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I'm trying to check wheter this function series is uniformly converges or not:$f_n(x)=\frac{2nx}{1+n^2x^2}$.

I tried to check $\lim_{n \to \infty} \sup |f_n(x)-f(x)|$ is 0 where $f(x)=\lim_{n \to \infty}f_n(x)$.

Let $r_n(x)=f_n(x)-f(x)$. I checked maximum points by $r_n'(x)=0$ and got $x=\frac{1}{n}$ as maximim, so $r_n(\frac{1}{n})=1$ so $\lim_{n \to \infty} \sup |f_n(x)-f(x)|=1 \neq0$ so I concluded that there's not uniform convergence. But then by simple inequalities: $|f_n(x)-f(x)|=\frac{2nx}{1+n^2x^2}\leq \frac{2}{\frac {1}{nx}+nx}\leq\frac{2}{n}\to0$, so there is a uniform convergence. whats wrong with my first try?

Edit:$x \in [1, \infty)$ Thanks a lot.

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up vote 3 down vote accepted

If you study the convergence on the interval $[1,\infty)$, then you're interested in $\sup\limits_{x\in[1,\infty)} |f_n(x)-f(x)|$. Since $\frac1n\notin[1,\infty)$, this is not what you computed in the first part.

I guess that if you have a look at your computations again, you'll find out that $r'_n(x)<0$ for $x>\frac1n$, which implies that the function $r_n(x)=f_n(x)-f(x)$ is decreasing on $[1,\infty)$. (In fact, it is decreasing already on a bigger interval $[1/n,\infty)$.)

Thus the maximal possible value is at the leftmost point of this interval, which is $r_n(1)=f_n(1)=\frac{2n}{1+n^2}$. This converges to zero.

So your first approach is fine, the only thing is you have to find the supremum of $|f_n(x)-f(x)|$ on the same interval, for which you want to test the uniform convergence.

EDIT: The rest is about the first version of the post, without the condition $x\in[1,\infty)$. I think the best thing is to keep it here too.

In your second solution you used the inequality which is not valid. You argued that: $$\frac1{nx}+nx \ge n \Rightarrow \frac2{\frac1{nx}+nx} \le \frac 2n.$$

In fact you only have $$\frac1{nx}+nx \ge 2 \Rightarrow \frac2{\frac1{nx}+nx} \le \frac 22=1,$$ which is consistent with your first solution.

There are many ways to see that for $t>0$ we have $$t+\frac1t\ge 2.$$ (This is the inequality I used above for $t=nx$.)

E.g. you can study the derivative of the function $f(t)=t+\frac1t$. Quite simple way to see this is using AM-GM inequality, which gives you: $$\frac{t+\frac1t}2\ge \sqrt{t\cdot\frac1t}=1.$$

It is known that we get the equality in AM-GM only when the summands are equal, which is in our case $t=\frac1t$, i.e. $nx=t=1$ and $x=\frac1n$; just the same as you found in your first approach.

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I forgot to mention that $x \in [1, \infty)$, I'm sorry. – Jozef Jan 28 '12 at 11:39

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