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$$\sum_{n=0}^{\infty}\frac{(-1)^{n-1}x^2}{(1+x^2)^n}$$

I want to know whether this function series in uniformly converges or not. I can recognize Leibniz here, so if I'll be able to prove that $\lim_{n \to \infty} f_n(x)=f(x)=0$ the function series will be uniformly converged to 0.

How should I do that?

Thanks.

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One should have $g(x) \sum_{n=0}^\infty h(x)^n = g(x)/(1-h(x))$ for all $x$ such that $|h(x)| < 1$. In this case $g(x) = -x^2$, and $h(x)=-1/(1+x^2)$. So you sure it converges to $0$ for all $x$? –  p.s. Dec 7 '11 at 10:05
    
@p.s. I´m sorry but I don´t think I have got your comment. May I ask you to be a little more explicit? In specific I don´t understand why you say: "One should have... ...for all $x$ such that $|h(x)|\leq 1$". –  Giovanni De Gaetano Dec 7 '11 at 10:16
    
This is an answer. I write it down first as a comment because I don´t feel confident in its exactness. As Jozef pointed out it is enough to prove the limit: $\lim_{n\rightarrow\infty} f_n(x)=0$ for each $x$, in order to have that the series is uniformly convergent (not to zero!). Then a computation shows that $max(f_n(x))= \frac{2}{n-2}(1+\frac{2}{n-2})^{-n} \leq \frac{1}{n}\cdot k \rightarrow 0$. Now, by Leibniz Criterion (it wouldn´t be true otherwise) the original series is uniformly convergent. Please give me a feedback, so (if correct) I can post it as an answer. –  Giovanni De Gaetano Dec 7 '11 at 10:44
    
@Student73 I'm sorry, I didn't see your comment. I don't think showing pointwise convergence is sufficient, though (?). I deleted my answer. If you like, you can post yours (your inequality shows the convergence is uniform) –  David Mitra Dec 7 '11 at 11:44
    
@David Mitra. I´m really sorry you deleted your answer! I didn´t mean to be ironic or sarcastic! Actually I think it is correct that you answer the question, for two reasons. 1) Who really answer the question is who knows that the answer is correct, I guessed but I didn´t know! 2) I´m always a little confused about different kinds of convergence and I don´t want take the risk to insert an error in the answer. Anyway I want to really thank you for your correctness and politeness! :) –  Giovanni De Gaetano Dec 7 '11 at 12:23
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1 Answer 1

up vote 4 down vote accepted

Your series is a geometric series; therefore the partial sums $s_n:=\sum_{k=0}^{n-1}\ldots$ can be computed explicitly. One obtains

$$s_n={x^2(1+x^2) \over 2+x^2}\ \left(1-\Bigl({-1\over 1+x^2}\Bigr)^n\right)\ ;$$

so it is enough to prove that $g_n(x):={x^2\over (1+x^2)^n}$ converges to $0$ uniformly with $n\to\infty$. To this end put $x^2=:u$ and investigate

$$h_n(u):=u(1+u)^{-n}\qquad (0\leq u<\infty)$$

instead. It is easy to verify that for $n\geq2$ the function $h_n$ assumes its maximum at $u_n={1\over n-1}$; and the value there is

$$h(u_n)={1\over n-1}\ \Bigl(1+{1\over n-1}\Bigr)^{-n}\ .$$

As the second factor on the right converges to ${1\over e}$ it follows that $h_n(u)\leq {C\over n}$ for some $C$ and all $u\geq 0$, $n\geq2$.

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