# 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:

• Fact that the sequence $$1/n$$ converges to $$0$$.
• Limit rules for sums, products, quotients.

# Series

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:

• 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.