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Consider the series $\sum_k\frac{(-1)^k}{k+x^2}$. Why does it converge uniformly on $[0,\infty)$? and why doesn't it converge absolutely, always on $[0,\infty)$?

The only thing that I noticed is that it converges by Leibniz.

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By "why" do you mean "give a proof", or "give me an intuitively satisfying reason"? – Alex Becker Jan 9 '12 at 7:02
Because for sufficiently big $k$ we $|\frac{(-1)^k}{k+x^2}|=\frac{1}{k+x^2}>\frac{1}{(1+c)k}$ for some $c>0$ depending on fixed $x$. Intuitively this series doesn't converge absolutely because series of modules is similar to harmonic series, which are doesn't converge. – Norbert Jan 9 '12 at 7:12
up vote 7 down vote accepted

It converges uniformly because for an alternating series $\sum_k(-1)^ka_k$, such that $a_k\geq 0$ for all $k$ and such that $a_k$ converges monotonically to $0$, the distance between the $n^\text{th}$ partial sum and the limit is less than or equal to $a_n$ (e.g., see Wikipedia), and because $\frac{1}{k+x^2}\leq\frac{1}{k}$ for all $x$. Therefore, for all $x$, you can show that the distance between the $n^\text{th}$ partial sum at $x$ and the limit at $x$ is bounded by the corresponding distance at $0$.

It does not converge absolutely anywhere because the harmonic series diverges, and because $\frac{1}{k+x^2}\geq \frac{1}{k(1+x^2)}=\frac{1}{1+x^2}\cdot\frac{1}{k}$ for all $k\geq 1$ and $x\geq 0$.

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This is just a slight expansion of Jonas Meyer's answer.

A series of functions converges uniformly on $I$ if and only if it is uniformly Cauchy on $I$; that is, if and only if for every $\epsilon>0$, there is a positive integer $N$ such that for all positive integers $k$ and $M\ge N$ $$ \tag{1}\Bigl|\,\sum_{n=M}^{M+k} f_n(x)\,\Bigr|<\epsilon $$ for all $x\in I$.

Your series has the form $\sum\limits_{i=1}^\infty (-1)^n f_n(x)$, where for each $x$, the sequence $\{f_n(x)\}$ satisfies

$\ \ \ $1) $f_n(x)$ is nonnegative for each $n$,

$\ \ \ $2) the sequence $f_n(x)$ is decreasing,

$\ \ \ $3) $\lim\limits_{n\rightarrow \infty}f_n(x)=0$.

One can show that if a series $\sum\limits_{i=1}^\infty (-1)^n f_n(x)$ satisfies 1), 2), and 3) for each $x\in I$, then for each $x\in I$ and for any positive integer $N$ we have $$ \Bigl|\,\sum_{n=N}^{N+k}(-1)^n f_n(x)\,\Bigr|\le f_N(x). $$ (In fact, this is a slight reformulation of a standard result in most introductory calculus courses.)

From the above, it follows that a series $\sum\limits_{n=1}^\infty (-1)^n f_n(x)$ satisfying 1), 2), and 3) for each $x$ in $I$ is uniformly Cauchy on $I$, and thus uniformly convergent on $I$, if and only if the sequence $\{f_n\}$ is uniformly convergent to $0$ on $I$.

In your series, $f_n(x)={1\over n+x^2}$ does converge uniformly to 0 on $[0,\infty)$ (on all of $\Bbb R$ in fact); thus $\sum\limits_{n=1}^{\infty}(-1)^n {1\over n+x^2}$ is uniformly convergent on $[0,\infty)$.

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