Calculus course: Sequences and Series of functions

A sum is the result of an addition. The notion of series is related to the sum of numbers.

For example, adding 1, 2, 3, and 4 gives the sum 10: 1+2+3+4=10. The decad represents the number ten.

“Ten is the very nature of number. All Greeks and all barbarians alike count up to ten, and having reached ten revert again to the unity. And again, Pythagoras maintains, the power of the number 10 lies in the number 4, the tetrad. This is the reason: If one starts at the unit (1) and adds the successive number up to 4, one will make up the number 10 (1 + 2 + 3 + 4 = 10). And if one exceeds the tetrad, one will exceed 10 too…. So that the number by the unit resides in the number 10, but potentially in the number 4.” (Aetius 1.3.8)

The Tetractys (greek Τετρακτύς) is a symbol composed of ten dots in an upward-pointing triangular formation.

The numbers being summed are called addends.

Given a set A, a sequence of  elements of A is a function F: N -> A; rather than using notation F(n) for the elements that have been selected from A, since the domain is always the natural numbers (N), we use the notation a_n=F(n) and denote the sequence:

{a_n} or a_1,a_2,a_3...

Given any sequence {c_n} of elements  of a set A, we have an associated sequence of nth partial sums:

{s_n} where s_n= \sum{c_k} with k=1…n

the symbol \sum{c_k} is called series (or infinite series).

We take A to be the set of real functions on R.

Pointwise convergence of sequence of functions.

Definition: let {f_n} be a sequence of functions defined on a set of real numbers A. We say that {f_n} converges pointwise to a function f on A for each x \in A, the sequence of real numbers {f_n(x)} converges to the number f(x). In other words, for each x \in A, we have:


Example: let f_n(x)=x^n, x \in [0,1] and let f(x)= 0 if 0 <=x<1, f(x)=1 if x=1.

Then {f_n(x)} converges to f pointwise on [0,1]

Suppose {f_n(x)} converges to f pointwise on A. Then given \epsilon >0, and given x \in A , there exists  N=N(x, \epsilon) \in I such that:

|f_n(x)-f(x)|< \epsilon for all n>=N

In general N depends on \epsilon a.w.a x.

Example: f_n(x)= \frac{x}{1+nx}.

We have f_n(x) <=1/n for all x \in [0, oo( and hence  for a given \epsilon >0, any N  with N>\frac{1}{\epsilon} will imply that |f_n(x)-0|< \epsilon for all  n>N and for x \in [0, oo(

Uniform  convergence of sequence of functions

Let { f_n(x)} be a sequence of function on A. We say that { f_n(x)} converges uniformly  to f on A if for given \epsilon >0, there exists  N=N(\epsilon) , depending on \epsilon , such that:

|f_n(x)-f(x)|< \epsilon for all n>=N and for all x \in A

If { f_n(x)} converges  uniformly to f on A, we write:  f_n -> f uniformly on A.

Remark: N depends only on \epsilon and not on x.

Example: f_n(x)= \frac{x}{1+nx} converges uniformly to 0 on [0, oo(

If f_n -> 0 uniformly on A and \epsilon >0 , then there exists  N>0 and all x \in A , |f_n(x)|< \epsilon. This implies  that for all n>N:

sup|f_n(x)|<= \epsilon and hence lim sup|f_n(x)|<= \epsilon.

Since \epsilon>0 is an arbitrary positive number, we conclude that:

if  f_n(x) ->0 uniformly on A, then lim_{x\to\infty} sup|f_n(x)|=0.

The converse is also true.


if  f_n(x) ->f uniformly on A, if and only if  lim_{x\to\infty} sup|f_n(x)-f(x)|=0.

Cauchy Criterion for Uniform Convergence. A sequence {f_n(x)} converges uniformly on A if and only if for a given \epsilon>0 , there exists N>0 such that for all n>=m>N and for all x\in A,

$latex| f_n(x)-f_m(x)|< \epsilon $

Theorem: If  {f_n(x)} is a sequence  of continuos functions on a bounded  and closed interval $ latex [a,b] and  {f_n(x)} converges pointwise to a continuous function f on [a,b], then f_n -> f uniformly on [a,b]

Consequences of uniform convergence

Theorem: if f_n -> f uniformly on [a,b], if f_n are continuous at c \in [a,b] , then f is continuous at c.

Corollary: if f_n -> f uniformly on [a,b],if f_n are continuous on [a,b], then f is continuous on [a,b].


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