Under what conditions is $\sum_{n\ge0}a_nx^n\ge\sum_{n\ge0}b_nx^n\Rightarrow a_n\ge b_n$ true?

Question

Under what conditions is $$\sum_{n\ge0}a_nx^n\ge\sum_{n\ge0}b_nx^n\Rightarrow a_n\ge b_n,\qquad x>0\tag1$$ true?

Context: I was trying to prove that the number of divisors of $n$ of the form $4k+1$ was greater than or equal to the number of divisors of $n$ of the form $4k+3$, and I thought to do so by comparing the generating functions $$f_{4,1}(q)=\sum_{n\ge0}\frac{q^{4n+1}}{1-q^{4n+1}},$$ and $$f_{4,1}(q)=\sum_{n\ge0}\frac{q^{4n+3}}{1-q^{4n+3}}.$$ Here, $f_{4,i}$ is the generating function for the number of divisors of $n$ in the form $4k+i$, $i=1,3$. It was easy enough to show that $f_{4,1}(q)>f_{4,3}(q)$ for $0<q<1$, but I was having trouble using that to show $[q^n]f_{4,1}(q)\ge [q^n]f_{4,3}(q)$.

Thoughts on the problem so far: It sort of makes sense intuitively, but clearly it only works in certain situations. One such condition is that the partial sums satisfy the inequality $$\sum_{n=0}^{m}a_nx^n\ge \sum_{n=0}^{m}b_nx^n$$ for all integers $m\ge0$. If this is the case, $$\begin{align} \sum_{n=0}^{m}a_nx^n&\ge \sum_{n=0}^{m}b_nx^n\\ \Rightarrow \sum_{n=0}^{m}a_nx^n-\sum_{n=0}^{m-1}a_nx^n&\ge \sum_{n=0}^{m}b_nx^n-\sum_{n=0}^{m-1}b_nx^n\\ \Rightarrow a_mx^m&\ge b_mx^m\\ \Rightarrow a_m&\ge b_m. \end{align}$$ Of course that works, but I feel like there are other, less restrictive ways. There might be a way to go about it by treating $A(x)=\sum_n a_nx^n$ and $B(x)=\sum_n b_nx^n$ as Taylor expansions, but I'm not super sure that would work.

Can you think of anything? Thanks :)

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EDIT: Apparently some find my question to be unclear. I am not looking for a different way to prove that the number of $4k+1$ divisors is at least the number of $4k+3$ divisors.

What, if any, restrictions must we put on the functions $A(x),B(x)$ in order to ensure that $$A(x)\ge B(x)\Rightarrow [x^n]A(x)\ge [x^n]B(x),\qquad x>0$$ is true?

Answer 1

It would be better to use number theory, and specifically modular arithmetic and cyclicity, in my opinion. A few pointers:

1. It suffices to consider only odd $n$. Any even number can be reduced to an odd one, as none of the even factors would contribute to the count.
2. In the prime factorization of $n$, any power of primes of the form $4k+1$ would be of the same form, while for primes of form $4k+3$, odd powers give same form while even powers give $4k+1$ form.
3. So, to construct a divisor of $4k+1$ form, you can pick up any even power of a $4k+3$ prime and multiply it with any power of a $4k+1$ prime. And, to construct a $4k+3$ divisor you can pick up any power of a $4k+1$ prime and multiply it with only odd powers of $4k+3$ primes. Now compare these two numbers. Hint for last pointer: Note that for an even number, number of even whole numbers is more than odd numbers, while for an odd number the number of even and odd whole numbers is equal. Can you continue now?