2. Sequences and series of functions

Sequences of functions, pointwise convergence

A sequence of functions is a sequence with functions as elements f_n\colon D\to\mathbb{R}, defined on a set D\subseteq \mathbb{R}. E.g.: f_n(x)=\frac{x}{n}.

We shall say that a sequence of functions converges pointwise to a function f\colon D\to \mathbb{R} (f_n\to f), if for all x\in D we have \lim_{n\to\infty} f_n(x)=f(x) (as a sequence of reals).

E.g.: f_n(x)=\frac{x}{n} converges pointwise to f(x)=0 na \mathbb{R}, because for all x\in \mathbb{R} we have \lim_{n\to\infty} \frac{x}{n}=0.

Uniform convergence

A sequence f_n converges uniformly on X\subseteq\mathbb{R} to f (f_n\rightrightarrows f), if:

    \[\forall_{\varepslion>0}\exists_{N\in\mathbb{N}}\forall_{n>N}\forall_{x\in X} |f_n(x)-f(x)|\leq\varepsilon.\]

Meaning that in every point it converges at least equally fast. In other words, a sequence of functions converges uniformly if the sequence \sup_{x\in X} |f_n(x)-f(x)| converges to zero.

E.g., sequence f_n(x)=\frac{x}{n} converges uniformly to [-1,1] to f(x)=0, because, if \varepsilon>0, then let N=\left\lceil\frac{1}{\varepsilon}\right\rceil and then for any n>N:


because |x|\leq 1.

The same sequence does not converge uniformly on the whole set \mathbb{R}, since if, e.g. \varepsilon=1, for any N, I can find x\in\mathbb{R}, such that |f_n(x)-0|>\varepsilon. Indeed, let x=2N. Then |f_n(x)-0|=\frac{2N}{N}=2>1=\varepsilon.

The following important theorem holds. If all f_n(x) are continuous on a set X and f_n\rightrightarrows f, then f is also continuous.

The above may not hold in the case of pointwise convergence. E.g. f_n(x)=x^n converges pointwise on [0,1] to

    \[f(x)=\begin{cases} 0 &\text{if }x\in[0,1)\\ 1 &\text{if }x=1\end{cases},\]

which is not continuous. Obviously, f_n is not uniformly convergent on X.

Series of functions

Similarly as in the case of series of reals we can create a series of functions out of a sequence of functions. A sum of such a sequence is as before the limit of sequence of partial sums. Therefore we can study sets of arguments on which a given series is convergent. E.g. series \sum_{n=0}^\infty \frac{x}{2^n} converges for any x\in\mathbb{R} to 2x.

Weierstrass criterion

This criterion seems quite clear: if there exists m\in\mathbb{N}, such that for any n\geq m and any x\in X\subseteq\mathbb{R} the following inequality holds: |f_n(x)|\leq a_n, where \sum_{n=1}^{\infty} a_n is convergent, then \sum_{n=1}^\infty f_n(x) is convergent on X absolutely and uniformly.

E.g. \sum_{n=1}^\infty \frac{\sin x}{n^2} converges on the whole \mathbb{R}, because \frac{|\sin x|}{n^2}\leq \frac{1}{n^2} and \sum_{n=1}^{\infty}\frac{1}{n^2} is convergent.

Power series

A series of functions of form \sum_{n=1}^{\infty} a_n (x-a)^n is called a power series. E.g. \sum_{n=1}^\infty\frac{(x-1)^n}{2^n}, is convergent for x\in [-1,3).

Radius of convergence

Given a series \sum_{n=1}^\infty a_n x^n, a real number R equal to supremum of the set of arguments for which this series is convergent is called the radius of convergence of this series. Actually, the radius of convergence up to two points describes the convergence of a power series, because the following fact.

If R is the radius of convergence of \sum_{n=1}^\infty a_n x^n then this series is absolutely convergent on (-R,R) and not convergent on (-\infty, -R)\cup (R,\infty). The theorem does not describe convergence for x=\pm R. Therefore to describe the set of convergence of a given power series, it suffices to calculate the radius of convergence and additionally check what happens for x=\pm R.

Calculating the radius of convergence

Given series \sum_{n=1}^\infty a_n x^n we can deduce from d’Alembert and Cauchy criterion the radius of convergence. The following two facts hold:

  • if \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lambda, then if \lambda >0, then R=\frac{1}{\lambda}, but if \lambda=0, then R=\infty,
  • if \lim_{n\to\infty}\sqrt[n]{\left|a_n\right|}=\lambda, then if \lambda >0, then R=\frac{1}{\lambda}, but if \lambda=0, then R=\infty.

E.g. for \sum_{n=1}^\infty \frac{x^n}{2^n} we have \lim_{n\to\infty}\sqrt[n]{\left|a_n\right|}=\lim_{n\to\infty}\sqrt[n]{\frac{1}{2^n}}=\frac{1}{2}, so R=\frac{1}{2}.