Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

So I screwed up a problem on my exam. I know that now. But pure mathematics is as difficult and terrifying as it is rewarding for me, and I can't let this go! If someone could tell me if the following is right or wrong, that would be much appreciated.

I was asked to prove the uniform convergence of the series $f_n(x)=\frac{x}{x+n}$ on $[1,\infty)$.

So I've taken $$f=\lim_{n\to \infty} \frac{x}{x+n}=0$$ and $$|f_n-f|=|\frac{x}{x+n}-0|=\frac{x}{x+n}<1, \forall x\in[1,\infty), \forall n\in\Bbb{N}$$

Then, I simply took $N>0$ and $n \ge N$, concluding that $f_n(x)$ converged uniformly to $1$.

This seems too easy to be right (so did the solution I put on my exam, God help me!). Is it?

Thanks!

share|improve this question

2 Answers 2

The sequence $f_n$ does not converge uniformly to $f$ since $$ \sup_{x\ge 1}\biggl|\frac x{x+n}\biggr|=1 $$ for each $n\ge1$.

share|improve this answer

Not really, sorry. You need to show, for given $\varepsilon>0$, that there is $N\in \mathbb{N}$, such that $$\frac{x}{x+n}< \varepsilon$$ for all $n\ge N$, independently of $x$. Which, I'm afraid, is not even true, because if you have any $N$ you can choose $x $ very large (larger than $N$) such that the fraction becomes $>\frac{1}{2}$, say. What can be shown is uniform convergence on compact sets, then you'd start out with an upper bound on $x$

share|improve this answer
    
How embarrassing... Clearly I misread the question in my sleep-deprived stupor... Thank you! –  user126291 Apr 14 at 6:56

Your Answer

 
discard

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.