Math 2710

Oct 28-Nov 1

More on limit rules

Proposition: If \(a_n\) converges to \(L\) then \(ca_n\) converges to \(cL\).

Definition: Divergence to infinity: A sequence \(a(n)\) diverges to infinity if, for any \(B\), there is an \(N\) so that \(a(n)>B\) if \(n\ge N\).

Remark on calculus trick for ratio of polynomials \(P(n)/Q(n)\): look at highest degree terms and if they are the same degree take ratio of leading coefficients.

Follows from:


A series is an infinite sum, but it is really a shorthand for a sequence. The series \[ a_0+a_1+a_2+\ldots \] is a short hand for the sequence of partial sums \((a_0,a_1+a_0,a_2+a_1+a_0,\ldots)\).

A series converges to a limit \(L\) means that the sequence of partial sums converges.

Key example is the geometric series \(\sum_{n=0}^{\infty} ar^{n}\).

Geometric series

Proposition: Suppose \(r\not=1\). The finite geometric series has sum \[ a+ar+ar^2+ar^3+\cdots+ar^n = a\frac{r^{n+1}-1}{r-1}. \]

Proof: By induction. If \(n=1\) then we get \(a+ar=a\frac{r^2-1}{r-1}=ar+r\) as desired. Suppose true for \(n=k\) Then \[ a+ar+ar^2+\cdots +ar^{k+1}=a\frac{r^{k+1}-1}{r-1}+ar^{k+1}=a\frac{r^{k+1}+r^{k+1}(r-1)}{r-1} \] and \[ a\frac{r^{k+1}-1+r^{k+2}-r^{k+1}}{r-1}=a\frac{r^{k+2}-1}{r-1}. \]

Infinite geometric seriens

Proposition: If \(|r|<1\), the infinite geometric series \[ a+ar+ar^2+\cdots \] converges to \[ \frac{a}{1-r}. \]

Proof: The partial sums are \(s(n)=a\frac{r^{n+1}-1}{r-1}\). Since \(\frac{a}{1-r}\) is a constant we can write this as \[ s(n)=\frac{a}{1-r}(1-r^{n+1}). \]

Now the sequence \(r^{n+1}\) converges to zero when \(r<1\) (proof?) and so \(1-r^{n+1}\) converges to \(1\) (proof) and so \(s(n)\) converges to \(a/(1-r)\).

Other examples

The harmonic series \[ 1+1/2+1/3+\cdots \] does not converge. For a proof, see problem 33 on page 105 of the Gilbert-Vanstone book where they ask you to show by induction that, for all \(n\), \[ 1+1/2+1/3+\cdots+1/2^{n}\ge 1+n/2 \] Therefore the sequence of partial sums diverges (slowly) to infinity.

Base ten decimals and the geometric series

An infinite decimal is shorthand notation for an infinite series (and thus a sequence). \[ .a_0a_1a_2\cdots = \sum_{i=0}^{\infty} a_{i}10^{-i} \]

An eventually repeating decimal is a decimal expansion such that there is an \(N\) and a \(k\) so that for all \(i\ge N\) we have \(a_{i+k}=a_{i}\). In other words there is a block of \(k\) digits \(a_N, a_{N+1},\cdots,a_{N+k}\) that repeat over and over.

Repeating decimals converge to rational numbers

Proposition: An eventually repeating decimal converges to a rational number.

Proof: First suppose our decimal begins repeating right at the decimal point, so it looks like \(a_1 a_2\cdots a_k a_1 a_2\cdots\). Let \(A\) be the integer \(a_1a_2\cdots a_k\). Then the decimal corresponds to the series \[ \frac{A}{10^{k}}(1+\frac{1}{10^{k}}+\cdots) \] This is a geometric series which we know converges to \[ \frac{A}{10^{k}}\frac{1}{1-(1/10)^{k}} \] which is a rational number.

Repeating decimals cont’d

If the decimal \(x\) has an initial, non repeating part, of length \(N\), with digits \(a_1 a_2 \ldots a_N\) followed by blocks of \(k\) digits \(b_1 b_2\ldots b_k\), we can split out the leading part and write \(x\) as the series:

\[ x=\frac{A}{10^{N}}+\frac{1}{10^{N+k}}(B + B/10^{k} + \cdots) \] where \(A\) and \(B\) are the integers made up of the initial digits and the repeating block respectively. The second part of this series converges to a rational number by the earlier work, and so the sum is a rational number.

Base \(r\)

There was nothing special about base \(10\) and the same would be true for repeating “decimals” in any base.

Every rational has a repeating decimal expansion

For the converse we make an inductive argument. We are given a fraction \(a/b\) and we want to construct a decimal expansion and prove that it is repeating. First, we use the division algorithm to write \(a=qb+r\) and see that \(a/b = q +r/b\). Now we proceed as follows. We multiply \(r\) by \(10\) and divide again by \(b\) using the division algorithm: \[ 10r = q_1b+r_1. \]

Repeating decimals continued

This tells us that \(r/b = q_1/10+r_1/10b\) Since \(r_1<b\), the fraction \(r_1/10b\) is less than \(1/10\). Also, since \(r_1<b\), \(10r_1<10b\) and so \(q_1<10\). Therefore \(q_1\) is the first decimal digit of \(r/b\). Now we repeat the process with \(r_1\): \[ 10r_1 = q_2b+r_2 \] which yields \(r_1/b = q_2/10 + r_2/(10b)\) or \[ r/b=q_1/10+q_2/100+r_2/(100b). \] Again, \(q_2<10\) and \(r_2/100b<1/100\).

Repeating decimals continued

Now we proceed by induction. Suppose we have \[ r/b = q_1/10+q_2/100+\cdots+q_n/10^{n}+\frac{r_n}{10^{n}b} \] with each \(q_i<10\) and \(r_n<b\) so that \(r_n/(10^{n}b)<1/10^{n}\). Define \(r_{n+1}\) by dividing by \(10r_{n}\) by \(b\): \[ 10r_{n}=q_{n+1}b+r_{n+1}. \] Again, \(q_{n+1}<10\) and \(r_{n+1}<b\). Also \[ r_{n}/b=\frac{q_{n+1}}{10}+\frac{r_{n+1}}{10b} \] so \[ r/b = \frac{q_1}{10}+\frac{q_2}{100}+\cdots+\frac{q_n}{10^n}+\frac{q_{n+1}}{10^{n+1}}+\frac{r_{n+1}}{10^{n+1}b}. \]

Finishing up

The key final observation in the proof is that the sequence of remainders \(r_{n+1}\) have two important properties:

These two facts force the sequence of remainders to ultimately repeat, because after considering \(b+1\) of them there have to be two of them which are equal; say \(r_{k}=r_{j}\) for some pair \(k,j\) with \(j>k\). But then \(r_{k+1}=r_{j+1}\), \(r_{k+2}=r_{j+2}\), and so on, and in fact \(r_{i+(j-k)}=r_{i}\) for any \(i\ge k\). Thus the sequence of \(r\)’s, and therefore the sequence of \(q\)’s, eventually becomes periodic.

Other bases: Notice that there was nothing special about \(10\) in the proof above, so rational numbers have periodic expansions in any base.