with series of exponentially growing numbers Let $(a_i)_{i \in \mathbb{N}}$ be a sequence of positive integers and suppose that 
$$
\lim\limits_{i \rightarrow \infty} a_i^{1/i} = \mu \in (0, \infty).
$$
Consider the function$$
f(\lambda) := \sum\limits_{i=0}^{\infty} a_i \lambda^i,
$$
which is finite for any $\lambda \in [0, 1/\mu)$ and infinite for any $\lambda \in (1/\mu, \infty)$. Whether $f(1/\mu)$ is finite or not depends on the sequence $(a_i)_{i \in \mathbb{N}}$.
Is it true that 
$$
\lim\limits_{\lambda \rightarrow 1/\mu^-} f(\lambda) = f(1/\mu),
$$
where the limit is taken from below? In other words, could it be that $f(1/\mu)$ is infinite but the limit is not infinite?
 A: Yes, $\lim_{\lambda \rightarrow 1/\mu^-} f(\lambda) = f(1/\mu)$. So if the right-hand-side is infinite, so is the left-hand-side. You can justify this by the monotone convergence theorem (MCT):
https://en.wikipedia.org/wiki/Monotone_convergence_theorem

In terms of general integrals (which can include sums by defining measures with "unit impulses" at nonnegative integers) the MCT is stated as:  
If $h_n(x)\geq 0$ for all $x \in \mathbb{R}$ and $h_n\nearrow h$, then $\int h_n \nearrow \int h$. 

The corresponding statement for sums can be stated as follows (see also the wiki link):
Suppose we have real values $h_n(i)$ and $h(i)$ for $i,n \in \{1, 2, 3, ...\}$ such that: 
i) $h_n(i) \geq 0$ for all $n, i \in \{1, 2, 3, ...\}$. 
ii) For each $i \in \{1, 2, 3, ...\}$ we have $h_n(i)\nearrow h(i)$ (the limit is as $n\rightarrow\infty$). 
Then 
$$\lim_{n\rightarrow\infty} \sum_{i=1}^{\infty} h_n(i) = \sum_{i=1}^{\infty} h(i) $$
In your case you can define 
$$ h_n(i) = a_i(1/\mu - (1/\mu)/n)^i \quad, \quad h(i) = a_i(1/\mu)^i$$
Then $h_n(i)$ are nonnegative and $h_n(i)\nearrow h(i)$ for all $i$.   Notice that $\sum_{i=1}^{\infty} h_n(i)$ is what you are calling $f(1/\mu - (1/\mu)/n)$. 
