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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!

QIRUN CONG
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1 Answers1

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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

$$\lim_{n\to \infty}\frac{L_a-A_n}n=\lim_{n\to \infty}\frac{L_a-A_{n+1}-L_a+A_n}{n+1-n}=\lim_{n\to \infty} -a_n \implies L_a-A_n \sim -na_n$$

$$\lim_{n\to \infty}\frac{L_b-B_n}n=\lim_{n\to \infty}\frac{L_b-B_{n+1}-L_b+B_n}{n+1-n}=\lim_{n\to \infty} -b_n\implies L_b-B_n \sim -nb_n$$

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