# Uniform convergence of $n\left(\sqrt{x+\frac{1}{n}}-\sqrt{x}\right)$

I need to say whether or not $f_n(x)=n\left(\sqrt{x+\frac{1}{n}}-\sqrt{x}\right)$ is uniformly convergent on $(0,\infty)$.

I've found that the function is locally convergent to $f(x)=\frac{1}{2\sqrt{x}}$ and was trying to find $\sup{|f_n(x)-f(x)|}$.

I got the derivative $f_n'(x)= \frac{2nx\left(x-\sqrt x\sqrt{x+\frac{1}{n}}\right)+\sqrt{x}\sqrt{x+\frac{1}{n}}}{...}$ and could not find $x$ so that $f_n'(x)=0$

Any ideas?

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Note that each of the functions $\displaystyle f_n(x) = n\biggl(\sqrt{x+\frac{1}{n}}-\sqrt{x}\biggr)$ is bounded on $(0,\infty)$, with $f_n(x) \leq \sqrt{n}$. Since $\displaystyle f(x) = \frac{1}{2\sqrt{x}}$ is unbounded on $(0,\infty)$, the sequence $\{f_n\}_{n=1}^\infty$ does not converge uniformly to $f$.

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How do I prove that this function is bounded? I've been trying to prove that there is $N(n)$ such that for each $x>N(n)$ the derivative is negative $f_n'(x)<0$ –  Michael May 27 '12 at 7:00

In fact, $f_n'(x)<0$ for all $x>0$. Thus, each $f_n$ is continuous, positive, and decreasing on $[0,\infty)$. It follows that $$\sup\{f(x):x>0\} = f(0) = \sqrt{n}.$$

As $f$ is unbounded, you can't have uniform convergence.

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Actually, I havn't been able to prove that $f_n'(x)<0$. For $x<\frac{1}{2n}$ the deriviative is positive. –  Michael May 26 '12 at 18:13

By the definition of uniform convergence, if they did uniformly converge to a limit function $f$ there would be some $N$ such that if $n \geq N$ then $|f_n(x) - f(x)| < 1$ for all $x$. In particular this would hold for $n = N$ itself. Since $f_N(x)$ is a bounded function, this means so is $f(x)$. So we can let $M$ be such that $|f(x)| < M$ for all $x$. Thus by the above, for all $n > N$ and all $x \in (0,\infty)$ we have $$|f_n(x)| \leq |f_n(x) - f(x)| + |f(x)|$$ $$< M + 1$$ But $f_n({1 \over n}) = (\sqrt{2} - 1)\sqrt{n}$. For $n$ large enough this will be greater than $M + 1$, a contradiction. So the functions don't converge uniformly.

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As already pointed out by several others, the convergence is not uniform on $(0,\infty)$ but it may be worth noting that it is uniform on every subinterval $[c,\infty)$ for $c > 0$.

This follows from the following computation:

$$\begin{split} |f_n(x) - f (x)| &= \left|\frac{1}{\sqrt{x+\dfrac1n}+\sqrt{x}} - \frac1{2\sqrt{x}}\right| \\ &= \left|\frac{\sqrt{x}-\sqrt{x+\dfrac1n}}{2\sqrt{x}\left( \sqrt{x+\dfrac1n}+\sqrt{x} \right)} \right|\\ &= \left|\frac{\dfrac1n}{2\sqrt{x}\left( \sqrt{x+\dfrac1n}+\sqrt{x} \right)^2} \right| \le \frac1n \cdot \frac{1}{8c\sqrt{c}}. \end{split}$$

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