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I'm being asked to show that a given series (of rational functions) converges uniformly on a given disc, and then and asked to use this fact to show that integrating its limit function (i.e. a summation of rational functions) along a given contour results in a given series.

In the solution provided, the arugment is that each of the rational function is continuous, thus the uniform convergence implies that the limit function is continuous too.

I don't understand the need to invoke the uniform-convergence-implies-continuous-limit theorem. Isn't a series of rational functions already automatically continuous?

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You need uniform continuity: For every $n$, $f_n(x) = x^{n}-x^{n+1}$ is a rational function on $[0,1]$, but $\sum_{n=0}^\infty f_n$ equals $1$ on $[0,1)$ and $0$ on $\{0\}$. – martini May 22 '13 at 18:27
@martini Thanks Martini. – Ryan May 22 '13 at 18:36
up vote 3 down vote accepted

Consider the following sequence:

$$f_1(x) = x \\ f_2(x) = x^2 - x \\ f_3(x) = x^3 - x^2 ...$$

Each function is a rational function (even a polynomial) and obviously continuous. Now look at the series $\sum_n f_n(x)$ in the interval $[0, 1]$. This converges (pointwise) to $0$ in $[0, 1)$ and to $1$ at $x=1$, which is not continuous.

Historical note: The following is taken from the Wikipedia article List of incomplete proofs.

In his Cours d'analyse of 1821, Cauchy "proved" that if a sum of continuous functions converges pointwise, then its limit is also continuous. However, Abel observed three years later that this is not the case. For the conclusion to hold, "pointwise convergence" must be replaced with "uniform convergence". There are many counterexamples. For example, a Fourier series of sine and cosine functions, all continuous, may converge to a discontinuous function such as a step function.

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Hmm okay, then how about a finite series of continuous terms? Surely that would be continuous, right?? – Ryan May 22 '13 at 18:36
The sum of two continuous functions is continuous, therefore (you can show this by induction) the sum of any finite amount of continuous functions is continuous. – Yoni Rozenshein May 22 '13 at 18:37
Whew! Thanks (for both answers). – Ryan May 22 '13 at 18:40
Oh, then does this mean that given an infinite series, without knowing whether it is uniformly convergent, I am NOT allowed to simply move the outside integral sign inside the summation sign, because the series may not even be integrable in the first place (cos its limit function may not be continuous) ? – Ryan May 22 '13 at 19:08
Yeah, exchanging summation and integration always requires you to explain why it is allowed. Uniform convergence is a good justification. Sometimes you don't have uniform convergence and it is still justified, but you must then use more powerful tools, such as DCT (Dominated Convergence Theorem). And sometimes it is just wrong. – Yoni Rozenshein May 22 '13 at 19:18

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