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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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    @David. Thank you for the precisations, they are useful to me!$k$is supposed to be a positive constant I didn´t want to compute, and I obtained my inequality as you did, i.e. by computing the first derivative!2011-12-07

1 Answers 1

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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 $u\geq 0$, $n\geq2$.