# The Unapologetic Mathematician

## Uniform Convergence of Series

Since series of anything are special cases of sequences, we can import our notions to series. We say that a series $\sum\limits_{n=0}^\infty f_n$ converges uniformly to a sum $f$ if the sequence of partial sums $s_n=\sum\limits_{k=0}^nf_k$ converges uniformly to $f$. That is, if for every $\epsilon>0$ there is an $N$ so that $n>N$ implies that $\left|f-\sum\limits_{k=0}^nf_k(x)\right|<\epsilon$ for all $x$ in the domain under consideration.

And we’ve got Cauchy’s condition: a series converges uniformly if for every $\epsilon>0$ there is an $N$ so that $m$ and $n$ both greater than zero implies that $\left|\sum\limits_{k=m}^nf_k(x)\right)<\epsilon$ for all $x$ in the domain.

Here’s a a great way to put this to good use: the Weierstrass M-test, which is sort of like the comparison test. Say that we have a positive bound for the size of each term in the series: $\left|f_n(x)\right| for all $x$ in the domain. And further assume that the series $\sum\limits_{n=0}^\infty M_n$ converges. Then the series $\sum\limits_{n=0}^\infty f_n(x)$ must converge uniformly.

Since the series of the $M_n$ converges, Cauchy’s condition for series of numbers tells us that for every $\epsilon>0$ there is some $N$ so that when $m$ and $n$ are bigger than $N$, $\sum\limits_{k=m}^nM_n<\epsilon$. But now when we consider $\left|\sum\limits_{k=m}^nf_n(x)\right|$ we note that it’s just a finite sum, and so we can use the triangle inequality to write

$\left|\sum\limits_{k=m}^nf_n(x)\right|\leq\sum\limits_{k=m}^n\left|f_n(x)\right|<\sum\limits_{k=m}^nM_n<\epsilon$

So Cauchy’s condition tells us that the series $\sum\limits_{n=0}^\infty f_n(x)$ converges uniformly in the domain under consideration.

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September 9, 2008 -

## 2 Comments »

1. […] some point so that for every point we have . And thus we have for all . Setting , we invoke the Weierstrass M-test — the series converges because is within the disk of convergence, and thus evaluation at […]

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