8.29 Sequences and series
1.Decimal expansions for real numbers
Decimal Expansion Theorem for Real Numbers
Let \(x\in R,x>0\).
Then there exists a sequence of digits \(a_0,a_1,a_2,a_3,...\)(one \(a_j ~for ~each~ j \in \N\)) and \(a_0\in N\) such that \(a_j\in \{0,1,2,3,4,5,6,7,8,9\}~~\forall j\in N\)
And \(\sum_{j=0}^{n}a_{j}10^{-j}\leq x<\sum_{j=0}^{n-1}a_{j}10^{-j}+\frac{a_{n}+1}{10^{n}}\) for every \(n\in N\)
More explicitly: \(a_{0}+\frac{a_{1}}{10}+\frac{a_{2}}{10^{2}}+\cdots+\frac{a_{n}}{10^{n}}\leq x<a_{n}+\frac{a_{1}}{10}+\frac{a_{2}}{10^{2}}+\cdots+\frac{a_{n}+1}{10^{n}}\)
For example
\(x=\frac{1}{4}=0.25\) \(a_0=0\)
\(0+\frac{2}{10}\leq x<0+\frac{3}{10}~~~a_{1}=2\)
\(0+\frac{2}{10}+\frac{5}{10^{2}}\leq x<0+\frac{2}{10}+\frac{5+1}{10^{2}}~~~a_{2}=5\)
\(0+\frac{2}{10}+\frac{5}{10^{2}}+\frac{0}{10^{3}}\leq x<0+\frac{2}{10}+\frac{5}{10^{2}}+\frac{1}{10^{3}}\)
\(when~x~has~finite~expansion,~we~get~a_{n}=0~~~\forall n\geq n_{0}.\)
Proof=9=
Take \(x\geq 0\). Then, taking the integer part of \(x\): \(\lfloor x\rfloor\leq x<\lfloor x\rfloor+1\)
Put \(a_{0}=\lfloor x\rfloor\). Consider \(x-a_{0}\in[0,1)\)
Then \(10\cdot(x-a_0)\in[0,10)\)
Put \(a_{1}=[10\cdot(x-a_0)]\in\{0,1,\cdots,9\}\)
\(a_{1}\leq10(x-a_{0})<a_{1}+1\\\frac{a_{1}}{10}\leq x-a_{0}<\frac{a_{1}+1}{10}\)
\(a_{0}+\frac{a_{1}}{10}\leq x<a_{0}+\frac{a_{1}+1}{10}\)
6.Integer part of x\in R(Also called "Floor of x")
Consider the set(for a given \(x\in R,x\geq0\))
\(A=\{m\in N/\exists \text{a collection of digits }a_{1},a_{2}\cdots a_{n}~and~a_{0}\in N/a_{0}+\frac{a_{1}}{10}+\frac{a_{2}}{10^{2}}+\cdots+\frac{a_{n}}{10^{n}}\leq x<a_{0}+\frac{a_{1}}{10}+\cdots+\frac{a_{n}+1}{10^{n}}\}\)
Using induction due to the whole m
We will show that \(A=N\) by induction
\(1\in A\) look what we just did above
Then, prove that \(n\in A\Rightarrow n+1\in A\)
If \(m\in A,\exists a_{0}\in N,a_{1},\cdots a_{n}~digits|a_{0}+\frac{a_{1}}{10}+...+\frac{a_{n}}{10^{n}}\leq x<a_{0}+\frac{a_{1}}{10}+...+\frac{a_{n}}{10^{n}}+\frac{1}{10^{n}}\)
\(0\leq\left(x-\sum_{j=0}^{m}a_{j}10^{-j}\right)<\frac{1}{10^{n}}\\0\leq10^{n+1}\left(x-\sum_{j=0}^{n}a_{j}10^{-j}\right)<10\)
Put \(a_{n+1}=\lfloor10^{n+1}(x-\sum_{j=0}^{n}a_{j}10^{-j})\rfloor\)
\(a_{n+1}\leq10^{n+1}\left(x-\sum_{j=0}^{n}a_{j}10^{-j}\right)<a_{n+1}+1\)
\(\sum_{j=0}^{n+1}a_{j}10^{-j}\leq x<\sum_{j=0}^{n}a_{j}10^{-j}+\frac{a_{n+1}+1}{10^{n+1}}\)
This implies that \(n+1\in A\). Then \(A=N\)
2.Sequences and series
Let \(X\) be a set
A sequence of elements from X is a function \(f: N\rightarrow X\)
\(f\left(1\right)=x_{1}\in X\\f\left(2\right)=x_{2}\in X\\f\left(3\right)=x_{3}\)
We write \((x_n)_{n\in N}\) to denote the sequence given by \(f:N\rightarrow X\)
For example
\((x_{n})_{n\in M}:x_{n}=\frac{1}{2^{n}}~~~(x_{n})_{n\in N}=\{\frac{1}{2},\frac{1}{4},\frac{1}{8},\cdots\}\)
\((a_{n})_{n\in N}:a_{n}=(-1)^{n}~~~a_{n}=\{-1,1,-1,1,-1,1,...\}\)
We will work with sequences of real numbers \(f:N\rightarrow R\)
Two graphical representation
-
As a function \(f:N\rightarrow R\)
\(m\mapsto1-\frac{1}{m}=\frac{m-1}{m}\)
-
As a set of points in \(R\)
Properties of sequences
Let \((a_n)_{n\in N}\) be a sequence
We say that \((a_n)\) is:
Bounded above: if \(\exists M\in R|a_n\leq M~~\forall n\in N\)
Bounded below: if \(\exists m\in R|m\leq a_n~~\forall n\in N\)
Bounded: if \(M\in R||a_{n}|\leq M~~~\forall n\in N\)
Increasing: if \(a_n\leq a_{n+1}~~\forall n\in N\)
Strictly increasing: if \(a_n< a_{n+1}~~\forall n\in N\)
Decreasing:if \(a_n\geq a_{n+1}~~\forall n\in N\)
Strictly decreasing:if \(a_n> a_{n+1}~~\forall n\in N\)
Monotonic: if \(a_n\) is increasing or decreasing
Examples
\(a_n=n^2\) Strictly increasing
Bounded below, not bounded above
\(b_n=7~~\forall n\in N\) Increasing and decreasing
Remark
We say that the sequence \(a_n\) is eventually increasing/decreasing/monotonic, if the property holds \(\forall n\geq n_0\in N\)
One interesting case
We can see that when \(n\) gets larger and larger, the value of \(a_n\) gets closer and closer to 1
3.Limit=9=
Now consider a sequence \(a_n\) such that
- \(a_n\) is bounded above: \(\exists M\in R: a_n\leq M~~\forall n\in N\)
- \(a_n\) is increasing: \(a_n\leq a_{n+1}~~~\forall n\in N\)
Two pictures
Note that \(a_n\) is actually bounded:
\(a_1\leq a_n\leq M\) \(\forall n\in N\)
The set \(A=\{a_{n}:n\in N\}\neq\emptyset\) and bounded
Since we are in R, the set A has a sup
Let us call \(L=sup(A)\)
Now, let \(\varepsilon >0\)(any fixed positive \(\varepsilon\))
By definition of \(sup(A)\), we know that \(\exists a\in A|L-\varepsilon<a\leq L\)
In our case a has to be of the form \(a_{n_0}\) for some \(n_0\in N\)
So we have \(L-\varepsilon<a_{n_{0}}\leq L\)
By monotonicity, we conclude that \(L-\varepsilon<a_{n}\leq L~~~~\forall n\geq n_{0}\)
In this case, we say that the sequence \(a_n\) converges to \(L\)
We write \(\lim_{n\to\infty}a_{n}=L\)
Exercise: Show that if \(a_n\) is a sequence such that decreasing and bounded below
Then we have \(\lim_{n\to\infty}a_{n}=\widetilde{L} =inf\{a_n:n\in N\}\)
We can write: \(\forall \varepsilon>0,\exists n_0\in N|\widetilde{L}\leq a_n<\widetilde{L}+\varepsilon~~~\forall n\geq n_0\) \(n_0=n_0(\varepsilon)\)
What happens with non-monotonic sequence
Example
\(a_{n}=1+\frac{(-1)^{n}}{n}\)
It look like \(a_n\) is getting closer to 1
How big has to be \(n_0\in N\)/ the difference between \(a_n\) and 1 is smaller than a given \(\varepsilon>0\)
Let \(\varepsilon>0\) be given (by the enemy).
Find \(n_0\in N\)/ \(|a_{n}-1|<\varepsilon~~~\forall n\geq n_{0}\)
\(\vert1+\frac{(-1)^{n}}{n}-1\vert=\frac{1}{n}<\varepsilon~if~n\geq\frac{1}{\varepsilon}\)
Choose \(n_0\in N\)/\(n_0>\frac{1}{\varepsilon}\Rightarrow |a_n-1|<\varepsilon ~~~\forall n\geq n_0\)
Definition
Let \((a_n)_{n\in N}\) be a sequence of real numbers
We say that: "\(a_n\) converges to \(L\in R\)" or "the limit of \(a_n\) when \(n\) goes to infinity is \(L\)"
Write: \(\lim_{n\to+\infty}a_{n}=L\) if \(\forall\varepsilon>0,\exists n_0(\varepsilon)/|a_{n}-L|<\varepsilon~~~\forall n\geq n_{0}\)
Two pictures
\(\vert a_{n}-L\vert<\varepsilon\)
\(-\varepsilon<a_{n}-L<\varepsilon\)
\(L-\varepsilon<a_{n}<\varepsilon +L\)
Example
\(a_{n}=\frac{3n-7}{2n+10}\neq\frac32\)
Find if it exists the \(\lim_{n\to\infty}a_{n}=L\)
If \(n\rightarrow+\infty\), then \(3n-7\sim 3n~and ~2n+10\sim2n\)
Candidate \(L=\frac{3}{2}\). Let's prove that \(\lim_{n\to\infty}a_{n}=\frac{3}{2}\)
Let \(\varepsilon>0\). Now, we find \(n_0||a_n-\frac{3}{2}|<\varepsilon\):
\(\left|\frac{3n-7}{2n+10}-\frac{3}{2}\right|=\left|\frac{3n-7-3(n+5)}{2(n+5)}\right|=\left|\frac{3n-7-3n-15}{2(n+5)}\right|=\frac{1}{2}\cdot \frac{22}{n+5}=\frac{11}{n+5}<\frac{11}{n}<\varepsilon\). if \(n>\frac{11}{\varepsilon}\) pick \(n_0\geq \frac{11}{\varepsilon}\)