# Prove that if $\sum_{n} p_n$ is convergent, then $\sum_{n} |a_n|$ is convergent.

Let {$$a_n$$} and {$$p_n$$} be two sequences with $$p_n>0$$ for all $$n\in \mathbb{N}$$.

(i) Suppose that for all $$n\in \mathbb{N}$$, $$a_n\neq 0$$, and $$\left|\frac{a_{n+1}}{a_n}\right|\leq \frac{p_{n+1}}{p_n}$$. Prove that if $$\sum_{n}p_n$$ is convergent, then $$\sum_{n}|a_n|$$ is convergent.

Initially, I thought Theorem 3.25 (a) in Baby Rudin, i.e.

If $$|a_n|\leq c_n$$ for $$n\geq N_0$$, where $$N_0$$ is some fixed integer, and if $$\sum_{n}c_n$$ converges, then $$\sum_{n}a_n$$ converges.

would work. However, I noticed it is difficult to use this to prove the absolute convergence like this problem. What approach or theorem does work?

Similarly,

(ii) Suppose that $$\lim\limits_{n\to\infty} \frac{|a_n|}{p_n}=l\in (0,\infty)$$. Prove that the series $$\sum_{n}|a_n|$$ and $$\sum_{n}p_n$$ are either both divergent or both convergent.

I have been looking for a theorem that can directly apply to this (ii) in the same text, however, I have no idea.

• (ii) is the well-known “limit comparison test” Nov 18, 2022 at 13:19
• Hint: $$\frac{|a_n|}{|a_1|}=\frac{|a_2|}{|a_1|}\cdots\frac{|a_n|}{|a_{n-1}|}\leq \frac{p_2}{p_1}\cdots \frac{p_n}{p_{n-1}}=\frac{p_n}{p_1}$$ Nov 18, 2022 at 13:20

Since$$|a_n|=\prod_{i=1}^{n-1} |\frac{a_{i+1}}{a_i} a_1|$$ By condition given,we get$$\prod_{i=1}^{n-1} |\frac{a_{i+1}}{a_i}| \le \prod_{i=1}^{n-1}\frac{p_{i+1}}{p_i}$$ Hence$$|a_n|\le p_n \frac{|a_1|}{p_1}$$ Since$$\sum_{i=1}^{\infty} p_i$$ is convergent,$$\sum_{i=1}^{\infty} |a_i|$$ is convergent by comparison test.