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Let $ a_n $ be a non-negative sequence, such that $ \sum_{n=1}^{\infty}\sqrt{a_{n}} $ converge. Is it true that $ \sum_{n=1}^{\infty}a_{n} $ converge?

I think that it is. But I want to make sure because it appeared in my final exam. Here is my reasoning:

Since $ \sum_{n=1}^{\infty}\sqrt{a_{n}} $ converges, $ \sqrt{a_{n}}\underset{n\to\infty}{\to}0 $ and thus $ a_{n}\underset{n\to\infty}{\to}0 $. So there exists some $ n_0 $ such that, for all $ n>n_0$, it follows that $ 0\leq a_{n}<\frac{1}{2} $. Thus, for each $ n>n_0 $ we have

$$ a_{n}<\sqrt{a_{n}} .$$

So, from the comparison test, we get the convergence of

$$ \sum_{n=1}^{\infty}a_{n}. $$

Do you agree?

RSpeciel
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FreeZe
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2 Answers2

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Your reasoning is correct, this is just a slick, slightly hidden comparison test. Note, however, that the converse does not hold! For example, with $a_n=\frac{1}{n^2}$, we have $$\sum_{n=1}^\infty a_n=\frac{\pi^2}{6}<\infty,$$ (this is the Basel Problem) but $$\sum_{n=1}^\infty \sqrt{a_n}=\sum_{n=1}^\infty\frac{1}{n},$$ which diverges.

RSpeciel
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Yes that proof is fine, as an alternative since

$$\sum a_n \le \left(\sum \sqrt{a_n}\right)^2$$

the series converges.

user
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