Chapter 6 Series

6.1 Finite series

Definition 6.1 (Finite series) Let $m,n$ be integers, and let $(a_n)_{n=m}^{\infty}$ be a finite sequence of real numbers, assigning a real number $a_i$ to each integer $i$ between $m$ and $n$ inclusive (i.e., $m\leq i\leq n$). Then we define the finite sum (or finite series) $\sum_{i=m}^na_i$ by the recursive formula $$\begin{aligned} \sum_{i=m}^na_i&:=0 \text{ whenever } n<m;\\ \sum_{i=m}^{n+1}a_i&:=\left(\sum_{i=m}^na_i\right)+a_{n+1} \text{ whenever } n\geq m-1. \end{aligned}$$

Strictly speaking, a series is an expression of the form $\sum_{i=m}^na_i$; this series if mathematically (but not semantically) equal to a real number, which is the sum of that series. Note that the variable $i$ (sometimes called the index of summation) is a bound variable (sometimes called a dummy variable); the expression $\sum_{i=m}^na_i$ does not actually depend on any quantity named $i$.

Lemma 6.1

• Let $m\leq n\leq p$ be integers, and let $a_i$ be a real number assigned to each integer $m\leq i \leq p$. Then we have $$\sum_{i=m}^na_i + \sum_{i=n+1}^pa_i=\sum_{i=m}^pa_i.$$
• Let $m\leq n$ be integers, $k$ be another integer, and let $a_i$ be a real number assigned to each integer $m\leq i \leq n$. Then we have $$\sum_{i=m}^na_=\sum_{i=m+k}^{n+k}a_{i-k}.$$
• Let $m\leq n$ be integers, and let $a_i, b_i$ be real numbers assigned to each integer $m\leq i \leq n$. Then we have $$\sum_{i=m}^n(a_i+b_i)=\sum_{i=m}^na_i+\sum_{i=m}^nb_i.$$
• Let $m\leq n$ be integers, and let $a_i$ be a real number assigned to each integer $m\leq i \leq n$, and $c$ be another real number. Then we have $$\sum_{i=m}^n(ca_i)=c\sum_{i=m}^na_i.$$
• (Triangle inequality fir finite series) Let $m\leq n$ be integers, and let $a_i$ be a real number assigned to each integer $m\leq i \leq n$. Then we have $$\left|\sum_{i=m}^na_i\right|\leq\sum_{i=m}^n|a_i|.$$
• (Comparison test for finite series) Let $m\leq n$ be integers, and let $a_i, b_i$ be real numbers assigned to each integer $m\leq i \leq n$. Suppose that $a_i\leq b_i$ for all $m\leq i\leq n$. Then we have $$\sum_{i=m}^na_i\leq\sum_{i=m}^nb_i.$$

Definition 6.2 (Summations over fintite sets) Let $X$ be a finite et with $n$ elements (where $n\in \mathbb{N}$), and let $f:X\to \mathbb{R}$ be a function from $X$ to the real numbers. Then we can define the finite sum $\sum_{x\in X}f(x)$ as follows. We first select and bijection $g$ from $\{i\in\mathbb{N}:1\leq i\leq n\}$ to $X$; such a bijection exists since $X$ is assumed to have $n$ elements. We then define $$\sum_{x\in X}f(x) :=\sum_{i=1}^nf(g(i)).$$

Proposition 6.1 (Finite summations are well-defined) Let $X$ be a finite et with $n$ elements (where $n\in \mathbb{N}$), and let $f:X\to \mathbb{R}$ be a function, and let $g:\{i\in\mathbb{N}:1\leq i\leq n\}\to X$ and $h:\{i\in\mathbb{N}:1\leq i\leq n\}\to X$ be bijections. Then we have $$\sum_{i=1}^nf(g(i))=\sum_{i=1}^nf(h(i)).$$

Proposition 6.2 (Basic properties of summation over finite sets)

• If $X$ is empty, and $f:X\to \mathbb{R}$ is a function, we have $$\sum_{x\in X}f(x) = 0.$$
• If $X$ consists of a single element, $X=\{x_0\}$, and $f:X\to \mathbb{R}$ is a function, we have $$\sum_{x\in X}f(x) = f(x_0).$$
• (Substitution, part I) If $X$ is a finite set, $f:X\to \mathbb{R}$ is a function, and $g:Y\to X$ is a bijection, then $$\sum_{x\in X}f(x) = \sum_{y\in Y}f(g(y)).$$
• (Substitution, part II) Let $n\leq m$ be integers, and let $X$ be the set $X:=\{i\in\mathbb{Z}:n\leq i\leq m\}$. If $a_i$ is a real number assigned to each integer $i\in X$, then we have $$\sum_{i=n}^m a_i=\sum_{i\in X}a_i.$$
• Let $X,Y$ be disjoint finite sets, and $f:X\cup Y\to \mathbb{R}$ is a function. Then we have $$\sum_{z\in X\cup Y}f(z)=\sum_{x\in X}f(x)+\sum_{y\in Y}f(y).$$
• (Linearity, part I) Let $X$ be a finite set, and let $f:X\to \mathbb{R}$ and $g:X\to \mathbb{R}$ be functions. Then $$\sum_{x\in X}(f(x)+g(x)) = \sum_{x\in X}f(x)+\sum_{x\in X}g(x).$$
• (Linearity, part II) Let $X$ be a finite set, and let $f:X\to \mathbb{R}$ be a function. Then $$\sum_{x\in X}cf(x) = c\sum_{x\in X}f(x).$$
• (Monotonicity) Let $X$ be a finite set, and let $f:X\to \mathbb{R}$ and $g:X\to \mathbb{R}$ be functions such that $f(x)\leq g(x)$ for all $x\in X$. Then we have $$\sum_{x\in X}f(x)\leq\sum_{x\in X}g(x).$$
• (Triangle inequality) Let $x$ be a finite set, and let $f:X\to \mathbb{R}$ be a function, then $$\left|\sum_{x\in X}f(x)\right|\leq\sum_{x\in X}|f(x)|.$$

The substitution rule in Proposition 6.2 indicates that we can rearrange the elements of a finite sequence at will and still obtain the same value.

Lemma 6.2 Let $X,Y$ be a finite set, and let $f:X\times Y\to \mathbb{R}$ be a function. Then $$\sum_{x\in X}\left(\sum_{y\in Y}f(x,y)\right) = \sum_{(x,y)\in X\times Y}f(x,y).$$

Corollary 6.1 (Fubini's theorem for finite series) Let $X,Y$ be finite sets, and let $f:X\times Y\to \mathbb{R}$ be a function. Then $$\begin{aligned} \sum_{x\in X}\left(\sum_{y\in Y}f(x,y)\right) &= \sum_{(x,y)\in X\times Y}f(x,y)\\ &= \sum_{(y,x)\in Y\times X}f(x,y)\\ &=\sum_{y\in Y}\left(\sum_{x\in X}f(x,y)\right) \end{aligned}$$

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Exercise 6.1 Let $X$ be a finite set, let $m$ be an integer and for each $x\in X$ let $(a_n(x))_{n=m}^{\infty}$ be a convergent sequence of real numbers. Show that the sequence $\sum_{x\in X}(a_n(x))_{n=m}^{\infty}$ is convergent, and $$\lim_{n\to \infty}\sum_{x\in X}a_n(x)=\sum_{x\in X}\lim_{n\to \infty}a_n(x).$$

6.2 Infinite series

Definition 6.3 (Formal infinite series) A (formal) series is any expression of the form $$\sum_{n=m}^{\infty}a_n,$$

where $m$ is an integer, and $a_n$ is a real number for any integer $n\geq m$.

Definition 6.4 (Convergence of series) Let $\sum_{n=m}^{\infty}a_n$ be a formal infinite series. For any integer $N \geq m$, we define the $N^{\text{th}}$ partial sum $S_N$ of this series to be $S_N:=\sum_{n=m}^Na_n$. If the sequence $(S_N)^{\infty}_{N=m}$ converges to some limit $L$ as $N\to\infty$, then we say that the infinite series $\sum_{n=m}^{\infty}a_n$ is convergent, and converges to $L$; we also write $L=\sum_{n=m}^{\infty}a_n$, and say that $L$ is the sum of the infinite series $\sum_{n=m}^{\infty}a_n$. If the partial sums $S_N$ diverge, then we say that the infinite series $\sum_{n=m}^{\infty}a_n$ is divergent, and we do not assign any real number value to that series.

Proposition 6.3 Let $\sum_{n=m}^{\infty}a_n$ be a formal series of real numbers. Then $\sum_{n=m}^{\infty}a_n$ converges iff for every real number $\varepsilon>0$. there exists an integer $N\geq m$ such that $$\left| \sum_{n=p}^{q}a_n\right|\leq \varepsilon \text{ for all }p,q\geq N$$

Corollary 6.2 (Zero test) Let $\sum_{n=m}^{\infty}a_n$ be a convergent series of real numbers. Then must have $\lim_{n\to\infty}a_n$. To put this another way, if $\lim_{n\to\infty}a_n$ is non-zero or divergent, then the series $\sum_{n=m}^{\infty}a_n$ is divergent.

Definition 6.5 (Absolute convergence) Let $\sum_{n=m}^{\infty}a_n$ be a formal series of real numbers. We say that this series is absolutely convergent iff the series $\sum_{n=m}^{\infty}|a_n|$ is convergent.

In order to distinguish convergence from absolute convergence, we sometimes refer to the former as conditional convergence.

Proposition 6.4 (Absolute convergence test) Let $\sum_{n=m}^{\infty}a_n$ be a formal series of real numbers. If this series is absolutely convergent, then it is also conditionally convergent. Furthermore, in this case we have the triangle inequality $$\left|\sum_{i=m}^{\infty}a_i\right|\leq\sum_{i=m}^{\infty}|a_i|.$$

Proposition 6.5 (Alternating series test) Let $\sum_{n=m}^{\infty}a_n$ be a sequence of real numbers which are non-negative and decreasing. Then the series $\sum_{n=m}^{\infty}(-1)^na_n$ is convergent iff the sequence $a_n$ converges to $0$ as $n\to\infty$.

Proposition 6.6 (Series laws)

• If $\sum_{n=m}^{\infty}a_n$ is a series of real numbers converging to $x$, and $\sum_{n=m}^{\infty}b_n$ is a series of real numbers converging to $y$, then $\sum_{n=m}^{\infty}(a_n+b_n)$ is also a convergent series, and converges to $x+y$. In particular, we have $$\sum_{i=m}^{\infty}(a_i+b_i)=\sum_{i=m}^{\infty}a_i+\sum_{i=m}^{\infty}b_i.$$
• If $\sum_{n=m}^{\infty}a_n$ is a series of real numbers converging to $x$, and $c$ is a real number, then $\sum_{n=m}^{\infty}(ca_n)$ is also a convergent series, and converges to $cx$. In particular, we have $$\sum_{i=m}^{\infty}(ca_i)=c\sum_{i=m}^{\infty}a_i.$$
• Let $\sum_{n=m}^{\infty}a_n$ be a series of real numbers, and $k\geq 0$ bean integer. If one of the two series $\sum_{n=m}^{\infty}a_n$ and $\sum_{n=m+k}^{\infty}a_n$ are convergent, then the other one is also, and we have the identity $$\sum_{n=m}^{\infty}a_n=\sum_{n=m}^{m+k-1}a_n+\sum_{n=m+k}^{\infty}a_n$$
• Let $\sum_{n=m}^{\infty}a_n$ be a series of real numbers converging to $x$, and let $k$ be an integer. Then $\sum_{n=m+k}^{\infty}a_{n-k}$ also converges to $x$.

Lemma 6.3 (Telescoping series) Let $(a_n)_{n=0}^{\infty}$ be a sequence if real numbers which converge to $0$. Then the series $\sum_{n=0}^{\infty}(a_n-a_{n+1})$ converges to $a_0$.

6.3 Sums of non-negative numbers

Proposition 6.7 Let $\sum_{n=m}^{\infty}a_n$ be a formal series of non-negative real numbers. Then this series is convergent iff there is a real number $M$ such that

$$\sum_{n=m}^{N}a_n\leq M \text{ for all integers } N\geq m.$$

:::{corollary name=“Comparision test”} Let $\sum_{n=m}^{\infty}a_n$ and $\sum_{n=m}^{\infty}b_n$ be two formal series of real numbers, and suppose that $|a_n|\leq b_n$ for all $n\geq m$. Then if $\sum_{n=m}^{\infty}b_n$ is convergent, then $\sum_{n=m}^{\infty}a_n$ is absolutely convergent, and in fact

$$\left|\sum_{n=m}^{\infty}a_n\right|\leq \sum_{n=m}^{\infty}|a_n|\leq\sum_{n=m}^{\infty}b_n.$$ :::

:::{.lemma name=“Geometric series} Let $x$ be a real number. If $|x|\geq 1$, the the series $\sum_{n=0}^{\infty}x^n$ is divergent. If however $|x|<1$, then the series is absolutely convergent and $$\sum_{n=0}^{\infty}x^n = 1/(1-x).$$ :::

Let $(a_n)_{n=1}^{\infty}$ be a decreasing sequence of non-negative real numbers (so $a_n\geq 0$ and $a_{n+1}\leq a_n$ for ll $n\geq 1$). Then the series $\sum_{n=1}^{\infty}a_n$ is convergent iff the series $$\sum_{k=0}^{\infty}2^ka_{2^k}=a_1+2a_2+4a_4+\dots$$ is convergent.

Let $q>0$ be a rational number. Then the series $\sum_{n=1}^{\infty}1/n^q$ is convergent when $q>1$ and divergent when $q\leq 1$.

Then quantity $\sum_{n=1}^{\infty}1/n^q$, when it converges, is called $\zeta(q)$, then Riemann-zeta function of $q$.

6.4 Rearrangement of series

Proposition 6.8 Let $\sum_{n=0}^{\infty}a_n$ be a convergent series of non-negative real numbers, and let $f:\mathbb{N}\to\mathbb{N}$ be a bijection. Then $\sum_{m=0}^{\infty}a_{f(m)}$ is also convergent, and has the same sum: $$\sum_{n=0}^{\infty}a_n=\sum_{m=0}^{\infty}a_{f(m)}.$$

Proposition 6.9 (Rearrangement of series) Let $\sum_{n=0}^{\infty}a_n$ be an absolutely convergent series o real numbers, and let $f:\mathbb{N}\to\mathbb{N}$ be a bijection. Then $\sum_{m=0}^{\infty}a_{f(m)}$ is also absolutely convergent, and has the same sum: $$\sum_{n=0}^{\infty}a_n=\sum_{m=0}^{\infty}a_{f(m)}.$$

A series which is conditionally convergent but bot absolutely convergent can in fact be rearranged to converge to any value.

6.5 The root and ratio tests

Theorem 6.1 (Root test) Let $\sum_{n=m}^{\infty}a_n$ be a series of real numbers, and let $\alpha:=\limsup_{n\to \infty}|a_n|^{1/n}$.

• If $\alpha<1$, then the series $\sum_{n=0}^{\infty}a_n$ is absolutely convergent (and hence conditionally convergent).
• If $\alpha>1$, then the series $\sum_{n=0}^{\infty}a_n$ is not conditionally convergent (and hence absolutely convergent).
• If $\alpha=1$, we cannot assert any conclusion.

Lemma 6.4 Let $(c_n)_{n=m}^{\infty}$ be a sequence of positive numbers. Then we have $$\liminf_{n\to\infty}\frac{c_{n+1}}{c_n} \leq \liminf_{n\to\infty}c_n^{1/n} \leq \limsup_{n\to\infty}c_n^{1/n} \leq \limsup_{n\to\infty}\frac{c_{n+1}}{c_n}.$$

Corollary 6.3 (Ratio test) Let $\sum_{n=m}^{\infty}a_n$ be a series of non-zero real numbers, and let $\alpha:=\limsup_{n\to \infty}\frac{|a_{n+1}|}{|a_n|}$.

• If $\alpha<1$, then the series $\sum_{n=0}^{\infty}a_n$ is absolutely convergent (and hence conditionally convergent).
• If $\alpha>1$, then the series $\sum_{n=0}^{\infty}a_n$ is not conditionally convergent (and hence absolutely convergent).
• If $\alpha=1$, we cannot assert any conclusion.

We have $\lim_{n\to \infty}n^{1/n}=1$.