Absolute convergence of $\sum_{n\ge 1}b_n$

Question

Let $\sum_{n\ge 1}a_n$ be an absolutely convergent series of complex numbers. Let $b_n=\begin{cases} a_n, &\quad\text{if } 1\leq n < 100\\ \frac{n+1}{n^2}a_n^2 &\quad\text{if } n\ge 100\\ \end{cases}$ Prove that $\sum_{n\ge 1}b_n$ is an absolutely convergent series.

My attempt:- Let $\sum_{n\ge 1}a_n$ be an absolutely convergent series of complex numbers. Hence, $\lim_{n\to \infty}|a_n|=0$. So, For $1>0$. There is a natural number $N(1)$: $\forall n\ge N(1)\implies |a_n|^2<1$. Consider the convergence of $\sum_{k=N(1)}^\infty |b_n|$.

Case 1. If $N(1)<100$ then avoid the terms before $100$. Run $k$ from $100$. Case 2. If $N(1)\ge 100$, then

$\sum_{k=N(1)}^\infty |b_k|\leq \sum_{k=N(1)}^\infty \frac{k+1}{k^2}$. If $b_n$ would ne non decreasing sequence I could use the result. Here I am helpless.

score 1 · Accepted Answer · answered Apr 28 '19 at 01:30

1

Hint: eventually we have $|a_n|<1$, so $|a_n|^2<|a_n|$. What do you know about the size of $\frac{n+1}{n^2}$? You can make the conclusion just using direct comparison test.

answered Apr 28 '19 at 01:30

Clayton

24,751

$\sum_{n=1}^\infty \frac{n+1}{n^2}=\sum_{n=1}^\infty \frac{1}{n} + \sum_{n=1}^\infty \frac{1}{n^2}$ – Apr 28 '19 at 02:14
Isn't it a divergent series? – Apr 28 '19 at 02:15
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@Unknown: $\frac{n+1}{n^2}<1$ so that $\frac{n+1}{n^2}|a_n|^2<|a_n|^2<|a_n|$. Can you finish? – Clayton Apr 28 '19 at 02:47
@Unknownx: See the message above – Clayton Apr 28 '19 at 03:49
Thank you verymuch:) – Apr 28 '19 at 05:07

Absolute convergence of $\sum_{n\ge 1}b_n$

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