Sum (Sigma) Notation

Hello. This post will be on sigma notation. Instead of having 1 + 2 + 3 + 4 + \dots + 100, you can use sigma notation such as \displaystyle \sum_{i = 1}^{100} i to save space, ink and paper.

When it comes to Sigma notation, one must have a keen eye in identifying patterns in numbers.


Table Of Contents


Finite Sums

To illustrate finite sums/series, an example will be shown.

    \[1 + 2 + 3 + 4 + 5 = \displaystyle \sum_{i = 1}^{5} i = 25 \text{ is an example of a finite sum/series. }\]

In a finite sum/series there are a finite number of terms.

In the sigma (sum) notation \displaystyle \sum_{i = 1}^{5} i we start from substituting i = 1 to get 1 as the first term. The next case is when i = 2 to get 2, then i = 3 to get 3 all the way to the upper limit of i = 5.

Another version of a finite sum is where the upper limit is an arbitrary positive integer. As an example we can have:

    \[1 + 2 + 3 + \dots + (n-1) + n= \displaystyle \sum_{i = 1}^{n} i\]

A more generalized version of a finite sum is where we add unknown integers from m to n (where m < n) such as:

    \[m + (m + 1) + (m + 2) + \dots + (n-1) + n= \displaystyle \sum_{i = m}^{n} i\]

Sometimes we deal with the sum of variables with increasing numeric subscripts. Consider this example:

    \[x_{1} + x_{2} + x_{3} + \dots + x_{n-1} + x_{n}= \displaystyle \sum_{i = 1}^{n}x_{i}\]


Infinite Sums

In an infinite sum/series, there is an infinite (never-ending) number of terms. As an example we can have:

2 + 4 + 6 + 8 + 10 + \dots = \displaystyle \sum_{i = 1}^{\infty} 2i = 2 \displaystyle \sum_{i = 1}^{\infty} i

The sum above computes to positive infinity and can be considered an infinite sum.


Some Properties

There are a lot of properties when it comes to sigma/summation notation. Here are a couple.

    \[1) \displaystyle \sum_{i = j}^{j} x_{i} = x_{j} \text{ (One term in the sum where i = j) }\]

    \[2) \displaystyle \sum_{i = m}^{n} x_{i} \text{ has (n - m + 1) terms in its sum}.\]

    \[3) \displaystyle \sum_{i = 1}^{n} kx_{i} = k \displaystyle \sum_{i = 1}^{n} x_{i} \text{ where k is a numeric constant. }\]

    \[4) \text{ Assuming k is a numeric constant we have: } \displaystyle \sum_{i = m}^{n} k = k \displaystyle \sum_{i = m}^{n} 1 = (n - m + 1)k\]

    \[5) \text{ (Splitting One Sum Into Two) } \displaystyle \sum_{i = m}^{n} x_{i} = \displaystyle \sum_{i = m}^{p} x_{i} + \displaystyle \sum_{i = p + 1}^{n} x_{i} \text{ for } m < p < n\]

    \[6) \displaystyle \sum_{i = m}^{n} (x_{i} + y_{i}) = \displaystyle \sum_{i = m}^{n} x_{i} + \displaystyle \sum_{i = m}^{n} y_{i}\]


Sum Formulas

This section will briefly look at sum formulas. A more extensive look at sum formulas can be found in a university level Calculus course which focuses on Sequences and Series.

Here are a few sum formulas.

    \[\displaystyle \sum_{i = 1}^{n} 1 = n\]

    \[\text{ Assuming k is a constant: }\displaystyle \sum_{i = 1}^{n} k = \displaystyle k \sum_{i = 1}^{n} 1 = nk\]

    \[\displaystyle \sum_{i = 1}^{n} i = \dfrac{n(n + 1)}{2}\]

    \[\displaystyle \sum_{i = 1}^{n} i^{2} = \dfrac{n(n + 1)(2n + 1)}{6}\]

    \[\displaystyle \sum_{i = 1}^{n} i^{3} = \Bigg(\dfrac{n(n + 1)}{2}\Bigg)^{2}\]


Examples

Example One

The sum 1 + 2 + 3 + 4 + \dots + 100 can be represented by \displaystyle \sum_{i = 1}^{100} i. Using the third formula from the previous section, this sum is equal to \dfrac{100(100 + 1)}{2} = \dfrac{(100)(101)}{2} = \dfrac{10100}{2} = 5050


Example Two

If we add even numbers starting from 2 up to 50 we have:

    \[2 + 4 + 6 + 8 + 10 + \dots + 48 + 50\]

.

Factoring out a 2 from each term we have:

    \[2(1 + 2 + 3 + 4 + 5 + \dots + 23 + 24 + 25)\]

In sigma notation, the above can be represented as:

    \[2 \displaystyle \sum_{i = 1}^{25} i\]

This sum evaluates to:

    \[2 \times \dfrac{25(26)}{2} = 25(26) = 650\]


Example Three (Fractions)

There are cases where we add fractions. Consider this example.

    \[1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} +\dots + \dfrac{1}{9} + \dfrac{1}{10}\]

From term to term, the denominator increases by one. The sum above can be represented as:

    \[\displaystyle \sum_{i = 1}^{10} \dfrac{1}{i}\]


Example Four (Alternating Sums)

This example is a more involved example (more for university calculus students).

When we think of sums adding is usually involved. However we can “add” negative numbers. There are cases where the sign is alternating. One such example is:

    \[1 - 2 + 3 - 4 + 5 - 6 + \dots + 19 - 20 = \displaystyle \sum_{i = 1}^{20} (-i)^{(i + 1)}\]

When we add positive numbers from 1 to 20 the (-i)^{(i + 1)} was needed. Since we have alternating sums we need the (-i)^{(i + 1)} part in the sum. Remember that (-1)^{\text{even exponent}} = 1. Since every second term is negative, we have (i + 1) in the exponent of (-i).

One can notice that we have two sums within this alternating signs sum. We have odd numbers and even numbers. Given a positive whole number k an odd number is the form of (2k + 1) (or 2k - 1) and an even number is of the form of (2k). Using these ideas in mind the sum can be separated into two.

The sum of odd numbers (1 + 3 + 5 + \dots + 17 + 19) can be represented as:

    \[\displaystyle \sum_{i = 1}^{10} (2i - 1)\]

The sum of (negative) even numbers (-2 -4 -6 -8 -10 - \dots - 18 - 20) can be represented as:

    \[\displaystyle \sum_{i = 1}^{10} (-2i) = -2 \displaystyle \sum_{i = 1}^{10} i\]

The original sum can be split into two sums. One sum has the positive odd numbers from 1 to 20 and the other sum has negative numbers from 1 to 20.

    \[\displaystyle \sum_{i = 1}^{20} (-i)^{(n + 1)} = \displaystyle \sum_{i = 1}^{10} (2i - 1) - 2 \displaystyle \sum_{i = 1}^{10} i\]

Using summation properties the above can be simplified and evaluated.

    \[\displaystyle \sum_{i = 1}^{10} (2i - 1) - 2 \displaystyle \sum_{i = 1}^{10} i = \displaystyle \sum_{i = 1}^{10} (2i - 1) + \displaystyle \sum_{i = 1}^{10} (-2i) = \displaystyle \sum_{i = 1}^{10} (2i -1 ) - 2i = \displaystyle \sum_{i = 1}^{10} (-1) = 10(-1) = -10\]

The sum 1 - 2 + 3 - 4 + 5 - 6 + \dots + 19 - 20 computes to -10. This makes sense since the first pair of 1 – 2 is -1, the next pair of 3 – 4 is -1. This happens eight more times so we add (-1) ten times to get -10.


Practice Problems

Here are some practice problems. The summation properties and formulas will come in handy. Hints and solutions are provided in the next sections.

1) Find the sum of the numbers from 1 to 15 (including 1 and 15).

2) Find the sum of the numbers from 1 to 500 (including 1 and 500).

3) Evaluate the sum 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 + 121 + 144 using summation notation, properties and/or formulas.

4) Find the sum of 1 + 8 + 27 + 64 + 125 + 216 + 343 + 512 using summation notation, properties and/or formulas.

5) Evaluate -2 + 4 - 6 + 8 - 10 + \dots - 22 + 24 using summation notation and properties.

6) Is \displaystyle \sum_{i = 1}^{10} (i + 3) greater than (-10 + 20 - 30 + 40 - 50 + 60 - 70 + 80)?

7) Evaluate \displaystyle \sum_{i = 1}^{16} (i^2 + 3i - 7).

8) Evaluate \displaystyle \sum_{i = 1}^{12} (2i^3 + 4i^2 - 2i + 1).


Hints To Practice Problems

1) Use the formula \displaystyle \sum_{i = 1}^{n} i = \dfrac{n(n + 1)}{2}.

2) Use the formula \displaystyle \sum_{i = 1}^{n} i = \dfrac{n(n + 1)}{2}.

3) Notice how each term in the sum is a squared number.

4) Each term in the sum is a cubed number.

5) This is an alternating signs case.

6) The first sum can be simplified and evaluated. The second sum is similar to the alternating signs example.

7) Use properties and formulas to evaluate the sum.

8) Use properties and formulas to evaluate the sum.


Solutions To Practice Problems

1) 120

2) 125250

3) 650

4) 1296

5)

    \[-2 + 4 - 6 + 8 - 10 + \dots - 22 + 24 = 2(-1 + 2 - 3 + \dots -11 + 12) = 2[ (-1 -3 - 5 -7 -9 -11) + (2 + 4 + 6 + 8 + 10 + 12)]\]

The original sum can be split into two sums. One sum has negative odd numbers and the other sum has positive even numbers. Here are the calculations.

    \[2 \displaystyle \sum_{i = 1}^{6} -(2i - 1) + 2 \displaystyle \sum_{i = 1}^{6} (2i)\]

    \[-2 \displaystyle \sum_{i = 1}^{6} (2i - 1) + 4 \displaystyle \sum_{i = 1}^{6} i\]

    \[-4 \displaystyle \sum_{i = 1}^{6} i + \displaystyle \sum_{i = 1}^{6} 2 + 4 \displaystyle \sum_{i = 1}^{6} i\]

    \[\displaystyle \sum_{i = 1}^{6} 2 = 6(2) = 12\]

6) The first sum \displaystyle \sum_{i = 1}^{10} (i + 3) computes to:

    \[\displaystyle \sum_{i = 1}^{10} (i + 3) = \displaystyle \sum_{i = 1}^{10} i + \displaystyle \sum_{i = 1}^{10} 3 = \dfrac{(10)(11)}{2} + 3(10) = 55 + 30 = 85\]

The second sum (-10 + 20 - 30 + 40 - 50 + 60 - 70 + 80) = 10(-1 + 2 -3 + 4 - 5 + 6 - 7 + 8) can be represented as:

    \[-10 \displaystyle \sum_{i = 1}^{4} (2i - 1) + 10 \displaystyle \sum_{i = 1}^{4} (2i)\]

The above can be simplified and evaluated as follows.

    \[-10 \displaystyle \sum_{i = 1}^{4} (2i) + \displaystyle \sum_{i = 1}^{4} 10 + 10 \displaystyle \sum_{i = 1}^{4} (2i) = \displaystyle \sum_{i = 1}^{4} 10 = 10(4) = 40\]

7)

    \[\displaystyle \sum_{i = 1}^{16} (i^2 + 3i - 7) = \displaystyle \sum_{i = 1}^{16} i^2 + 3 \displaystyle \sum_{i = 1}^{16} i - \displaystyle \sum_{i = 1}^{16} 7 = \dfrac{(16)(17)(33)}{6} + 3 \dfrac{(16)(17)}{2} - (16)(7) = 1792\]

8) This question is similar to number seven.

    \[\displaystyle \sum_{i = 1}^{12} (i^3 + 2i^2 - i - 5) = \Bigg(\dfrac{(12)(13)}{2}\Bigg)^{2} + 2\dfrac{(12)(13)(25)}{6} - \dfrac{(12)(13)}{2} - 60 = 7246\]


Notes

There are more topics related to sigma notation, series and other types of sums which are not covered here. (i.e Arithmetic series, geometric series, Taylor series, etc.)

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