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The task is to specify convergence of infinite function series (pointwise, almost uniform and uniform convergence):

a) $\displaystyle \sum_{n=1}^{+\infty}\frac{\sin(n^2x)}{n^2+x^2}, \ x\in\mathbb{R}$

b) $\displaystyle\sum_{n=1}^{+\infty}\frac{(-1)^n}{n+x}, \ x\in[0;+\infty)$

c) $\displaystyle\sum_{n=0}^{+\infty}x^2e^{-nx}, \ x\in(0;+\infty)$

I know basic facts about these types of convergence and Weierstrass M-test, but still have problems with using this in practise..

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"almost uniform"? –  David Mitra Apr 10 '12 at 13:18
    
@David: uniform except for a set of measure $0$? (at least that is how i know the literal german translation) –  example Apr 10 '12 at 13:26
    
I was taught (hence I had to explain this concept into English and sorry I didn't clarify) that almost uniform is a weaker version of uniform convergence. The series converges almost uniformly when it converges uniformly after cutting the domain to any interval. Especially uniform convergence implies almost uniform convergence. –  xan Apr 10 '12 at 13:53

1 Answer 1

up vote 2 down vote accepted

First observe that each of the series converges pointwise on its given interval (using standard comparison tests and results on $p$-series, geometric series, and alternating series.

Towards determining uniform convergence, let's first recall the Weierstrass $M$-test:

Suppose $(f_n)$ is a sequence of real-valued functions on the set $I$ and $(M_n)$ is a sequence of positive real numbers such that $|f_n(x)|\le M_n$ for $x\in I$, $n\in\Bbb N$. If the series $\sum M_n$ is convergent then $\sum f_n$ is uniformly convergent on $I$.

It is worthwhile to consider the heart of the proof of this theorem:

Under the given hypotheses, if $m>n$, then for any $x\in I$ $$\tag{1} \bigl| f_{n+1}(x)+\cdots+f_m(x)\bigr| \le| f_{n+1}(x)|+\cdots+|f_m(x)\bigr| \le M_{n+1}+\cdots M_n. $$ So if $\sum M_n$ converges, we can make the right hand side of $(1)$ as small as we wish. Noting that the right hand side of $(1)$ is independent of $x$, we can conclude that $\sum f_n$ is uniformly Cauchy on $I$, and thus uniformly convergent on $I$.


Now on to your problem:

To apply the $M$-test, you have to find appropriate $M_n$ for the series under consideration. Keep in mind that the $M_n$ have to be positive, summable, and bound the $|f_n|$. Sometimes they are easy to find, as in the series in a). Here note that for any $n\ge 1$ and $x\in\Bbb R$, $$ \biggl| {\sin(n^2x)\over n^2+x^2}\biggr|\le {1\over n^2}. $$ So, take $M_n={1\over n^2}$ and apply the $M$-test. The series in a) converges uniformly on $\Bbb R$.


Sometimes finding the $M_n$ is not so easy. This is the case in c). Crude approximations for $f_n(x)=x^2e^{-nx}$ will not help. However, we could try to find the maximum value of $f_n$ over $(0,\infty)$ and perhaps this will give us what we want. And indeed, doing this (using methods from differential calculus), we discover that the maximum value of $f_n(x)=x^2e^{-nx}$ over $(0,\infty)$ is ${4e^{-2}\over n^2}$. And now the road towards using the $M$-test is paved...


Sometimes the $M$-test doesn't apply. This is the case for the series in b), the required $M_n$ can't be found (at least, I can't find them). However, here, the proof of the $M$-test gives us an idea. Since the series in b) is alternating (that is, for each $x\in[0,\infty)$, the series $\sum\limits_{n=1}^\infty{(-1)^n\over x+n}$ is alternating), perhaps we can show it is uniformly Cauchy on $[0,\infty)$.

Indeed we can:

For any $m\ge n$ and $x\ge0$ $$\tag{2} \Biggl|\,{(-1)^n\over n+x}+{(-1)^{n+1}\over (n+1)+x}+\cdots+{ (-1)^m\over m+x}\,\Biggl|\ \le\ {1\over n+x}\le {1\over n}. $$ The term on the right hand side of $(2)$ is independent of $x$ and can be made as small as desired. So, the series in b) is uniformly Cauchy on $[0,\infty)$, and thus uniformly convergent on $[0,\infty)$.

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thank you very much :-) I've analyzed your answer very carefully and understand everything. You've helped me very much, thank you! –  xan Apr 10 '12 at 18:15
    
@xan You're welcome; glad to help. –  David Mitra Apr 10 '12 at 18:48

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