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 \[ \begin{aligned} \sum_{x\in X}\left(\sum_{y\in Y}f(x,y)\right) &= \sum_{(x,y)\in X\times Y}f(x,y)\\ &= \sum_{(y,x)\in Y\times X}f(x,y)\\ &=\sum_{y\in Y}\left(\sum_{x\in X}f(x,y)\right) \end{aligned} \]

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 \(\lim_{n\to \infty}n^{1/n}=1\).