Series convergence
Series convergence
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.
$$\mathcal{S} = \sum_{n=1}^{\infty} n a_n \sin \frac{1}{n}$$
also convergent? Give a brief explanation.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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Re: Series convergence
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.
\(\displaystyle{\left|n\,a_n\,\sin\,\dfrac{1}{n}\right|=n\,a_n\,\left|\sin\,\dfrac{1}{n}\right|\leq n\,a_n\,\dfrac{1}{n}=a_n}\)
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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