Infinite divergent series - does the convergence of quotient imply convergence of difference?

Question

I have two series with all terms positive,

$$\sum_{i=1}^{n}a_i \equiv A_n >0,\;\; \sum_{i=1}^{n}b_i\equiv B_n>0.$$

Each series diverge as $n\to \infty$. We also have $A_n \leq B_n, \forall n$.

Assume that

$$\lim_{n\to \infty}\frac{A_n}{B_n} \to 1$$

Does this imply that

$$\lim_{n\to \infty}\left(B_n-A_n\right) \to 0,\;\;\;???$$

I suspect not, because, clumsily,

$$B_n-A_n = B_n\cdot \left (1-\frac{A_n}{B_n}\right)$$

so it appears we are examining the limit

$$\lim_{n\to \infty}\left[B_n\cdot \left (1-\frac{A_n}{B_n}\right)\right]$$

and one of the terms goes to infinity while the other goes to zero, so some condition related to the speed of convergence against the speed of divergence appears to be needed. Am I right in this?

Answer 1

Your thoughts are correct. To form our intuition, consider the two functions $f(x)=x^2+2x$ and $g(x)=x^2$. Clearly $\lim_{x\to\infty}f(x)/g(x)=1$ while $f(x)-g(x)=2x\to\infty$ as $x\to\infty$. Formalizing this to series is not too hard: define $$a_i=2i-1\qquad\text{and}\qquad b_i=2i.$$ Then we can show $A_n=n^2$ while $B_n=n^2+n$ and the problem is essentially the same as how we treated $f(x)$ and $g(x)$ above.

Answer 2

This is not true. Suppose that $$A_n=1+n(n+1)$$ $$B_n=n(n+1)$$ Then $$\lim_{n\to\infty} \frac{1+n(n+1)}{n(n+1)}=1$$ But $$\lim_{n\to\infty} (n(n+1)-(1+n(n+1)))=-1\ne0$$ A valid pair of sequences that give these $A_n,B_n$ is $$a_i=\begin{cases}3&i=1\\2i&\text{otherwise}\end{cases}$$ $$b_i=2i$$