Series

Definition

Let $(a_n)$ be a sequence of Complex Numbers. Let

$$S_n:=a_1+a_2+\cdots +a_n=:\sum _{k=1}^n{a}_k$$

is called the partial sums. We denote by $a_1+a_2+a_3+\cdots$ the formal sum

$$\sum _{k=1}^{\infty }{a}_k\text{ or }\sum _{n\in \mathbb{N}}{a}_k\text{ or }\sum _k{a}_k$$

and call it a series.

• The series converges if $s_n\to s$ for some $s\in \mathbb{C}$. We write ${\sum }_{k=1}^{\infty }{a}_k=s$.

Lemma (Geometric Series)

Let $a\in \mathbb{R}\backslash \{0\},r\in \mathbb{R}$. Then

$$\sum _{n\in \mathbb{N}\cup \{0\}}a{r}^n\text{ converges }⟺|r|<1$$

In which case,

$$\sum _{n\in \mathbb{N}\cup \{0\}}a{r}^n=\frac{a}{1-r}$$

Proof: If $r=1$, then

$$\begin{array}{rl}{s}_n& =a{r}^0+a{r}^1+\cdots a{r}^{n-1}\\ & =a+\cdots +a\\ & =na\end{array}$$

so $s_n=na$, which diverges. If $r\ne 1$,

$$\begin{array}{rl}{s}_n& =a(1+r+r^2+\cdots +r^{n-1})\\ & =a\left(\frac{1-r^n}{1-r}\right)\end{array}$$

Now, if $|r|<1$ then $r^n{\to }_{n\to \infty }0$ by special sequences. So by algebra of sequences, we have

$$s_n=a\left(\frac{1-r^n}{1-r}\right)\to \frac{a}{1-r}$$

If $|r|>1$ then it does not converge (diverges to $\pm \infty$ or fluctuates to $\pm \infty$). If $|r|=1$ then $r=1,-1$ and clearly these do not work. Then it must converge.

Cauchy Criterion for Sequences

We say a series $\sum a_k$ satisfies a Cauchy Criterion if the sequence of partial sums $(s_n)$ is a Cauchy Sequence.

Remark

Recall that $(s_n)$ is Cauchy if $\forall \epsilon >0,\exists N\in \mathbb{N}$ such that $\forall n\ge N,|s_n-s_m|<\epsilon$. We note that

$$\begin{array}{rl}{s}_n& =a_1+a_2+\cdots +a_n\\ s_m& =a_1+a_2+\cdots +a_m\end{array}$$

so if $m\ge n\ge N$, then

$$\begin{array}{rl}{s}_m-s_n& =a_{n+1}+a_{n+2}+\cdots +a_m\\ & =\sum _{k=n+1}^m{a}_k\end{array}$$

so a sum ${\sum }_n{a}_n$ satisfies a Cauchy Criterion $⟺$ $\forall \epsilon >0,\exists N\in \mathbb{N}$ such that $\forall m\ge n\ge N$, $|s_m-s_n|=\left|{\sum }_{k=n}^m{a}_k\right|<\epsilon$.

Remark (Equivalent Convergence)

$$\begin{array}{rl}\sum a_k\text{ converges}& ⟺(s_n)\text{ converges}\\ & ⟺(s_n)\text{ is Cauchy}\\ & ⟺\sum a_k\text{ satisfies a Cauchy Criterion}\end{array}$$

Corollary (Divergence Test)

Let $\sum a_n$ be a convergent series. Then $a_n{\to }_{n\to \infty }0$.

Proof: Since $\sum a_n$ converges, then it satisfies a Cauchy Criterion. Let $\epsilon >0$. Then $\exists N\in \mathbb{N},\forall m\ge n\ge N$,

$$\left|\sum _{k=n}^m{a}_k\right|<\epsilon$$

Let $m=n\ge N$. Then

$$\left|\sum _{k=n}^m{a}_k\right|=\left|\sum _{k=n}^n{a}_k\right|=|a_n|<\epsilon$$

The converse is not true. Consider the Harmonic Series.

Lemma (Convergence, Boundness)

Let $a_n\ge 0$. Then $\sum a_n$ converges $⟺$ $(S_n)=(a_1+a_2+\cdots +a_n)$ is bounded.

Proof: Since $a_n\ge 0$. Then

$$S_n=a_1+a_2+\cdots +a_n\le (a_1+\cdots +a_n)+a_{n+1}=S_{n+1}$$

so $(S_n)$ is monotonically increasing. So by MCT, $(S_n)$ converges $⟺$ $(S_n)$ is bounded.

Lemma (Removal of Absolute)

Let $(z_n)$ be a sequence of Complex Numbers. Assume

$$\sum _{n\in \mathbb{N}}|z_n|$$

converges. Then

$$\sum _n{z}_n$$

converges.

Proof: By Triangle Inequality.

The converse is not true. Consider $\sum \frac{(-1{)}^n}{n}$.

Definition (Absolute Convergence)

Let $(a_n)$ be a sequence of $\mathbb{C}$ numbers. We say ${\sum }_n{a}_n$ is absolutely convergent if $\sum |a_n|$ converges.

Lemma (Comparison Test)

Let $(a_n),(b_n),(c_n)\subset \mathbb{R}$. Suppose $a_n\le b_n\le c_n$ for all $n\ge N_0$. If $\sum a_n$ and $\sum c_n$ converge, then $\sum b_n$ converge.

Proof: We are given that for all $n\ge N_0$:

$$a_n\le b_n\le c_n$$

Since $\sum a_n,\sum c_n$ converges, then $\exists N_1,N_2\in \mathbb{N}$ such that $\forall m\ge n\ge N_{1,2}$

$$\left|\sum _{k=n}^m{a}_k\right|<\epsilon \left|\sum _{k=n}^m{c}_k\right|<\epsilon$$

Let $N=\max \{N_0,N_1,N_2\}$. Then $\forall m\ge n\ge N$, for each $n\le k\le m$, that $a_k\le b_k\le c_k$. Thus,

$$-\epsilon <\sum _{k=n}^m{a}_k\le \sum _{k=n}^m{b}_k\le \sum _{k=n}^m{c}_k<\epsilon$$

Indeed,

$$\left|\sum _{k=n}^m{b}_k\right|<\epsilon$$

Corollary (Convergence by Above/Below)

• Suppose $0\le b_n\le c_n$ for $n\ge N_0$. If $\sum c_n$ converges then $\sum b_n$ converges.
• If $\sum b_n$ diverges, then $\sum c_n$ diverges.

Theorem (Cauchy Condensation)

Let $a_1\ge a_2\ge \cdots \ge 0$. Then the series

$$\sum _{n=1}^{\infty }{a}_n\text{ converges}⟺\sum _{k=0}^{\infty }{2}^k{a}_{2^k}\text{ converges}$$

Proof: Assume ${\sum }_n{a}_n$ converges.

Corollary (P-Test)

1. $\sum 1/n^p$ converges $⟺$ $p>1$
2. ${\sum }_{n=2}^{\infty }1/(2\log n{)}^p$ converges $⟺$ $p>1$ Proof: Use Cauchy Condensation

Proposition (Root Test)

Let $(a_n)$ be a sequence. Put $\alpha =\overline{\lim }\sqrt[n]{|a_n|}$.

1. If $\alpha <1$, then $\sum a_n$ converges.
2. If $\alpha >1$, then $\sum a_n$ diverges.
3. If $\alpha =1$, then no conclusion.

Proof (1): Since $\alpha <1$, by checkable criterion we have that $\exists N\in \mathbb{N}$ such that $\forall n\ge N$,

$$\sqrt[n]{|a_n|}<r⟹|a_n|<r^n$$

for all $n\ge N$. Since $r<1$ then $\sum r^n$ convergences by geometric series. So by comparison test, $\sum a_n$ is absolutely convergent. Lastly, by Remark (Equivalent Convergence) it is convergent.

Proof (2): Suppose $\alpha >1$. Then by definition, $\alpha =\sup (\mathcal{A})$, then $\exists$ subsequence

$$n_1<n_2<\cdots$$

such that

$$\sqrt[n]{|a_{n_i}|}=\alpha >1$$

so infinitely many $n_i\in \mathbb{N}$ with $|a_{n_i}|>1$ and so $|a_{n_i}|\nrightarrow 0$ and $a_{n_i}\nrightarrow 0$ and by divergence test, it must diverge.

Proposition (Ratio Test)

Let $(a_n)$ be a sequence of nonzero real numbers.

1. If

$$\beta :=\overline{\lim }\left|\frac{a_{n+1}}{a_n}\right|<1$$

then $\sum a_n$ converges. 2. If $\exists n_0$ such that

$$\left|\frac{a_{n+1}}{a_n}\right|\ge 1$$

, $\forall n\ge n_0$ then $\sum a_n$ diverges.

Remark (Ratio - Root Relationship)

The ratio test is not perfect. Indeed, If the ratio test shows convergence of a series, then the root test must also. The converse is not necessarily true (Rudin 3.35).

Indeed, if the root test implies divergence, then the ratio test will also imply divergence. This is because for any sequence $\{c_n\}$ of positive numbers,

$$\begin{array}{rl}\underset{n\to \infty }{\mathrm{liminf}}\frac{c_{n+1}}{c_n}& \le \underset{n\to \infty }{\mathrm{liminf}}\sqrt[n]{c_n}\\ \underset{n\to \infty }{\mathrm{limsup}}\sqrt[n]{c_n}& \le \underset{n\to \infty }{\mathrm{limsup}}\frac{c_{n+1}}{c_n}\end{array}$$

Proof: We prove the second inequality. Let $\alpha ={\mathrm{limsup}}_{n\to \infty }\frac{c_{n+1}}{c_n}$. If $\alpha =+\infty$, we are done. The $\alpha$ is finite, choose $\beta >\alpha$. By the checkability lemma, $\exists N\in \mathbb{Z}$ such that

$$\frac{c_{n+1}}{c_n}\le \beta$$

for $n\ge N$. In particular, for any $p>0$,

$$c_{N+k+1}\le \beta c_{N+k}(k=0,1,\dots ,p-1)$$

such that by inducting on $k$ to $p$,

$$c_{N+p}\le {\beta }^p{c}_N$$

As $n\ge N$ then for $n=N+p$,

$$\begin{array}{rl}{c}_n& \le c_N{\beta }^{n-N}\\ \sqrt[n]{c_n}& \le \beta \sqrt[n]{c_N{\beta }^{-N}}\\ \underset{n\to \infty }{\mathrm{limsup}}\sqrt[n]{c_n}& \le \beta \end{array}$$

The last line is by special sequences. Since this is true for every $\beta >\alpha$ then

$$\underset{n\to \infty }{\mathrm{limsup}}\sqrt[n]{c_n}\le \alpha$$

The proof is similar for $\mathrm{liminf}$.

Summation by Parts

Given two Sequences $\{a_n\}$ and $\{b_n\}$, put

$$A_n=\sum _{k=0}^n{a}_k$$

if $n\ge 0$ put $A_{-1}=0$. Then if $0\le p\le q$, we have

$$\sum _{n=p}^q{a}_n{b}_n=A_q{b}_q-A_{p-1}{b}_p+\sum _{n=p}^{q-1}{A}_n(b_n-b_{n+1})$$

this useful for:

Theorem (Multiplicative Convergence)

Suppose

1. the partial sums $A_n$ of $\sum a_n$ form a bounded sequence
2. $b_0\ge b_1\ge b_2\ge \cdots$
3. ${\lim }_{n\to \infty }{b}_N=0$ Then $\sum a_n{b}_n$ converges.

Operations on Series

Lemma

If $A=\sum a_n,B=\sum b_n$ are convergent, then

1. $\sum (a_n+b_n)=A+B$
2. $\sum (c{a}_n)=cA,\forall c\in \mathbb{R}$
3. Adding, removing finitely many elements to $\sum a_n$ does not change convergence.