Series convergence

Real Analysis
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Riemann
Posts: 176
Joined: Sat Nov 14, 2015 6:32 am
Location: Melbourne, Australia

Series convergence

#1

Post by Riemann »

Let $\{a_n\}_{n \in \mathbb{N}}$ be a sequence of positive real number such that $\sum \limits_{n=1}^{\infty} a_n$ converges. Is the series

$$\mathcal{S} = \sum_{n=1}^{\infty} n a_n \sin \frac{1}{n}$$

also convergent? Give a brief explanation.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
Papapetros Vaggelis
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Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: Series convergence

#2

Post by Papapetros Vaggelis »

For every \(\displaystyle{n\in\mathbb{N}}\) holds

\(\displaystyle{\left|n\,a_n\,\sin\,\dfrac{1}{n}\right|=n\,a_n\,\left|\sin\,\dfrac{1}{n}\right|\leq n\,a_n\,\dfrac{1}{n}=a_n}\)

and \(\displaystyle{\sum_{n=1}^{\infty}a_n<\infty}\).

So, the series \(\displaystyle{\sum_{n=1}^{\infty}n\,a_n\,\sin\,\dfrac{1}{n}}\) converges since

it converges absolutely.
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