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

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 $$\sum a_n$$

• I believe this is a hint: $\sum a_n$ converging implies $a_n\to0$ as $n\to\infty$.Thus, $1-a_n\to1$ as $n\to\infty$. – Clayton Jul 20 at 11:06
• $\sum_{n=1}^{\infty}a_n(1-a_n)$ is an expression, not a statement. You cannot prove or disprove it. – Martin R Jul 20 at 11:13
• 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. – runway44 Jul 20 at 11:22

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.
• 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? – zhw. Jul 20 at 19:10
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 $$\log (1-x) < -x$$: $$\sum_{k=1}^{\infty}e^{-a_k} < \int_{1}^{\infty}e^{-x}dx<\infty$$