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For the series $$f(x) = \sum_{n=0}^\infty \frac{1}{n(1+nx^2)}$$ my lecture notes use the Weierstrass M-test to show that this converges uniformly on any interval of the form $(-\infty,-a)$ or $(a,\infty)$ for $a>0$ and says that it converges non uniformly on $(0,a)$ or $(-a,0)$ for any $a>0$.

Isn't this a contradiction? I.e. can't you find two such $a$'s where it converges uniformly and non-uniformly.

Also how does one show that there is some interval where this converges non-uniformly?


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Do you mean $(-\infty,-a)$? –  robjohn Apr 5 '12 at 11:39
ah yes thank you –  user26069 Apr 5 '12 at 11:40
Just notice that the harmonic series $\sum_{n\geq 1}\frac{1}{n}$ diverges. Now for any $m$, if $S_m$ denotes the partial sum of $m$ terms of the harmonic series, notice that by taking $x$ sufficiently close to zero, one can ensure that the partial sum of $m$ terms of $f(x)$ is close to $S_m$. Now use divergence of $S_m$ as $m\rightarrow\infty$. –  William Apr 5 '12 at 11:45
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3 Answers

up vote 1 down vote accepted

First, you need to start your sum at $n=1$, not $n=0$.

To address your first question:

There is no contradiction. If a series converges uniformly on a set $O$, then it converges uniformly on any subset of $O$; but, the series does not necessarily converge uniformly on a set $U\supset O$.

You have uniform convergence on a set $(a,\infty)$, $a>0$ and non-uniform convergence on a larger set $(0,\infty)$, which is ok...

To, hopefully, address your second question:

In your example, you have uniform convergence on sets of the form $(a,\infty)$, $a>0$. You have pointwise convergence on the set $(0,\infty)$. Moreover, since the series diverges at $x=0$, this is the largest interval, unbounded on the right, for which you have pointwise convergence. This is then the natural candidate for an interval over which you could have non-uniform convergence. Moreover, things would have to go awry near $x=0$.

But is this indeed the case?

Let's look at the graphs of the first few partial sums $s_n=\sum\limits_{k=1}^n {1\over k(1+kx^2). }$

enter image description here

We are led to suspect that the series does not converge uniformly on $(0,\infty)$. (look at how the difference between the $s_k$ grows larger as $x\rightarrow0^+$).

In fact, given $N$, we may choose $M>N$ so that $\sum\limits_{n=N}^M{1\over n}\ge 1$. Then $$ \lim_{x\rightarrow0^+}\sum_{n=N}^M {1\over n(1+nx^2)} =\lim_{x\rightarrow0^+}\sum_{n=N}^M {1\over n}\ge1. $$ So the series $\sum\limits_{n=1}^\infty {1\over n(1+nx^2)}$ is not uniformly Cauchy on $(0,\infty)$ and thus not uniformly convergent on $(0,\infty)$.

Since $x=0$ is the "bad point" as far as uniform convergence is concerned, the series does not converge uniformly on any interval of the form $(0,a)$.

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Define $$ f_k(x)=\sum_{n=0}^{k-1}\frac{1}{n(1+nx^2)}\tag{1} $$ For $|x|>a$, we have $$ \begin{align} |f(x)-f_k(x)| &=\sum_{n=k}^\infty\frac{1}{n(1+nx^2)}\\ &<\sum_{n=k}^\infty\frac{1}{n(1+na^2)}\\ &<\frac{1}{a^2}\sum_{n=k}^\infty\frac{1}{n^2}\\ &<\frac{1}{a^2}\sum_{n=k}^\infty\left(\frac{1}{n-1/2}-\frac{1}{n+1/2}\right)\\ &=\frac{1}{a^2(k-1/2)}\tag{2} \end{align} $$ Estimate $(2)$ says that $f_k\to f$ uniformly on $\{x:|x|>a\}$.

However, $$ \begin{align} |f(x)-f_k(x)| &=\sum_{n=k}^\infty\frac{1}{n(1+nx^2)}\\ &\ge\sum_{n=k}^{1/x^2}\frac{1}{n(1+nx^2)}\\ &\ge\frac12\sum_{n=k}^{1/x^2}\frac{1}{n}\\ &\ge-\frac12\log(x^2(k+1))\tag{3} \end{align} $$ That is, for any positive integer $k$, if $0<x<\frac{1}{e\sqrt{k+1}}$, then $|f(x)-f_k(x)|\ge1$. Thus, the convergence is not uniform on any neighborhood of $x=0$.

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There is no contradiction. Note that $a>0$ is arbitrary. This series converges uniformly on any interval of the form $(-\infty,-a)$ or $(-\infty,-b]$ or $(a,\infty)$ or $[b,\infty)$ where $a>0,b>0$. Due to

$$|\frac{1}{n(1+nx^2)}|\leq \frac{1}{n(1+na^2)}$$

(the series $\sum\frac{1}{n(1+na^2)}$ converges), so by Weierstrass M-test, the form series converges uniformly. For the other hand, give you any interval $0\in(a,b)$, take $x=0$ , then the series becomes $\sum\frac{1}{n}$, obviously, this series does not converge.

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