# Show a series of integrals is convergent

Let $\{r_1, r_2, r_3,\ldots \}$ be an enumeration of the rationals in $[0,1]$, and let $(a_k)$ be a sequence such that $\sum_{k=1}^{\infty}{a_k}$ converges absolutely. Set $f_k(x) = a_k\,|x-r_k|^{-1/2}$. I'd like to show that $$\sum_{k=1}^{\infty}{\int_{[0,1]}{|f_k(x)|\,dx}}<\infty\,.$$

I've rewritten the sum as $$\sum{|a_k|\,\int_0^1{\dfrac{1}{|x-r_k|^{1/2}}\,dx}}\,.$$ For each $k$, I broke up the integral into two integrals to account for the absolute value. After integrating, I got $2(\sqrt{r_k}+\sqrt{1-r_k})$. (If anyone is willing to double check that... that'd be awesome.)

But now how do I show that $$2\sum{|a_k|\,(\sqrt{r_k}+\sqrt{1-r_k})}$$ converges? That's where I'm stuck.

EDIT: You know, I typed all this stuff up, and then as soon as I submitted, I realized that $\sqrt{r_k}$ and $\sqrt{1-r_k}$ are bounded by $1$. So $\sum{|a_k|\,(\sqrt{r_k}+\sqrt{1-r_k})}\le \sum{2|a_k|}$, and we know that $\sum{|a_k|}$ is convergent. It's a simple comparison. (Someone please tell me I'm correct, haha.)

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you are right and in fact you can get a tighter bound of $\sqrt{2}$ instead of $2$ –  user17762 Nov 25 '10 at 3:46

## 1 Answer

$\textbf{HINT:}$

Note that $\sqrt{r_k} + \sqrt{1-r_k} \leq \sqrt{2}$, $\forall r_k \in \mathbb{Q} \cap [0,1]$.

Prove the above by Arithmetic mean - Geometric Mean Inequality

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