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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?


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

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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 $x$

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

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