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?


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):


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 $f(1/\mu - (1/\mu)/n)$.

  • $\begingroup$ I notice that my summation index $i$ starts at $i=1$, while yours starts at $i=0$. That is not a crucial distinction. $\endgroup$ – Michael Aug 11 '18 at 20:18

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