Sums and sigma notation

Integral calculus is about summing lot's of numbers. So let's talk about sums, and a convenient notation for expressing sums. We want to sum over many numbers, so let's start with lists of numbers.

Definition 1

A list of nn numbers in a given order a1,a2,...,ana_1, a_2, ..., a_n

is called a sequence, and the numbers are often called terms. The small number attached to each letter aa is called the index of aa and tells you the position of the number in the sequence.

These numbers can be totally arbitrary, such as

−2,100.1,−23.243,2/3,10000.01-2,100.1, -23.243, 2/3, 10000.01

(here it is n=5n=5). However, often there is some pattern present:

Example 1

Consider "the first seven even numbers": 2,4,6,8,10,12,142,4,6,8,10,12,14. In this case we could also write

ak=2k, where k=1,2,...,7a_k = 2k, \text{ where } k=1,2,...,7

So using this formula we have

a1=2⋅1=2a2=2⋅2=4a3=2⋅3=6a4=2⋅4=8a5=2⋅5=10a6=2⋅6=12a7=2⋅7=14\begin{array}{lll} a_1 & = & 2\cdot 1 = 2\\ a_2 & = & 2\cdot 2 = 4\\ a_3 & = & 2\cdot 3 = 6\\ a_4 & = & 2\cdot 4 = 8\\ a_5 & = & 2\cdot 5 = 10\\ a_6 & = & 2\cdot 6 = 12\\ a_7 & = & 2\cdot 7 = 14\\ \end{array}

Now let's introduce a notation for adding the terms of a sequence.

Definition 2

Given is a sequence of nn terms a1,a2,...,ana_1, a_2, ..., a_{n}. We use the [sigma notation} notation to indicate the sum of those nn numbers:

Equation 1
∑k=1nak=a1+a2+...+an\sum_{k=1}^{n} a_k = a_1+a_2+... +a_{n}

Sigman notation for adding numbers

Note that the sigma notation does not help to find the sum of the values. It is pure notation, helping to write an expression of the form a1+a2+...+ana_1+a_2+... +a_{n} a bit shorter.

Note 1

The "k=1k=1" beneath and the "nn" above the sigma indicate that start term and the end term of the sequence over which we form the sum. We can change this. For example, if we want to form the sum starting with the second term and ending with the 55-th term, that is a2+a3+a4+a5a_2+a_3+a_4+a_{5}, we can write this using the sigma notation as follows:

∑k=25ak\sum_{k=2}^{5} a_k
Example 2

Consider the sequence

a1=2a2=4a3=5a4=−3a5=1a6=0\begin{array}{lll} a_1 &=& 2\\ a_2 &=& 4\\ a_3 &=& 5\\ a_4 &=& -3\\ a_5 &=& 1\\ a_6 &=& 0\\ \end{array}

Then we have

∑k=16ak=a1+...+a6=2+4+5−3+1+0=9∑k=36ak=a3+...+a6=5−3+1+0=3∑k=23ak=a2+a3=4+5=9∑k=22ak=a2=4\begin{array}{lll} \sum_{k=1}^6 a_k &=& a_1+...+a_6\\ &=& 2+4+5 -3 + 1 + 0 = 9\\ \sum_{k=3}^6 a_k &=& a_3+...+a_6\\ &=& 5 -3 + 1 + 0 = 3\\ \sum_{k=2}^3 a_k &=& a_2+a_3\\ &=& 4+5 = 9\\ \sum_{k=2}^2 a_k &=& a_2 = 4\\ \end{array}
Example 3

Consider the sequence of numbers x1=12,x2=22,...,x5=52x_1=1^2,x_2=2^2,...,x_{5}=5^2, and the function f(x)=2x−1f(x)=2x-1. Thus

∑k=15f(xk)=∑k=15f(k2)=∑k=15(2k2−1)=1+7+17+31+49\sum_{k=1}^{5} f(x_k) = \sum_{k=1}^{5} f(k^2) = \sum_{k=1}^{5} (2k^2-1)= 1+7+17+31+49
Exercise 1

  1. Write the terms of the sequence give by the following rule:

    1. ak=k2a_k=k^2\quad (where k=1,2,...,10k=1,2,...,10)
    2. bn=n+1b_n = n+1\quad (where n=4,5,...,10n=4,5,...,10)
    3. cl=1/(1−l)c_l = 1/(1-l)\quad (where l=10,11,...,15l=10,11,...,15)
    4. dk=kd_k=k\quad (where k=0,...,5k=0,...,5)
    5. ek=2k2+k+1e_k = 2k^2+k+1\quad (where k=0,1,2k=0,1,2)

  2. Determine the following sums:

    1. ∑k=03(10k+1)\sum_{k=0}^3 (10k+1)
    2. ∑u=47u\sum_{u=4}^7 u
    3. ∑s=24(s2+1)\sum_{s=2}^4 (s^2+1)
    4. ∑k=05(−1)k\sum_{k=0}^5 (-1)^k

  3. Write with the help of the sigma notation:

    1. 1+2+3+4+5+61+2+3+4+5+6
    2. 4+9+16+254+9+16+25
    3. 1+12+13+141+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}

  4. Consider the sequence xk=k⋅14x_k=k\cdot \frac{1}{4} for k=0,1,...,4k=0,1,...,4, and the function f(x)=x2f(x)=x^2. Determine the sum

    ∑k=13f(xk)\sum_{k=1}^3 f(x_k)

  5. Sum of first nn natural numbers. Prove, that

    ∑k=1nk=1+2+3+...+n=n(n+1)2\sum_{k=1}^{n} k = 1+2+3+...+n = \frac{n(n+1)}{2}

    Hint:

  6. Sum of first nn square numbers. Prove, that

    ∑k=1nk2=12+22+32+...+n2=n(n+1)(2n+1)6\sum_{k=1}^{n} k^2 = 1^2+2^2+3^2+...+n^2 = \frac{n(n+1)(2n+1)}{6}
Solution
    1. 1,4,9,16,25,36,49,64,81,1001,4,9,16,25,36,49,64,81,100 (first ten squares)

  2. 5,6,7,8,9,10,115,6,7,8,9,10,11

  3. 1−9,1−10,1−11,1−12,1−13,1−14\frac{1}{-9},\frac{1}{-10},\frac{1}{-11},\frac{1}{-12},\frac{1}{-13},\frac{1}{-14}

  4. 0,1,2,3,4,50,1,2,3,4,5

  5. 1,4,111,4,11

    1. ∑k=03(10k+1)=1+11+21+31=64‾\sum_{k=0}^3 (10k+1)=1+11+21+31=\underline{64}

  8. ∑u=47u=4+5+6+7=22‾\sum_{u=4}^7 u = 4+5+6+7=\underline{22}

  9. ∑s=24(s2+1)=5+10+17=32‾\sum_{s=2}^4 (s^2+1)=5+10+17=\underline{32}

  10. ∑k=05(−1)k=1−1+1−1+1−1=0‾\sum_{k=0}^5 (-1)^k=1-1+1-1+1-1=\underline{0}

    1. 1+2+3+4+5+6=∑k=16k1+2+3+4+5+6=\sum_{k=1}^6 k

  12. 4+9+16+25=∑k=25k24+9+16+25=\sum_{k=2}^5 k^2

  13. 1+12+13+14=∑k=141k1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\sum_{k=1}^4 \frac{1}{k}

∑k=13f(xk)=f(14)+f(24)+f(34)=(14)2+(12)2+(34)2=78‾\begin{array}{lll} \sum_{k=1}^3 f(x_k) &= & f(\frac{1}{4})+f(\frac{2}{4})+f(\frac{3}{4})\\ & = & \left( \frac{1}{4} \right)^2 + \left( \frac{1}{2} \right)^2 + \left( \frac{3}{4} \right)^2\\ &=&\underline{\frac{7}{8}} \end{array}
  1. See hint. The number of squarea areas is
1+2+3+4+51+2+3+4+5

and note that

n22+n2=n2+n2=n(n+1)2\begin{array}{lll}\frac{n^2}{2}+\frac{n}{2} &= &\frac{n^2+n}{2}\\ &=&\frac{n(n+1)}{2}\\ \end{array}
  1. See hint. The number of cubes depicted is
12+22+321^2+2^2+3^2

and note that

13n(n+1)(n+12)=n(n+1)3⋅(n+12)⋅22=n(n+1)3⋅2(n+12)2=n(n+1)3⋅2n+12=n(n+1)(2n+1)6\begin{array}{lll} \frac{1}{3}n(n+1)(n+\frac{1}{2}) &= &\frac{n(n+1)}{3}\cdot (n+\frac{1}{2})\cdot\frac{2}{2}\\ &= &\frac{n(n+1)}{3}\cdot \frac{2(n+\frac{1}{2})}{2}\\ &=& \frac{n(n+1)}{3}\cdot \frac{2n+1}{2}\\ &=& \frac{n(n+1)(2n+1)}{6}\\ \end{array}