Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 2 down vote accepted

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 $f$.

share|cite|improve this answer
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.

share|cite|improve this answer
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.

share|cite|improve this answer

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:

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.