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I have to use the Dominated Convergence Theorem to show that $\lim \limits_{n \to \infty}$ $\int_0^1f_n(x)dx=0$ where $f_n(x)=\frac{n\sqrt{x}}{1+n^2x^2}$.

I did the following:
$$\frac{n\sqrt{x}}{1+n^2x^2} <\frac{n\sqrt{x}}{n^2x^2} = \frac{x^{-\frac{3}{2}}}{n}\leq x^{-\frac{3}{2}} $$
But $$\int_0^1x^{-\frac{3}{2}}dx=\frac{-2}{\sqrt{x}}\biggr|_0^1$$ which doesn't seem right. Any help will be appreciated.

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    I imagine there should be a limit as $n\to\infty$ on the left side of your original statement? – 2017-02-05
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    Thanks for the edit. It should be like you said, but the question in the book I was solving stated it like that. So I didn't notice. – 2017-02-05
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    no worries. I will give this some thought. I just took my qualifying exams in analysis, so this would be good practice. – 2017-02-05

3 Answers 3

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Let $f_n(x)=\frac{n\sqrt{x}}{1+n^2x^2}$. Then, differentiating with respect to $n$ reveals

$$\begin{align} \frac{df_n(x)}{dn}&=\frac{\sqrt{x}}{1+n^2x^2}-\frac{2n^2x^{5/2}}{(1+n^2x^2)^2}\\\\ &=\frac{\sqrt{x}(1-n^2x^2)}{(1+n^2x^2)^2}\tag1 \end{align}$$

From $(1)$, it is easy to see that $\frac{df_n(x)}{dn}=0$ when $n=1/x$. Moreover, we can see from $(1)$ that $f_n(x)$ is a maximum when $n=1/x$. Therefore, we have

$$\sup_{n} \left(\frac{n\sqrt{x}}{1+n^2x^2}\right)=\frac{1}{2\sqrt{x}}$$

whereby we have a dominating function that is integrable on $[0,1]$.

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    This is brilliant. However, I have a doubt. Won't we have to check if $\frac{d^2f_n(x)}{dn^2}<0$ for it to be a maxima? – 2017-02-06
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    We only need to look at the first derivative to the left and right of $n=1/x$ to see it's a maximum. – 2017-02-06
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Use for example:

$$\frac{n\sqrt x}{1+n^2x^2}=\frac{1}{\sqrt x}\cdot\frac{1}{\frac{1}{nx}+nx}\le \frac{1}{2\sqrt x}$$ since $u+\frac{1}{u}\ge 2$ for $u>0$

And $\int_0^1\frac{1}{2\sqrt x}=1<\infty$

(this is equivalent to AM-GM inequality on the denominator $1+n^2x^2\ge2nx$)

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    That's a really nice trick you used there. Thanks! – 2017-02-05
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    Beat me to it. This is a great answer. +1 – 2017-02-05
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    You are welcome. Glad it helped. – 2017-02-05
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A worse bound, a slightly different approach: $$ \frac{n\sqrt x}{1+n^2x^2} \le \frac{n\sqrt x}{\sqrt{1+n^2x^2}} = n \sqrt{ \frac{x}{1+n^2x^2}} = \frac{1}{\sqrt{x}} \sqrt{\frac{1}{1+\frac{1}{n^2x^2}}} \le \frac{1}{\sqrt{x}} $$

Where we used tha facts that $x \ge \sqrt{x} $ if $x \ge 1$ and $\sqrt{\frac{1}{1+x}} \le 1 $ on $[0,+\infty)$.