Could you tell me why the series $\displaystyle{\sum _k \frac{1}{1+k^2x^2}}$ doesn't converge uniformly on $(0,1]$?
Why doesn't this series converge uniformly?
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0Apply Cauchy criterion and consider $x_k=1/k$ – 2012-01-09
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0When you say "tell me why the series ... doesn't converge uniformly" do you mean "prove that the series ... doesn't converge uniformly" or something else? – 2012-01-09
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1The glib answer is that it doesn't converge uniformly on $(0,1]$ because it does not converge at all at $x=0$. – 2012-01-09
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0If $x$ is close to $0$, we have to go a ridiculously long way out to have the partial sum anywhere close to the full sm. – 2012-01-09
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0@Norbert: You meant this: like in the answer we should have that $\frac{1}{1+N^2x^2}<\varepsilon$ (for a fixed $\varepsilon<\frac{1}{2}$), but if we take $x=\frac{1}{N}$ we obtain a contradiction. Is this you argument? – 2012-01-09
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0@John If we want to prove that series doest converge uniformly we must to find $\varepsilon>0$ such that for all $N\in\mathbb{N}$ there exist $n,p\in\mathbb{N}$, $x\in(0,1]$ such that $|\sum\limits_{i=n}^{n+p}a_i(x)|\geq\varepsilon$. Now take $\varepsilon=\frac{1}{2}$, and for all $N\in\mathbb{N}$ take $n=N$, $p=0$ $x=\frac{1}{N}$. Thus we obtain our desired contradiction. – 2012-01-09
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0Is there some differences from what I said? – 2012-01-09
2 Answers
Because, if $x\rightarrow 0$ for fixed $k$, the individual terms tend to $1$.
For uniform convergence you need to have that, independently of the choice of $x$, for each $\varepsilon >0$ there is $N\in \mathbb{N}$ such that $|\sum_{k=n}^{m} a_k(x) | < \varepsilon$ for $n,m \ge N $. Suppose such an $N$ exists. Now choose $\varepsilon = 1/2$ and look at $$ \sum_{N}^{N}a_k = \frac{1}{1+N^2 x^2}.$$ If $x\rightarrow 0$ this tends to $1 > \varepsilon$.
A necessary condition to establish the uniform convergence of a series of functions is for its general term, a sequence of functions, to converge uniformly.
Let's let $\:E=(0,1];\:f_n(x)=1/\left(1+n^2x^2\right).$
We have $$\forall x\in E\:\:\forall\varepsilon>0\:\:\exists N_{x,\varepsilon}>0\:\:\forall n\geqslant N_{x,\varepsilon}\implies|f_n(x)-0|<\varepsilon,\\\text{where }\:N_{x,\varepsilon}>\frac{\sqrt{1-\varepsilon}}{x}.$$
Although, $$\exists\varepsilon\in(0,1/2]\:\:\forall N_{\varepsilon}>0\:\:\exists x_n=\frac{1}n\in E\:\:\:\exists n\geqslant N_{\varepsilon}\implies|f_n(1/n)-0|\geqslant\varepsilon\\$$
$$\implies f_n\overset{\text{unif.}}{\not\to}f\equiv 0.$$