Series

Consider the following sequence of numbers (odd numbers):

1,3,5,...1,3,5,...

What is the sum of the 33 first terms of the sequence? Of course 1+3+5=91+3+5=9. But what is the sum of the 1000010\,000 first terms of the sequence? This is more difficult to answer. Is there a formula that lets us determine this sum? This question leads to the so-called series.

Definition 1

Given a sequence aa:

(an)=(a1,a2,a3,...,an,...)(a_n)=(a_1,a_2, a_3, ..., a_n, ...)

Based on this sequence, a new sequence ss can be formed,

(sn)=(s1,s2,s3,...,sn,...)(s_n)=(s_1, s_2, s_3, ..., s_n, ...)

the so-called series of the sequence aa. The nn-th series element sns_n is defined as the sum of the nn first terms of the sequence of aa:

s1=a1s2=a1+a2s3=a1+a2+a3...sn=a1+a2+a3+...+an\begin{array}{lll} s_1 &=& a_1\\ s_2 &=& a_1+a_2\\ s_3 &=& a_1+a_2+a_3\\ ... & &\\ s_n &=& a_1+a_2+a_3+...+a_n\\ \end{array}
Example 1

Determine s3s_3 of the sequence (an)=1,1.2,1.4,1.5,...(a_n)=1,1.2, 1.4, 1.5, ....

Solution

s3=1+1.2+1.4=3.6s_3=1+1.2+1.4=3.6

Extra: Sigma notation

Before we go on, we want to introduce the sigma notation. It helps to avoid long expressions like

sn=a1+a2+a3+...+ans_n=a_1+a_2+a_3+...+a_n

a more compact notation is introduced:

Definition 2

Given a sequence aa:

(an)=(a1,a2,a3,...)(a_n)=(a_1,a_2, a_3, ...)

The sigma notation

Σk=uvak\Sigma_{k=u}^{v} a_k

means "add the terms aka_k, where kk goes from uu to vv, i.e. au+...+ava_u+...+a_v. The symbol Σ\Sigma (Greek S) stands for "sum".

Some examples:

Example 2
Σk=35ak=a3+a4+a5Σk=11ak=a1Σk=12ak=a1+a2\begin{array}{lll} \Sigma_{k=3}^5 a_k &=& a_3+a_4+a_5\\ \Sigma_{k=1}^1 a_k &=& a_1\\ \Sigma_{k=1}^2 a_k &=& a_1+a_2\\ \end{array}

If we have an explicit rule, it can be used directly in the sigma notation. For example, if the sequence aa is given by ak=k2a_k=k^2 (k=1,2,...), then we have

Σk=13k2=12+22+32=14\Sigma_{k=1}^3 k^2 = 1^2+2^2+3^2=14
Exercise 1

Write as a sum without the Σ\Sigma:

  1. Σk=24(2k1)\Sigma_{k=2}^4 (2k-1)

  2. Σi=131i\Sigma_{i=1}^3 \frac{1}{i}

  3. Σw=68(3w)\Sigma_{w=6}^8 (3-w)

Solution
  1. Σk=24(2k1)=3+5+7\Sigma_{k=2}^4 (2k-1)=3+5+7
  2. Σi=131i=1+12+13\Sigma_{i=1}^3 \frac{1}{i}=1+\frac{1}{2}+\frac{1}{3}
  3. Σw=68(3w)=3+(4)+(5)=345\Sigma_{w=6}^8 (3-w) = -3+(-4)+(-5)=-3-4-5

Back to the series. With the help of the sigma notation, we can thus write for the nn-th term sns_n:

sn=Σk=1nak=a1+...+ans_n=\Sigma_{k=1}^n a_k=a_1+...+a_n
Example 3
s4=Σk=14ak=a1+a2+a3+a4s_4=\Sigma_{k=1}^4 a_k=a_1+a_2+a_3+a_4