Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
4  
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
add comment

2 Answers 2

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 $x\geq 0$.

share|improve this answer
add comment

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 $[0,\infty)$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.