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Prove or disprove two statements about absolute convergence

Prove or disprove the following two statements.

  1. If \sum a_n is absolutely convergent, then

        \[ \sum \frac{a_n^2}{(1+a_n^2)} \]

    is absolutely convergent.

  2. If \sum a_n is absolutely convergent, and if a_n \neq -1 for all n, then,

        \[ \sum \frac{a_n}{1+a_n} \]

    is absolutely convergent.


  1. Proof. Since \sum a_n converges absolutely, we know \sum |a_n| converges. By a previous exercise (Section 10.20, Exercise #50) we know that this implies that \sum a_n^2 converges. Then we have,

        \[ \frac{a_n^2}{1+a_n^2} \leq a_n^2 \]

    for all n since 1+ a_n^2 \geq 1. Hence, by comparison with \sum a_n^2 we have the convergence of

        \[ \sum_{n=1}^{\infty} \frac{a_n^2}{1+a_n^2}. \qquad \blacksquare \]

  2. Proof. Since \sum a_n converges absolutely, we know \sum |a_n| converges. Then, using the limit comparison test we have,

        \begin{align*}  \lim_{n \to \infty} \frac{|a_n|}{\left| \frac{a_n}{1+a_n} \right|} &= \lim_{n \to \infty} |1 + a_n| \\[9pt]  &= 1. \end{align*}

    The final equality follows since \sum a_n converges means we must have \lim_{n \to \infty} a_n = 0. Hence, the series \sum \left| \frac{a_n}{1+a_n} \right| converges; and thus,

        \[ \sum \frac{a_n}{1+a_n} \]

    converges absolutely (as long as a_n \neq -1 so that all of the terms are defined). \qquad \blacksquare

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