1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ (ii) is the well-known “limit comparison test” $\endgroup$
    – Martin R
    Nov 18, 2022 at 13:19
  • 3
    $\begingroup$ Hint: $$\frac{|a_n|}{|a_1|}=\frac{|a_2|}{|a_1|}\cdots\frac{|a_n|}{|a_{n-1}|}\leq \frac{p_2}{p_1}\cdots \frac{p_n}{p_{n-1}}=\frac{p_n}{p_1}$$ $\endgroup$
    – Thomas Andrews
    Nov 18, 2022 at 13:20

1 Answer 1

Reset to default
3
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.