Real analysis/Series

Definition

If $(a_n)$ is a sequence, then we express the associated series as

$\sum _{n=1}^{\mathrm{\infty }}{a}_n$.

We have the n-th partial sum

$s_n:=a_1+\cdots +a_n$,

and hence an associated sequence of partial sums, $\{s_n\}$. If the sequence of partial sums converges, then we define the infinite sum to be the limit:

$\sum _{n=1}^{\mathrm{\infty }}{a}_n=\underset{n\to \mathrm{\infty }}{\lim }{s}_n$,

and say that the infinite series converges. Otherwise we shall say that it diverges, in which case no meaning for the symbol $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ exists.

Examples

We'll start with some concrete examples, of which there are many:

• Geometric Series: Any series of the form $\sum _{n=0}^{\mathrm{\infty }}a{r}^n$ is called a geometric series. As we'll see later, these converge for $|r|<1$ and are easy to sum. The general formula is $\sum _{n=1}^{\mathrm{\infty }}a{r}^n=\frac{a}{1-r}$.

• Telescoping Series: Any series of the form $\sum _{n=1}^{\mathrm{\infty }}{a}_n-a_{n-1}$ is called a telescoping series. These are named after their "telescoping property", that the partial sums depend only on the first and last terms. As an exercise, prove that $s_n=a_n-a_0$. Thus $\sum _{n=1}^{\mathrm{\infty }}(a_n-a_{n-1})=\underset{n\to \mathrm{\infty }}{\lim }{s}_n=(\underset{n\to \mathrm{\infty }}{\lim }{a}_n)-a_0$.

• Alternating Series: Any series of the form $\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n{a}_n$, where for all i, $a_i\ge 0$ is called an alternating series.

• p-series: Any series of the form $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^p}$ is called a p-series. This kind of series is generally harder to evaluate than the two previously discussed. When p=2, for example, we get the amazing identity $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}=\frac{{\pi }^2}{6}$. That takes considerable effort to prove, so for the moment we will content ourselves with showing that $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^p}$ diverges for p $\le$ 1 and converges for p>1.

• Decimal Expansions: Although these are intimately familiar to us, we have still not given a rigorous justification for their use. Such a justification can be given, as we will see later. The basic fact is that, up to a ambiguity involving repeated 9s, every real number $a$ can be uniquely expressed in the form $a=\sum _{n=0}^{\mathrm{\infty }}\frac{a_n}{10^n}$

• Power Series: These should be familiar to anyone who has taken calculus. A power series is actually a function $f$ that takes a real number $x$ and returns a series, $f(x)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$. These, along with Fourier Series, will be examined in detail in a later chapter of the book, once we have covered sequences and series of functions.

• Fourier Series: Like power series, these are functions that take a real number and return a series. However, the terms are trigonometric functions rather than powers of x. The general expression for a Fourier series is: $f(x)=a_0+\sum _{n=1}^{\mathrm{\infty }}{a}_{n}cos(nx)+b_{n}sin(nx)$.

Basic Facts

Here are some fairly straightforward facts about infinite series that require verifying. First, we need to make sure that series, like sequences, behave properly when we add them together, multiply by constants, and the like.

Theorem (Algebraic Operations)

If $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ and $\sum _{n=1}^{\mathrm{\infty }}{b}_n$ converge to a and b, respectively, then:

• So does $\sum _{n=1}^{\mathrm{\infty }}c{a}_n$, and $c\sum _{n=1}^{\mathrm{\infty }}{a}_n=\sum _{n=1}^{\mathrm{\infty }}c{a}_n$ for any real c.
• So does $\sum _{n=1}^{\mathrm{\infty }}(a_n+b_n)$, and $\sum _{n=1}^{\mathrm{\infty }}{a}_n+\sum _{n=1}^{\mathrm{\infty }}{b}_n=\sum _{n=1}^{\mathrm{\infty }}(a_n+b_n)$

Proof

• Since $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges to a, $\mathrm{\forall }ϵ>0:\mathrm{\exists }N:\mathrm{\forall }n\ge N:|\sum _{k=1}^n{a}_n-a|<\frac{ϵ}{|c|}$.

Thus $|c||\sum _{k=1}^n{a}_n-a|<|c|\frac{ϵ}{|c|}⟹|\sum _{k=1}^{n}c{a}_n-ca|<ϵ$

• $\sum _{n=1}^{\mathrm{\infty }}(a_n+b_n)=\underset{n\to \mathrm{\infty }}{\lim }\sum _{k=1}^n(a_n+b_n)$$=\underset{n\to \mathrm{\infty }}{\lim }(a_1+b_1+a_2+b_2+\dots +a_n+b_n)$$=\underset{n\to \mathrm{\infty }}{\lim }(a_1+a_2+\dots +a_n+b_1+b_2+\dots b_n)$$=\underset{n\to \mathrm{\infty }}{\lim }(\sum _{k=1}^n(a_n)+\sum _{k=1}^n(b_n))$$=\underset{n\to \mathrm{\infty }}{\lim }\sum _{k=1}^n(a_n)+\underset{n\to \mathrm{\infty }}{\lim }\sum _{k=1}^n(b_n)=a+b$

The following is merely a restatement of Cauchy's Criterion in the language of infinite series:

Theorem (Cauchy's Criterion for Series)

$\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges if and only if $\mathrm{\forall }ϵ>0\mathrm{\exists }N:\mathrm{\forall }m>n\ge N:|a_n+a_{n+1}+\cdots +a_{m-1}|<ϵ$.

Proof

$\sum _{n=1}^{\mathrm{\infty }}{a}_n$ = $\underset{n\to \mathrm{\infty }}{\lim }{s}_n$, which converges if and only if $\mathrm{\forall }ϵ>0\mathrm{\exists }N:\mathrm{\forall }m>n\ge N:|s_n-s_m|=|a_n+a_{n+1}+\cdots +a_{m-1}|<ϵ$. $◻$

Although the following theorem is stated in terms of convergence, it actually gives a useful criterion for divergence. Namely, if the terms of a series do not have limit 0, the series must diverge.

Theorem (Terms Have Limit 0)

For any converging series $\sum _{n=1}^{\mathrm{\infty }}{a}_n$, $\underset{n\to \mathrm{\infty }}{\lim }(a_n)=0.$

Proof

By the Cauchy Criterion for series $\mathrm{\forall }ϵ>:\mathrm{\exists }N:\mathrm{\forall }m,n\ge N:|a_n+\dots a_{m-1}|<ϵ$. In particular, $\mathrm{\forall }n\ge N,|a_n|<ϵ$. By definition, $(a_n)\to 0$.

Positive and Alternating Series

We need to have ways of checking for convergence and divergence of various series. The following types of series arise quite often, and it is easy to verify when they converge and diverge:

Theorem (Positive Series Converge)

If $a_n>0$ for all n, then $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ either converges or diverges to infinity.

Proof

Either the partial sums are bounded or unbounded. If they are bounded, $s_{n+1}=s_n+a_{n+1}>s_n$, so they form a monotone sequence, which must converge by the monotone convergence theorem. If they are unbounded, then since all terms are positive, $\underset{n\to \mathrm{\infty }}{\lim }{s}_n=\mathrm{\infty }$. In other words, $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ diverges to infinity.

Theorem (Alternating Series Test)

If $(a_n)$ is a monotone sequence of positive real numbers such that $\underset{n\to \mathrm{\infty }}{\lim }{a}_n=0$, then the alternating series $\sum _{n=0}^{\mathrm{\infty }}(-1{)}^{n+1}{a}_n$ converges.

Proof

Let $s_n=\sum _{k=1}^n(-1{)}^{k+1}{a}_k$, and consider the sequence of intervals $I_n=[s_{2n},s_{2n+1}]$.

By assumption, the following inequalities hold:

$s_{2n+2}=s_{2n}+a_{2n+1}-a_{2n+2}>s_{2n}$,

$s_{2n+2}=s_{2n+1}-a_{2n}<s_{2n+1}$,

$s_{2n+3}=s_{2n+1}-a_{2n+2}+a_{2n+3}<s_{2n+1}$,

and $s_{2n+3}=s_{2n+2}+a_{2n+3}>s_{2n+2}$

Thus $(I_n)$ is a nested sequence of intervals, and $\underset{n=1}{\overset{\mathrm{\infty }}{\cap }}{I}_n$ must contain at least one point, $x_0$.

Now, $|s_{2n}-x_0|<|s_{2n}-s_{2n+1}|=|a_{2n+1}|<|a_{2n}|$ and $|s_{2n+1}-x_0|<|s_{2n}-s_{2n+1}|=|a_{2n+1}|$.

Since $(a_n)\to 0$, given $ϵ>0:\mathrm{\exists }N:\mathrm{\forall }n\ge N:|s_n-x_0|<|a_{n+1}|<ϵ$. Thus $\sum _{n=0}^{\mathrm{\infty }}(-1{)}^{n+1}{a}_n=x_0$. $◻$

Comparison Test

To determine whether or not a series converges, sometimes it is necessary to compare it term-by-term with another series whose convergence or divergence we have previously ascertained. To do this rigorously, we need the following theorem:

Theorem (Comparison Test)

If $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ and $\sum _{n=1}^{\mathrm{\infty }}{b}_n$ are two series with $a_n\ge b_n\ge 0$, then:

• If $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges, then $\sum _{n=1}^{\mathrm{\infty }}{b}_n$ converges.
• If $\sum _{n=1}^{\mathrm{\infty }}{b}_n$ diverges, then $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ diverges.

Proof

• $\mathrm{\forall }n:a_n>b_n>0$, so the sequence of partial sums for $a_n$ is monotone and $\sum _{k=1}^n{b}_n<\sum _{k=1}^n{a}_n<\sum _{n=1}^{\mathrm{\infty }}{a}_n$. Since the sequence of partial sums for $b_n$ is monotone and bounded above, $\sum _{n=1}^{\mathrm{\infty }}{b}_n$ converges as well.

• Since positive series either converge or diverge to infinity, this is just the contrapositive of the first statement.

Theorem (Limit Comparison Test)

If $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ and $\sum _{n=1}^{\mathrm{\infty }}{b}_n$ are two series with $0<lim\Vert \frac{a_n}{b_n}\Vert <\mathrm{\infty }$, then $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges absolutely if and only if $\sum _{n=1}^{\mathrm{\infty }}{b}_n$ converges.

Proof

• Suppose the limit of the absolute value of their ratios converges to some number 0<r<$\mathrm{\infty }$. Then there exists an N such that if n>N, that $\Vert \frac{a_n}{b_n}\Vert -r<\frac{r}{2}$, so $\frac{r}{2}<|\frac{a_n}{b_n}\Vert <\frac{3r}{2}$. This means that $\frac{r}{2}\Vert b_n\Vert <\Vert a_n\Vert <\frac{3r}{2}\Vert b_n\Vert$. Hence, by the comparison test, if $\sum _{n=1}^{\mathrm{\infty }}{b}_n$ converges, then $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ also converges. Dividing by $\frac{r}{2}$, one can also see by the comparison test that if $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges, then $\sum _{n=1}^{\mathrm{\infty }}{b}_n$ also converges.

Absolute Convergence

Frequently, we find that series with positive terms are easiest to work with. For one thing, these series always converge or diverge, as we have already seen. So, it is sometimes useful to compare a series to a corresponding positive-termed series, the series of absolute values:

We say a series $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ is absolutely convergent if the series of the absolute values of its terms, $\sum _{n=1}^{\mathrm{\infty }}|a_n|$, converges.

Theorem (Absolute Convergence)

If $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges absolutely, then it converges.

Proof

If the series converges absolutely, then the sequence of partial sums $s_n=\sum _{k=1}^n|a_n|$ converges.

Then, by the Cauchy Criterion, $\mathrm{\forall }ϵ>0:\mathrm{\exists }N:\mathrm{\forall }n,m\ge N:||a_n|+\dots +|a_{m-1}||<ϵ$.

By the triangle inequality $|a_n+\dots +a_{m-1}|\le |a_n|+\dots +|a_{m-1}|=||a_n|+\dots +|a_{m-1}||<ϵ$.

Again using the Cauchy Criterion, we see that $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges.

Comparison with Geometric Series

A clever thing to do is to think of infinite series as geometric series with a variable ratio. If this ratio is "mostly" less than 1, the series converges, and if it is "mostly" greater than 1, the series diverges. This is expressed rigorously in the following two theorems. [Note: to avoid disrupting the format of the article, we are assuming without proof here that all geometric series with ratio less than 1 converge. For a proof of this, see the Examples with Proof section]

Theorem (Ratio Test)

If $r=\underset{n\to \mathrm{\infty }}{\lim }\frac{|x_n|}{|x_{n+1}|}<1$, then $\sum _{n=1}^{\mathrm{\infty }}{x}_n$ converges absolutely. If r > 1, then $\sum _{n=1}^{\mathrm{\infty }}{x}_n$ diverges.

Proof

Let $ϵ=1-r>0$.

Since $\underset{n\to \mathrm{\infty }}{\lim }\frac{|x_n|}{|x_{n+1}|}=r$,$\mathrm{\exists }N:\mathrm{\forall }n\ge N:|\frac{|x_n|}{|x_{n+1}|}-r|<\frac{ϵ}{2}$.

So, $\mathrm{\forall }n\ge N:|x_n|/|x_{n+1}|<r+\frac{ϵ}{2}$.

Thus $\sum _{n=1}^{\mathrm{\infty }}|x_n|$ = $\sum _{n=1}^{N-1}|x_n|+\sum _{n=N}^{\mathrm{\infty }}|x_n|$

$=\sum _{n=1}^{N-1}|x_n|+\sum _{n=N}^{\mathrm{\infty }}|x_N|\frac{|x_{N+1}|}{|x_N|}\frac{|x_{N+2}|}{|x_{N+1}|}\cdots \frac{|x_n|}{|x_{n-1}|}$

$<\sum _{n=1}^{N-1}|x_n|+\sum _{n=N}^{\mathrm{\infty }}|x_N|(r+\frac{ϵ}{2}{)}^{n-N}$, which converges since it is a constant plus a geometric series with ratio $r+\frac{ϵ}{2}<1$.

By the comparison test, $\sum _{n=1}^{\mathrm{\infty }}|x_n|$ converges.

Thus $\sum _{n=1}^{\mathrm{\infty }}{x}_n$ converges absolutely. $◻$

The next theorem, the root test, is stronger than the ratio test in the sense that it works whenever the ratio test works (and returns the same number r), and it sometimes works even when the ratio test does not. This is because it uses the limit superior, which always converges for bounded sequences, and does not involve division. On the other hand, the ratio test uses regular limits, which do not always converge, and sometimes involves division by 0.

Theorem (Root Test)

Let $R=\underset{n\to \mathrm{\infty }}{\lim }sup(|a_n{|}^{\frac{1}{n}})$. If R < 1, then the series $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges absolutely. If R > 1, then it diverges.

Proof

If R<1, let $R<\rho <1$. Since $\underset{n\to \mathrm{\infty }}{\lim }sup(|a_n{|}^{\frac{1}{n}})=R$ $\mathrm{\exists }N:\mathrm{\forall }n\ge N:|sup\{|a_k{|}^{\frac{1}{k}}|k>n\}-R|<\rho -R$.

That is, $\mathrm{\forall }n\ge N:sup\{|a_k{|}^{\frac{1}{k}}|k>n\}<\rho$

Thus $\sum _{n=1}^{\mathrm{\infty }}|a_n|$$=\sum _{n=1}^{\mathrm{\infty }}(|a_n{|}^{\frac{1}{n}}{)}^n$$=\sum _{n=1}^{N-1}|a_n|+\sum _{n=N}^{\mathrm{\infty }}(|a_n{|}^{\frac{1}{n}}{)}^n$$<\sum _{n=1}^{N-1}|a_n|+\sum _{n=N}^{\mathrm{\infty }}(sup\{|a_k{|}^{\frac{1}{k}}|k>n\}{)}^n$$<\sum _{n=1}^{N-1}|a_n|+\sum _{n=N}^{\mathrm{\infty }}(\rho {)}^n$, which converges since it is a constant plus a geometric series.

By the comparison test, $\sum _{n=1}^{\mathrm{\infty }}|a_n|$ converges as well. Thus $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges absolutely.

If R > 1, then $sup\{|a_n{|}^{\frac{1}{n}}|k>n\}>1$. Thus there are infinitely many n such that $|a_n{|}^{\frac{1}{n}}>1⟹|a_n|>1$.

Thus $(a_n)\to ̸0$, which implies that $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ diverges. $◻$

Sums of Products

We'll often be asked to consider series of the form $\sum _{n=1}^{\mathrm{\infty }}{a}_n{b}_n$. The following theorems give criteria for convergence of these series.

Theorem (Dirichlet's Test)

If the partial sums of $\sum _{n=1}^{\mathrm{\infty }}{x}_n$ are bounded, and $(y_n)$ is a decreasing sequence with $\underset{n\to \mathrm{\infty }}{\lim }{y}_n=0$, then $\sum _{n=1}^{\mathrm{\infty }}{x}_n{y}_n$ converges. [Note: $\sum _{n=1}^{\mathrm{\infty }}{x}_n$ need not converge]

Proof

Let $s_n=\sum _{m=1}^n{x}_m$ be the $n$th partial sum, so that there exists $B>0$ such that $|s_n|\le B$.

We can write

$\sum _{n=1}^N{x}_n{y}_n=x_1{y}_1+\sum _{n=2}^N(s_n-s_{n-1})y_n=x_1{y}_1+\sum _{n=2}^N{s}_n{y}_n-\sum _{n=2}^N{s}_{n-1}{y}_n$.

Changing the index of summation in the last sum, this becomes

$\sum _{n=1}^N{x}_n{y}_n=x_1{y}_1+\sum _{n=2}^N{s}_n{y}_n-\sum _{n=1}^{N-1}{s}_n{y}_{n+1}=x_1{y}_1+\sum _{n=2}^{N-1}{s}_n(y_n-y_{n+1})+s_N{y}_N$.

The sum on the right-hand side is bounded absolutely by the telescoping sum

$\sum _{n=2}^{N-1}|s_n(y_n-y_{n+1})|\le B\sum _{n=2}^{N-1}(y_n-y_{n+1})=B(y_2-y_N)\le B{y}_2$;

here we have used the fact that $(y_n)$ is positive and decreasing. It follows that

$\sum _{n=2}^{\mathrm{\infty }}{s}_n(y_n-y_{n+1})$ is absolutely convergent, hence convergent.

Notice that $\underset{N\to \mathrm{\infty }}{\lim }{s}_N{y}_N=0$ since $s_N$ is bounded and $(y_n)$ tends to 0 as $n$ tends to infinity. Therefore we can take limits as $N$ goes to infinity:

$\sum _{n=1}^{\mathrm{\infty }}{x}_n{y}_n=x_1{y}_1+\sum _{n=2}^{\mathrm{\infty }}{s}_n(y_n-y_{n+1})+\underset{N\to \mathrm{\infty }}{\lim }{s}_N{y}_N=x_1{y}_1+\sum _{n=2}^{\mathrm{\infty }}{s}_n(y_n-y_{n+1})$,

which proves that the left-hand side is convergent.

Abel's Test can be regarded as a special case of Dirichlet's Test, once some modifications have been made:

Theorem (Abel's Test)

If $\sum _{n=1}^{\mathrm{\infty }}{x}_n$ converges and $(y_n)$ is a positive, decreasing sequence, then $\sum _{n=1}^{\mathrm{\infty }}{x}_n{y}_n$ converges.

Proof

Because $(y_n)$ is a bounded, monotone sequence, it converges to some limit y.

Let $z_n=y_n-y$. Then $\sum _{n=1}^{\mathrm{\infty }}{x}_n$ and $(z_n)$ satisfy the conditions for Dirichlet's test.

$\sum _{n=1}^k{x}_n{y}_n$$=y\sum _{n=1}^k{x}_n+\sum _{n=1}^k{x}_n{z}_n$. By Dirichlet's test, $\sum _{n=1}^k{x}_n{z}_n$ converges as $k\to \mathrm{\infty }$.

Since both sums on the right converge, $\sum _{n=1}^{\mathrm{\infty }}{x}_n{y}_n$ converges as well. $◻$

Examples with Proof

Now we'll do the computations promised in the Examples, plus a few extra.

Theorem (Geometric Series)

If $|r|<1$, the geometric series $\sum _{n=1}^{\mathrm{\infty }}a{r}^n=\frac{a}{1-r}$. If $|r|\ge 1$, the series diverges.

Proof

In this case, it is best to explicitly compute the partial sums and take the limit. Without loss of generality, we'll consider the case $a=1$. Then we can apply the theorem on algebraic operations to obtain the general result.

Note that $s_n(1-r)=(1+r+r^2+...+r^n)(1-r)=(1+r+r^2+...+r^n)-(r+r^2+...+r^{n+1})=1-r^{n+1}$

Thus $s_n=\frac{1-r^n}{1-r}$. Taking the limit (and remembering some basic facts about sequences):

If $|r|<1$, $\sum _{n=1}^{\mathrm{\infty }}{r}^n=\underset{n\to \mathrm{\infty }}{\lim }{s}_n=\underset{n\to \mathrm{\infty }}{\lim }\frac{1-r^n}{1-r}=\frac{1}{1-r}$.

If $|r|\ge 1$, the sequence of partial sums diverges to infinity, and thus by definition the series diverges.

This proof seems to work except for the fact that this series will converge to $\frac{a}{1-r}$ $◻$

Theorem(p-series)

The p-series $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^p}$ converges for $p>1$ and diverges for $0<p\le 1$. [Note: We have technically only defined this series for p rational. However, the theorem is still valid when p is irrational, for the same reasons]

Proof

First we'll consider the special cases $p=1,p=2$, and then obtain the general result from these.

• If p = 1, let $x_n$ be the greatest power of 2 less than $\frac{1}{n}$. That is, $(x_n)=(\frac{1}{2},\frac{1}{4},\frac{1}{4},\frac{1}{8},\frac{1}{8},\frac{1}{8},\frac{1}{8},\dots )$.

Grouping like terms together, $\sum _{i=1}^{\mathrm{\infty }}{x}_n$$=\sum _{i=0}^{\mathrm{\infty }}(\sum _{j=2^i}^{2^{i+1}-1}{x}_j)$$=\sum _{i=0}^{\mathrm{\infty }}(\sum _{j=2^i}^{2^{i+1}-1}\frac{1}{2^{i+1}})$$=\sum _{i=1}^{\mathrm{\infty }}\frac{1}{2}$, which diverges.

By definition, $x_n<\frac{1}{n}$, so by the comparison test $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}$ diverges.

• If p = 2, let $x_n=\frac{1}{n(n-1)}=\frac{1}{n-1}-\frac{1}{n}$ if n>1 and 1 if n=1.

Using the theorem on telescoping series, $\sum _{n=1}^{\mathrm{\infty }}{x}_n$$=1+\sum _{n=2}^{\mathrm{\infty }}(\frac{1}{n-1}-\frac{1}{n})$$=1+1-\underset{n\to \mathrm{\infty }}{\lim }(1/n)=2$

By definition, $x_n>\frac{1}{n\cdot n}=\frac{1}{n^2}$, so by the comparison test $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}$ converges as well.

• If 0<p<1, then $\frac{1}{n^p}>\frac{1}{n}$. By the comparison test $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^p}$ diverges.

Theorem(Decimal Expansions)

Given any real number x, $0<x<1$there is a unique sequence $(x_n):\sum _{n=0}^{\mathrm{\infty }}\frac{x_n}{10^n}=x$, $0\le x_n<10$ and $\mathrm{\forall }N:\mathrm{\exists }n\ge N:x_n=9$

Proof

Inductively, assume that there exist $x_0,x_1,\dots ,x_k$ such that $\sum _{n=0}^k\frac{x_n}{10^n}\le x<\sum _{n=0}^k\frac{x_n}{10^n}+\frac{1}{10^k}$.

Rearranging, we see that $0\le x-\sum _{n=0}^k\frac{x_n}{10^n}<\frac{1}{10^k}$, or $0\le (x-\sum _{n=0}^k\frac{x_n}{10^n})10^{k+1}<10$.

Let $x_{k+1}$ by the greatest integer such that $x_{k+1}\le (x-\sum _{n=0}^k\frac{x_n}{10^n})10^{k+1}$.

Then $\frac{x_{k+1}}{10^{k+1}}\le x-\sum _{n=0}^k\frac{x_n}{10^n}<\frac{x_{k+1}+1}{10^{k+1}}$(otherwise, $x_{k+1}$ would not be the greatest).

Adding $\sum _{n=0}^k\frac{x_n}{10^n}$, we see that $\sum _{n=0}^{k+1}\frac{x_n}{10^n}\le x<\sum _{n=0}^{k+1}\frac{x_n}{10^n}+\frac{1}{10^{k+1}}$.

Given $ϵ>0$ pick N such that $ϵ>\frac{1}{10^N}$.

Thus $|\sum _{n=0}^k\frac{x_n}{10^n}-1|<\frac{1}{10^k}<ϵ$ for all $k>N$. That is, $x=\sum _{n=0}^{\mathrm{\infty }}\frac{x_n}{10^n}$.

TODO: eliminate 9s, uniqueness.

Topics to come: Rearrangement of terms, Alternating Series Test, Sums of products(i.e. Abel's Test and Dirichlet's Test), Multiple Summations, Infinite Products, decimal expansions, zeta function