Can asymptotic approximations of terms be used to give asymptotic approximations of a sum?

Question

Let $a_n \sim b_n$,

More formally I mean, $\displaystyle{} \lim_{n \rightarrow \infty}\frac{a_n}{b_n} = 1$

Would that then imply that for some $n_0$, $\displaystyle{} \lim_{T \rightarrow \infty} \cfrac{\sum_{n = n_0}^{T} a_n}{\sum_{n = n_0}^{T} b_n}$ = 1,

Answer 1

Consider $a_n=\frac{1}{n(n+1)},\ b_n=\frac{1}{n^2}.$ If your idea worked, then the two infinite sums from some $n_0$ onward would be equal (since each sum converges). However the first sum telescopes to a rational number, while the second sum differs from $\frac{\pi^2}{6}$ by a rational number.

Edit: Even easier: since for each $n$ we have $a_n<b_n$ the sums from some point on cannot be equal.

Answer 2

Consider $a_n=1/n^2$, $b_n=a_n+ 1/n^4$

They satisfy your pre-condition, but $ \sum a_n = c_1 \neq c_2 = \sum b_n$