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Given a convergent sum $\sum_{n=1}^{\infty}a_n \ $, prove/disprove: $\sum_{n=1}^{\infty}a_n(1-a_n)$ is convergent

My Attempt:

By dividing the question into cases, as for the first case; $\sum_{n=1}^{\infty}a_n$ is definitely converges, then it's pretty easy to prove that $\sum_{n=1}^{\infty}(a_n)^2$ converges, therefore $\sum_{n=1}^{\infty}a_n - \sum_{n=1}^{\infty}(a_n)^2$ converges and we're done.

In the second case, $\sum_{n=1}^{\infty}a_n$ is conditionally convergent, and now it's not clear that $\sum_{n=1}^{\infty}(a_n)^2$ converges. for example, let $a_n = \frac{(-1)^n}{\sqrt{n}}$, then $\sum a_n$ converges, but $\sum (a_n)^2 = \sum \frac{1}{n}$ which diverges.

I've made another attempt and tried to use Abel theorem.

$\sum {a_n}$ converges, then I've tried to prove that the sequence $\{(1-a_n)\}_{n=1}^{\infty}$ is monotonic and booundedn. clearly, $\{1-a_n\}$ is bounded as $\lim_{n\to \infty} (1-a_n) = 1$, but I have no idea if it's even possible to prove that this sequence is monotonic, as there is no information given on $\{a_n\}$ positivity/ negativity, but only that $\sum a_n$

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  • $\begingroup$ I believe this is a hint: $\sum a_n$ converging implies $a_n\to0$ as $n\to\infty$.Thus, $1-a_n\to1$ as $n\to\infty$. $\endgroup$ – Clayton Jul 20 at 11:06
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    $\begingroup$ $\sum_{n=1}^{\infty}a_n(1-a_n)$ is an expression, not a statement. You cannot prove or disprove it. $\endgroup$ – Martin R Jul 20 at 11:13
  • $\begingroup$ You can't prove $(1-a_n)$ is monotonic. It may not be. In fact, you even gave an example where it isn't monotonic, namely $a_n=(-1)^n/\sqrt{n}$. In fact, since you showed (in that case) $\sum a_n$ converges but the sum of squares $\sum a_n^2$ diverges, haven't you provided an example of $\sum a_n(1-a_n)$ diverging? So you've already found a disproof. $\endgroup$ – runway44 Jul 20 at 11:22
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I think you already have a complete answer.

For the example of $a_n=\frac{(-1)^n}{\sqrt n}$, we know $$\sum_{n=1}^Ma_n(1-a_n)=\sum_{n=1}^Ma_n-\sum_{n=1}^M(a_n)^2$$ The first sum is bounded by $$L=\sum_{n=1}^\infty a_n<\infty$$ but $\sum_{n=1}^M(a_n)^2$ tends to infinity as $M\to\infty$. Thus as $M\to\infty$ we have $$\sum_{n=1}^Ma_n(1-a_n)\to-\infty$$ So in this case (convergent, but not absolutely convergent) it is disproved.

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    $\begingroup$ I don't think "The first sum is bounded" is what you want to say. Why not just say $\sum a_n$ converges and $\sum a_n^2$ diverges? $\endgroup$ – zhw. Jul 20 at 19:10
  • $\begingroup$ @zhw Yes, you're right. For some reason I was thinking the sum was alternating starting with a positive term, not a negative. But it is still bounded (and convergent), either of which is sufficient. $\endgroup$ – Especially Lime Jul 20 at 20:43
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Consider $b_n = e^{\log a_n(1-a_n)} = e^{\log a_n + \log (1-a_n)}$. Since $a_n \to 0,$ the first term is upper-bounded by $1$. Again, since $a_n \to 0$, the second term can be upper-bounded using Maclaurin series, since $\log (1-x) < -x$: $$ \sum_{k=1}^{\infty}e^{-a_k} < \int_{1}^{\infty}e^{-x}dx<\infty $$

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