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Let $$f_n(x)=\frac{x^2}{(1+x^2)^n}$$ be a sequence of functions from $[0,\infty)$ to $\mathbb{R}$.

Does the series of these functions converge uniformly on $[0,\infty)$?

What about $$f_n(x)=\frac{(-1)^{n-1}x^2}{(1+x^2)^n}$$


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tomerg, you should think of accepting answers to your questions. – xen Dec 10 '11 at 18:05
thank you for the advising! – tomerg Dec 10 '11 at 18:08

1 Answer 1

The first series, $ \sum\limits_{n=0}^\infty{x^2\over (1+x^2)^n}$, converges pointwise to 0 for $x=0$. For $x\ne0$, the series is geometric and converges pointwise to $$f(x)=x^2\cdot{1\over 1-{1\over 1+x^2}}=1+x^2.$$

Note, the limit function is not continuous (at $x=0$ , in particular).

Since the uniform limit of a series of continuous functions is continuous, the series $ \sum\limits_{n=0}^\infty{x^2\over (1+x^2)^n}$ does not converge uniformly on $[0,\infty)$.

The second series ($\sum\limits_{n=0}^\infty{(-1)^{n-1}x^2\over (1+x^2)^n}$) has been asked about here, recently: Prove: the function series, $\sum_{n=0}^{\infty}\frac{(-1)^{n-1}x^2}{(1+x^2)^n}$ uniformly converges

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