An easy assertion about series

Task number: 2936

Show that if the series \(\displaystyle \sum_{n=1}^{\infty}a_n \) converges absolutely, then the series \(\displaystyle \sum_{n=1}^{\infty}a_n^2 \) also converges.

Does this assertion hold even for convergent series that do not converge absolutely?

  • Resolution

    Because the series \(\displaystyle \sum_{n=1}^{\infty}a_n \) converges, there exists some \(n_0\) such that \(a_n\le 1\) whenever \(n>n_0\).

    We can bound the given series by the convergent series \(\displaystyle \sum_{n=n_0}^{\infty}a_n^2\le \sum_{n=n_0}^{\infty}|a_n| \).

    (The first \(n_0-1\) terms do not influence the convergence of the series.)

    Absolute convergence is required; for example, the series \(\displaystyle \sum_{n=1}^{\infty}(-1)^nn^{-\frac12} \) converges but not absolutely, and the series \(\displaystyle \sum_{n=1}^{\infty}\left((-1)^nn^{-\frac12}\right)^2= \sum_{n=1}^{\infty}n^{-1}\) diverges of course.

Difficulty level: Easy task (using definitions and simple reasoning)
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