What Are Series in Math?

Series is a powerful tool in finance, among other fields. Banks use series to calculate loans, savings and investments, as well as the values of cash flows. By understanding the basic principles of series, you will gain more insight into the inner workings of finance.

Theory

Series

A series is a sequence of numbers where you exchange the comma with a plus or a minus. Series generally look like this:

a_{1} + a_{2} + a_{3} + \dots + a_{n} + \dots

Here, $n$ is the number of the term and $a_{n}$ is the actual number in that term.

When you calculate the sum of a very long series, it can get pretty tiresome to write out the entire series. For that reason, mathematicians have managed to find a way to do this much easier by introducing the Greek letter sigma: $\sum$ . You can write the sum of a series in the following way:

Theory

The Summation Symbol $\sum$

The sum of the first $n$ terms of a series of numbers can be written in the following way:

S_{n} = \sum_{i = 1}^{n} a_{i} = a_{1} + a_{2} + \dots + a_{n}

Here, $S_{n}$ is the sum of $n$ terms. $i = 1$ tells you that you will begin counting from term number 1, $n$ shows you which term to stop at, and $a_{i}$ is the formula that describes term number $i$ .

Series do not have to be finite. There are also infinite series. When working with infinite series, the most frequent question is: what happens to the sum of the series? Will the terms of the series become so small that in the end, no matter how many terms you add, the sum will still become a specific number, or will the terms of the series be so large that their sum becomes infinitely large? Mathematically, these two cases are called convergence and divergence respectively. In general, you have:

Theory

Convergence and Divergence

Convergence:: The sum of the series tends towards a specific number when $n \to \infty$ .
Divergence:: The sum of the series does not converge towards a specific number, often because it tends to $\pm \infty$ when $n \to \infty$ .

Example 1

You have a series where
$\begin{array}{l} a_{1} & = 3, a_{2} = 6, a_{3} = 9, \\ \dots, a_{n} = 3 n, \dots \end{array}$

$a_{1} = 3, a_{2} = 6, a_{3} = 9, \dots, a_{n} = 3 n, \dots$

Find the sum of the first ten factors. Check what happens with the sum when $n \to \infty$ .

In order to find the sum of the first ten terms, you need to use the summation symbol $\sum$ with $n = 10$ and $a_{i} = 3 i$ . That will give you

\begin{array}{l} S_{10} & = \sum_{i = 1}^{10} 3 i \\ = 3 \cdot 1 + 3 \cdot 2 + 3 \cdot 3 + 3 \cdot 4 \\ + 3 \cdot 5 + 3 \cdot 6 + 3 \cdot 7 \\ + 3 \cdot 8 + 3 \cdot 9 + 3 \cdot 10 \\ = 3 + 6 + 9 + 12 + 15 + 18 \\ + 21 + 24 + 27 + 30 \\ = 165 . \end{array}

\begin{array}{l} S_{10} & = \sum_{i = 1}^{10} 3 i \\ = 3 \cdot 1 + 3 \cdot 2 + 3 \cdot 3 + 3 \cdot 4 + 3 \cdot 5 \\ + 3 \cdot 6 + 3 \cdot 7 + 3 \cdot 8 + 3 \cdot 9 + 3 \cdot 10 \\ = 3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30 \\ = 165 . \end{array}

In order to check what happens to the sum when

n \to \infty

, it can be worth checking out the formula for the terms. In this case,

a_{n} = 3 n

. These terms are multiples of

3

, which are just going to get larger and larger, making the sum grow larger and larger as well. That means you can conclude that the sum will just increase by more and more in such a manner that it goes beyond all boundaries when

n \to \infty

. For that reason, the sum of this series diverges.

Example 2

You have a series where

$\begin{array}{l} a_{1} & = \frac{1}{2} a_{2} = \frac{1}{4} a_{3} = \frac{1}{8} \\ \dots a_{n} = \frac{1}{2^{n}} \dots \end{array}$

$a_{1} = \frac{1}{2}, a_{2} = \frac{1}{4}, a_{3} = \frac{1}{8}, \dots, a_{n} = \frac{1}{2^{n}}, \dots$

Find the sum of the first five terms. Check what happens to the sum when $n \to \infty$ .

In order to find the sum of the first five terms, you need to use the summation symbol $\sum$ with $n = 5$ and $a_{i} = \frac{1}{2^{i}}$ . That will give you

\begin{array}{l} S_{5} & = \sum_{i = 1}^{5} \frac{1}{2^{i}} \\ = \frac{1}{2^{1}} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + \frac{1}{2^{4}} + \frac{1}{2^{5}} \\ = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} \\ = \frac{1 \cdot 16}{2 \cdot 16} + \frac{1 \cdot 8}{4 \cdot 8} + \frac{1 \cdot 4}{8 \cdot 4} + \frac{1 \cdot 2}{16 \cdot 2} + \frac{1}{32} \\ = \frac{16 + 8 + 4 + 2 + 1}{32} \\ = \frac{31}{32} . \end{array}

In order to check what happens with the sum when $n \to \infty$ , it can be worth taking a look at the formula for the terms. In this case $a_{n} = \frac{1}{2^{n}}$ . These terms become smaller and smaller the further out in the series you get. The fact that the terms become smaller and smaller does not necessarily mean that the sum of the series converges, but it means that there is a chance that the sum of the series converges. You will see how to decide whether it converges or not by looking at the entry about geometric series. In this particular case, I can tell you that this series decreases fast enough for the sum of the series to converge towards a number, which happens to be exactly 1. Then you can conclude that the sum of the series converges when $n \to \infty$ . Mathematically, it will look like this:

S = \sum_{n = 1}^{\infty} \frac{1}{2^{i}} = 1 .

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Series

The Summation Symbol ∑ ⁡

Convergence and Divergence

The Summation Symbol $\sum$