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.

Sequences of numbers

A list of nn numbers in a given order,

a1,a2,...,an\boxed{a_1, a_2, ..., a_n}

is called a sequence, and the numbers are often called terms. 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, here we are primarily interested in sequences whose terms arise from a rule, such as "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}

The letter kk in the sequence ak=2ka_k=2k is called an index. As the index describes the position of the term in the sequence (first term, second term, ....), it makes sense to request that kk is a natural number 1,2,3...1,2,3.... However, in principle we could start the sequence with any other value, such as

a4,a5,a6,...a_4, a_5, a_6,...

or even

a−2,a−1,a0,a1,...a_{-2}, a_{-1}, a_0, a_1, ...

But what we definitely do not allow is an index which is not an integer, such as a∗1.5,a∗2.5,...a*{1.5}, a*{2.5}, ... .

Sometimes we have a sequence

x1,x2,...,xnx_1, x_2, ..., x_n

and use those as the input for some function ff to create another sequence

f(x1),f(x2),...,f(xn)\boxed{f(x_1), f(x_2), ..., f(x_n)}

For example, say the sequence is

1,1.5,2,2.5,31, 1.5, 2, 2.5, 3

and the function we apply is

f(x)=x2f(x)=x^2

Thus we get the new sequence

12,1.52,22,2.52,321^2, 1.5^2, 2^2, 2.5^2, 3^2 1,2.25,4,6.25,91, 2.25, 4, 6.25, 9

Let us exercise this a bit.

Exercise 1
Q1.1

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)
Q1.2

Find the formula that generates the following sequence:

  1. −3,−4,−5-3,-4,-5
  2. 1,3,5,7,91,3,5,7,9
  3. 1,12,13,14,15,16,171, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \frac{1}{7}
  4. 10,15,20,25,3010, 15, 20, 25, 30
  5. 2,0.2,0.02,0.002,0.0002,0.000022, 0.2, 0.02, 0.002, 0.0002, 0.00002
  6. 3,6,12,24,483,6,12,24,48
Q1.3

Determine the new sequence by applying function ff:

  1. 1,2,3,4,51,2,3,4,5 and f(x)=−xf(x)=-x

  2. 1,1.1,1.2,1.3,1.41, 1.1, 1.2, 1.3, 1.4 and f(x)=2x+1f(x)=2x+1

  3. xk=2k x_k=\frac{2}{k}\, (where k=1,2,...,5k=1,2,...,5)  \,and  f(x)=3x\,f(x)=\frac{3}{x}

Solution
A1.1
  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
A1.2

There are often several possibilities, here we give just one:

  1. ak=−ka_k=-k\quad (where k=3,4,5k=3,4,5)
  2. ak=2k+1a_k=2k+1\quad (where k=0,1,2,3k=0,1,2,3)
  3. ak=1ka_k=\frac{1}{k}\quad (where k=1,2,..,7k=1,2,..,7)
  4. ak=10+5(k−1)a_k=10+5(k-1)\quad (where k=1,2,...,5k=1,2,...,5)
  5. ak=2⋅10−ka_k=2\cdot 10^{-k}\quad (where k=0,1,..,5k=0,1,..,5)
  6. ak=3⋅2ka_k=3\cdot 2^k\quad (where k=0,1,...,4k=0,1,...,4)
A1.3
  1. −1,−2,−3,−4,−5-1,-2,-3,-4,-5
  2. 3,3.2,3.4,3.6,3.83, 3.2, 3.4, 3.6, 3.8
  3. x1=21=2,x2=22=1,x3=23,x4=24=0.5,x5=25=0.4x_1=\frac{2}{1}=2, x_2=\frac{2}{2}=1, x_3=\frac{2}{3}, x_4=\frac{2}{4}=0.5, x_5=\frac{2}{5}=0.4, f(x1)=32=1.5,f(x2)=31=3,f(x3)=32/3=4.5,f(x4)=30.5=6,f(x5)=30.4=7.5f(x_1)=\frac{3}{2}=1.5, f(x_2)=\frac{3}{1}=3, f(x_3)=\frac{3}{2/3}=4.5,f(x_4)=\frac{3}{0.5}=6, f(x_5)=\frac{3}{0.4}=7.5.

Summing numbers

Given a sequence of nn terms

a1,a2,...,ana_1, a_2, ..., a_{n}

we use the following symbol to indicate that we form the sum (also called a series) of these terms:

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

Sometimes we will also use a slightly smaller version:

Σk=1nak\Sigma_{k=1}^n a_k

This is called the sigma-notation. The symbol σ\sigma (say "sigma") is the capital letter SS in the greek language. 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 we write

∑k=25ak=a2+a3+a4+a5\sum_{k=2}^{5} a_k = a_2+a_3+a_4+a_{5}
Example 1

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}

If the sequence if given by a rule, we typically write this rule after the sigma. For example, consider the sequence of odd numbers,

ak=2k+1,where k=0,1,...a_k = 2k+1,\quad\text{where }k=0,1,...

then instead of writing

∑k=09ak=1+3+...+19\sum_{k=0}^9 a_k=1+3+...+19

we write

∑k=09(2k+1)=1+3+...+19\sum_{k=0}^9 (2k+1) = 1+3+...+19

Also, if we have a sequence

x1,x2,...,xnx_1, x_2, ..., x_n

and apply a function ff to it to get the sequence

f(x1),...,f(xn)f(x_1), ..., f(x_n)

we write

∑k=1nf(xk)=f(x1)+f(x2)+...+f(xn)\sum_{k=1}^n f(x_k)= f(x_1)+f(x_2)+...+f(x_n)

for summing the terms of the new sequence.

Again some exercises ...

Exercise 2
Q2.1

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
Q2.2

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}
Q2.3

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)
Q2.4 (famous sums, part 1)

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: uncollapse (picture from here)

Show
Q5 (famous sums, part 2)

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}

Hint: uncollapse (picture from here)

Show

And uncollapse to see the solutions.

Solution
A2.1
  1. ∑k=03(10k+1)=1+11+21+31=64‾\sum_{k=0}^3 (10k+1)=1+11+21+31=\underline{64}
  2. ∑u=47u=4+5+6+7=22‾\sum_{u=4}^7 u = 4+5+6+7=\underline{22}
  3. ∑s=24(s2+1)=5+10+17=32‾\sum_{s=2}^4 (s^2+1)=5+10+17=\underline{32}
  4. ∑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}
A2.2
  1. 1+2+3+4+5+6=∑k=16k1+2+3+4+5+6=\sum_{k=1}^6 k
  2. 4+9+16+25=∑k=25k24+9+16+25=\sum_{k=2}^5 k^2
  3. 1+12+13+14=∑k=141k1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\sum_{k=1}^4 \frac{1}{k}
A2.3
∑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}
A2.4

See hint. The number of squares depicted are

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}
A2.5

See hint. The number of cubes depicted are

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}