Compact Sets
Compactness is super important in both real analysis and topology.
A collection \mathcal{A} of subsets of a space X are a cover of X if the union of the elements of \mathcal{A} is equal to X.
And
Let A be a set in some metric space X. If every open cover of A contains a finite subcollection that also covers A.
More explicitly, a set A is compact in some space X if for every open cover $$
Let (a_n) be a sequence of real numbers. We notate the partial sum from m to n by
\sum^n_{k=m} a_k
We may also save time by letting
\sum a_n := \sum^{\infty}_{n=0} a_n \text{ or } \sum^{\infty}_{n=1} a_n
where are index starts at 0 or 1, whichever makes sense in context.
Let (a_n) be a sequence of real numbers. We define a sequence (s_n) of partial sums by
s_n = \sum^n_{k=1} a_k = a_1 + a_2 + \cdots + a_n.
We refer to the sequence (s_n) as an infinite series.
We define (s_n) as a sequence of partial sums in order to formalize the series into a sequence, which we’ve already studied. All of our previous theorems will apply to (s_n), and therefore we can prove many theorems essentially for free!
Convergence
If (s_n) converges to a real number s, we say the series \sum a_n is convergent and write
\sum^\infty_{n=1}a_n = s.
If (s_n) does not converge, the series \sum^\infty_{n=1}a_n is divergent. If
\begin{align*} \lim_{n \to \infty} s_n &= +\infty \quad \text{then} \quad \sum^\infty_{n=1}a_n = +\infty \quad \text{ and}\\ \lim_{n \to \infty} s_n &= -\infty \quad \text{then} \quad \sum^\infty_{n=1}a_n = -\infty. \end{align*}
The question of whether or not a series converges is of great interest to mathematicians.
The harmonic series is defined
\sum^\infty_{n=1} \frac{1}{n}
The sequence (s_n) is divergent, and thus the harmonic series is divergent. This is
The series
\sum^\infty_{n=1} \frac{1}{n(n+1)}
has the partial sum
s_n = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \cdots + \frac{1}{n(n+1)}
Let \sum a_n = s and \sum b_n = t. Then
(a) \sum (a_n + b_n) = s + t and
(b) \sum (ka_n) = ks for every k \in \mathbb{R}.
(a) By converting our series to limits, we obtain \sum (a_n + b_n) = \lim (s_{a_n} + s_{b_n}) = s + t.
(b) Similarly, we obtain
\sum (ka_n) = \lim ks_{a_n} = k\lim s_{a_n} = ks.
If \sum a_n is a convergent series, then \lim_{n \to \infty} a_n = 0.
If \sum a_n is a convergent series, then the sequence of partial sums (s_n) must have a finite limit.
Since a_n = s_n - s_{n-1}, we have \lim a_n = \lim s_n - \lim s_{n-1} = 0.
The infinite series \sum a_n converges if and only if for each \varepsilon > 0, there exists a natural number N such that if n \geq m \geq N, then
|a_m + a_{m+1} + \cdots + a_{n}| = |s_n - s_{m-1}| < \varepsilon.
Let \sum a_n be a convergent series. Then the sequence (s_n) of partial sums converges. By the Cauchy Convergent Criterion, (s_n) is Cauchy. Thus, for any \varepsilon > 0, there exists N \in \mathbb{N} such that m,n \geq N implies |s_n - s_m| < \varepsilon. Hence, if n \geq m \geq N + 1, then m - 1 \geq N, and |s_n - s_{m-1}| < \varepsilon as desired.
Conversely, for all \varepsilon > 0 let N exists such that n \geq m \geq N implies |s_n - s_{m-1}| < \varepsilon. Then for n > m \geq N we have m + 1 > N, so that |s_n - s_m| < \varepsilon. This implies (s_n) is Cauchy, and therefore convergent.
Recap
Let \sum a_n = s and \sum b_n = t. Then
(a) \sum (a_n + b_n) = s + t and
(b) \sum (ka_n) = ks for every k \in \mathbb{R}.
If \sum a_n is a convergent series, then \lim_{n \to \infty} a_n = 0.
The infinite series \sum a_n converges if and only if for each \varepsilon > 0, there exists a natural number N such that if n \geq m \geq N, then
|a_m + a_{m+1} + \cdots + a_{n}| = |s_n - s_{m-1}| < \varepsilon.