Suppose I have an infinite sum like $$ \sum_{n=0}^{\infty} 2^n $$ which diverges.
I want to know what is the limit of the following sum $$ \lim_{N\to\infty} \frac{\sum_{n=0}^{N-1} 2^n}{2^N-1}$$
One may argue that $$ \lim_{N\to\infty} \frac{\sum_{n=0}^{N-1} 2^n}{2^N-1} = \lim_{N\to\infty} \frac{2^N - 1}{2^N-1} = \lim_{N\to\infty} 1 = 1$$ However, if we write $$ \lim_{N\to\infty} \sum_{n=0}^{N-1} \frac{2^n}{2^N-1} $$ The sum fails to pass the ratio test because $2^{n+1}/2^n = 2 > 1$ which suggests that series is divergent.
How should I rigorously understand this?
As others have pointed out, you cannot apply the ratio test to a finite sum. In the limit, the ratio test would be:
$$\lim_{n \to \infty}\dfrac{\displaystyle \lim_{N \to \infty}\tfrac{2^{n+1}}{2^N-1}}{\displaystyle \lim_{N \to \infty} \tfrac{2^n}{2^N-1}} \to \dfrac{0}{0}\text{ which is undefined}$$
You were trying to take the limit of the finite ratios, which is not the same as the limit of the limit ratio. In other words, the Ratio Test does not apply/cannot be used for this series.