I have trouble solving the following problem:
The positive series $\sum_{n=0}^\infty a_n$ converges, and $\lim_{n\to \infty}\frac{b_n}{a_n}=1$. Prove that the series $\sum_{n=0}^\infty b_n$ converges ($\sum_{n=0}^\infty b_n$ isn't necessarily a positive series), and $$\lim_{n\to \infty}\frac{\sum_{k=n}^\infty b_k}{\sum_{k=n}^\infty a_k}=1$$
I tried to use the Stolz-Cesaro theorem, and also tried if it's available to solve this problem like limit of quotient of two series, but I couldn't make it. Any help would be apprecicated, thanks!
For the first part the proof reduces to limit comparison test which is discussed for example here
For the second part by Stolz-Cesaro we have
$$\frac{\sum_{k=n}^\infty b_k}{\sum_{k=n}^\infty a_k}=\frac{\sum_{k=0}^\infty b_k-\sum_{k=0}^{n-1} b_k}{\sum_{k=0}^\infty a_k-\sum_{k=0}^{n-1} a_k}=\frac{L_b-B_n}{L_a-A_n}=\frac{\frac{L_b-B_n}n}{\frac{L_a-A_n}n}\sim \frac{-nb_n}{-na_n}\to 1$$
using that
$$\lim_{n\to \infty}\frac{L_a-A_n}n=\lim_{n\to \infty}\frac{L_a-A_{n+1}-L_a+A_n}{n+1-n}=\lim_{n\to \infty} -a_n \implies L_a-A_n \sim -na_n$$
$$\lim_{n\to \infty}\frac{L_b-B_n}n=\lim_{n\to \infty}\frac{L_b-B_{n+1}-L_b+B_n}{n+1-n}=\lim_{n\to \infty} -b_n\implies L_b-B_n \sim -nb_n$$