Swapping the order of limits in combinatoric?

Part $A$

Let a power series be $ \sum_{r=1}^\infty x^{a_r}$

Now, we are interested square of the power series with the condition:

$$ \sum_{m=1}^\infty \sum_{n=1}^\infty x^{a_m + a_n} = \sum_{r=3}^\infty C_r x^{2r} $$

where $C_r \neq 0$ and we want the limit to approach infinity as slow as possible:

$$ \sum_{r=1}^\infty x^{a_r} \sim G(x,a_1,a_2,\dots)$$

$x \to 1^-$

Part $B$:

We note that the minimum number of terms required to construct all the even numbers less than $n$ is $O(\sqrt{n})$. This is also true when $n = \infty$ $(?)$. Hence, we construct a ploynomial such that:

$$ (\sum_{r=1}^n x^{a_r})^2 = \sum_{r=3}^{2n} C_r x^{2r}$$

Taking $\lim_{x \to 1^-}$

$$ \lim_{x \to 1^-} \sum_{r=1}^n x^{a_r} = O(\sqrt{n}) $$

By taking $ n \to \infty $

$$ \lim_{x \to 1^-} \sum_{r=1}^n x^{a_r} = O(\sqrt{n}) $$

We note $n \to \infty$

$$ \lim_{x \to 1^-} \sum_{r=1}^n x^{a_r} = O(\sqrt{n}) < \lim_{n \to \infty} \lim_{x \to 1^-} \sum_{r=1}^n x^{p_r} \sim \frac{n}{\ln(n)}$$

Part $C$:

Upon closer examination of part $B$ no matter how large $n$ may become $n < \infty$. Hence, it will never include all the even numbers and taking $n \to \infty$ (as rightly noticed by Thomas in the comments). The right way to proceed would be:

$$ \lim_{x \to 1^-} \lim_{n \to \infty} \sum_{r=1}^n x^{a_r} $$

Question

Can someone construct an explicit example of $a_n$

where,

$$ (\sum_{r=1}^n x^{a_r})^2 = \sum_{r=3}^{2n} C_r x^{2r} $$

and $C_r \neq 0$

And its limit approachs $\infty$ much slower than

$$ \lim_{n \to \infty} \sum_{r=1}^n x^{a_r} < \lim_{n \to \infty} \sum_{r=2}^n x^{p_r} \sim \frac{1}{(x-1)\ln(1-x)}$$

where $x \to 1^-$

where $p_r$ is the $r$’th prime.

Note: If Goldbach’s conjecture is true then:

$$ (\sum_{r=2}^n x^{p_r})^2 = \sum_{r=3}^\infty C_r x^{2r} $$

P.S: I’ve tried asking this before however I got estimates in $f(n)$ and not $f(x)$. Hence, I decided to include part B and part C

Solutions Collecting From Web of "Swapping the order of limits in combinatoric?"

Erdos’s construction in http://ftp.math-inst.hu/~p_erdos/1956-17.pdf gives an expected $\sum_{n\in S} x^s = \sum_{n \ge 1} C \sqrt{\frac{\log n}{n}}x^n$, which if I’m not mistaken, is asymptotic to $$O\big(\sqrt{\frac{|\log (1-x)|}{1-x}}\big)$$ as $x\to 1^-$. This is obviously quite close to optimal, however it is a probabilistic construction, and you asked for an explicit sequence.

An explicit sequence is given by the set

$$S = \{n \ge 0: \text{the base-$3$ representation of $n$ consists of digits $0,1$}\}.$$ Every nonnegative number, even or odd, can be represented as a sum of two elements of $S$ (you could restrict to odd values only, by applying the transform $x\mapsto 2x+1$, it’s really not important).

The number of elements of $S$ up to $3^k$ is $2^k$, so the number of elements of $S$ up to $n$ is $\asymp n^{\log 2/\log 3} \approx n^{0.631}$. Note this is an order of magnitude, not an asymptotic (there is no asymptotic counting function for $S$ as it goes through long gaps without any elements) but it is still much, much thinner than the primes. Consequently, I would expect the order of growth for $\sum_{n \in S} x^S$ to be about equal to $O((1-x)^{-0.631})$.

I suggest you read up on Tauberian theorems as a tool for converting between estimates for the counting function of a set and estimates for the growth rate of the generating function. I think this is the concept you are struggling to describe adequately — hopefully someone else can provide a better reference.