Oct 28-Nov 1

**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:

- Fact that the sequence \(1/n\) converges to \(0\).
- Limit rules for sums, products, quotients.

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 \(\sum_{n=0}^{\infty} ar^{n}\).

**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 \[
a\frac{r^{k+1}-1+r^{k+2}-r^{k+1}}{r-1}=a\frac{r^{k+2}-1}{r-1}.
\]

**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 \(a/(1-r)\).

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.

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.

**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.

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.

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

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: \[ 10r = q_1b+r_1. \]

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 \(r_2/100b<1/100\).

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 \[ r/b = \frac{q_1}{10}+\frac{q_2}{100}+\cdots+\frac{q_n}{10^n}+\frac{q_{n+1}}{10^{n+1}}+\frac{r_{n+1}}{10^{n+1}b}. \]

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

- First, they are all between \(0\) and \(b\), and second
- Knowing \(r_{n}\) determines \(r_{n+1}\).

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.