Basic Ideas

Definition 6.1.1  
A series is the addition of the terms of an infinite sequence $(a_k)_k$, denoted by

$$S = \sum_{k=0}^{\infty} a_k$$ (6.1.1)

Often, $S_n$ will be denoted by $\sum a_k$. This sum can either be finite or infinite. If the sum doesn't exist or is infinite the series is called divergent. If the sum exists and is a real number, then it is called convergent.
The partial sum of the series is defined as

$$S_n = \sum_{k=0}^na_k$$ (6.1.2)

It is clear that $$S = \lim_{n \to \infty}S_n$$.

Remark 6.1.2 (Arithmetic Series)  
The simplest class of series is the arithmetic series. A series is called arithmetic iff its sequence is of the form

$$a_k = a_0 + k d$$ (6.1.3)

In other words, the difference of two consecutive elements of the sequence is a constant. This immediately leads to the fact that a series is an arithmetic series iff

$$a_k : = \frac{a_{k-1}+a_{k+1}}{2}$$ (6.1.4)

The partial sum of an arithmetic series is

$$\begin{aligned}S_n &= \sum_{k=0}^na_k \\ &= \sum_{k=0}^na_0 + k... ...d \sum_{k=0}^nk \\ &= (n+1)a_0 + d \frac{n(n+1)}{2} \end{aligned}$$

It is easy to come up with a couple of examples of arithmetic series:

1. $(a_k)_k = 1,2,3,4,5,6,7,8,9,\ldots$
2. $(a_k)_k = 1,4,7,10,13,16,19,\ldots$
3. $(a_k)_k = 1,-1,-3,-5,-7,-9,-11,\ldots$

The partial sum is easy to obtain by plugging in the correct $a_0$ and $d$ into (6.1.5). Note that any sum of an arithmetic series is infinite. Therefore all arithmetic series are divergent.

Remark 6.1.3 (Geometric Series)  
A geometric series has a sequence of the form

$$a_k = a_0 r^k$$ (6.1.6)

where $r$ is called the ratio of the geometric series. So every element in the sequence is the previous element in the sequence times a certain number. We can also remark that a series is a geometric series iff

$$a_k = \sqrt{a_{k-1} \cdot a_{k+1}}$$ (6.1.7)

The partial sum for a geometric series is a bit more difficult to calculate:

$$\begin{aligned}S_n &= \sum_{k=0}^na_0 r^k \\ r S_n &= \sum_{k=0... ...0 (r^{n+1} - 1) \\ S_n &= a_0 \frac{r^{n+1}-1}{r-1} \end{aligned}$$

The sum is then simply the limit of $S_n$. But this limit doesn't always exist. If $\vert r\vert<1$, we have

$$S = \lim_{n \to \infty}S_n = \lim_{n \to \infty}\frac{r^{n+1}-1}{r-1} = \frac{a_0}{1-r}$$ (6.1.9)

otherwise, the limit diverges to infinity. So a geometric series is convergent iff the ratio is strictly between $-1$ and $1$ and then the sum converges to $\frac{a_0}{1-r}$. A few examples of geometric series are

1. $(a_k)_k = 1,3,9,27,81,243,729,\ldots$
2. $(a_k)_k = 1,-1,1,-1,1,-1,1,-1,\ldots$
3. $(a_k)_k = 1,\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},\frac{1}{32},\ldots$

The first two clearly diverge, because the ratio is outside $]-1,1[$, respectively ($3$ and $-1$), while the sum of the third series converges to $2$.

Remark 6.1.4  
Usually, when looking at a series we will concentrate on dealing with the sum, i.e. establishing if the partial sums converge or not, and not so much with what the actual value of each partial sum is. Nevertheless, here we present a few partial sums:

Table of a Few Partial Sums

Series	Partial Sum
$1+2+3+4+5+\ldots + n$	$$\frac{n(n+1)}{2}$$
$1+3+5+7+9+\ldots + (2n-1)$	$n^2$
$2+4+6+8+10+\ldots + 2n$	$n(n+1)$
$1^2+2^2+3^2+4^2+\ldots + n^2$	$$\frac{n(n+1)(2n+1)}{6}$$
$1^3+2^3+3^3+4^3+\ldots + n^3$	$$\frac{n^2(n+1)^2}{4}$$
$1^2+3^2+5^2+7^2+9^2+\ldots + (2n-1)^2$	$$\frac{n(4n^2-1)}{3}$$
$1^3+3^3+5^3+7^3+9^3+\ldots + (2n-1)^3$	$n^2(2n^2-1)$

Note that for all these series, the partial sum diverges.

Properties 6.1.5  

1. If the series $(S_n)_n$ is convergent, then $(a_k)_k$ converges to 0.
2. If $(a_k)_k$ diverges or converges to any other number than 0, $(S_n)_n$ is divergent.
3. If $\sum a_n$ is convergent, then $$\sum_{k=0}^{\infty}c a_n = c \sum_{k=0}^{\infty}a_n$$
4. If $\sum a_n$ and $\sum b_n$ are convergent, then $$\sum_{k=0}^{\infty}a_n + \sum_{k=0}^{\infty}b_n = \sum_{k=0}^{\infty}(a_n+b_n)$$
5. If $\sum a_n$ and $\sum b_n$ are convergent, then $$\sum_{k=0}^{\infty}a_n - \sum_{k=0}^{\infty}b_n = \sum_{k=0}^{\infty}(a_n-b_n)$$
6. If $a_n \leq b_n \leq c_n$ and $\sum a_n$ and $\sum c_n$ are converging to the same number, then so is $\sum b_n$.

Proof.

1. Let $S_n = a_0 + \ldots + a_n$. Then $a_n = S_n - S_{n-1}$. Taking limits on the left and the right hand side then gives

   $$\lim_{n \to \infty}a_n = \lim_{n \to \infty}S_n - S_{n-1} = \lim_{n \to \infty}S_n - \lim_{n \to \infty}S_{n-1} = 0$$ (6.1.10)

   
   

2. This is just the converse of the previous statement.
3. The scalar multiplication, sum and difference of two convergent series is convergent because of the corresponding equality for limits.
4. The last statement is essentially the squeeze lemma (2.0.17) for series.

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