Sigma Notation

Sigma notation is used to hold all the terms of a series on one small space on a page.

Take for example the sequence 5, 9, 13, 17, 21, \dots. This sequence has general term t_n=4n+1. The notation:


is the instruction to add together the first five terms of the sequence 4n+1. It reads ‘sum the terms of the sequence 4n+1 starting at n=1 and ending at n=5.’

Writing out the sum in full we have

    \[\sum_{n=1}^{5}(4n+1)=5 + 9+ 13+17+21\]

We can also understand this notation as representing the number or expression which the sum is equal to. In this case, 5+9+13+17+21=65. Therefore,


Evaluate Sigma Notation

There are three ways to calculate the value of (evaluate) a sum represented with sigma notation.

  1. Write out all the terms and add them together.
  2. If you know a formula for the kind of sequence, use the formula.
  3. Use a scientific calculator, or an online sigma calculator.

Using formulae

Example 1 Arithmetic Series


In this case, the sequence 4n+1 is arithmetic, and so we can calculate the sum of the first 100 terms using the formula for arithmetic series: S_n=\dfrac{n}{2}\left(2a+(n-1)d\right) where a is the first term and d is the common difference.

The sequence t_n=4n+1 has first term t_1=4(1)+1=5 and common difference d=4.

    \[\sum_{n=1}^{100}(4n+1)=\dfrac{100}{2}\left(2\times 5 + 99 \times 4\right)=20300\]

It is clear that having a formula offers a significant advantage over writing out all the terms and adding them together.

Try typing example 1 into this sigma calculator.

Example 2: A partial sum


In this example, our sequence is still arithmetic, however we are beginning at the 7th term and finishing at the 20th term. That is a total of 14 terms. (Calculate: 20 - 7 +1 We ‘+1‘ because the 7th term is included).

Now that we have established n=14, let’s find the first term in this summation and the last term, so that we can use the other formula for summing an arithmetic sequence which is S_n=\dfrac{n}{2}\left(t_1+t_n\right)

First term: when n=7, 4n+1=4(7)+1=29.

Last term: when n=20, 4n+1=81.



Example 3 A geometric series

A geometric sequence has general term t_n=a(r)^{n-1}. We read here that the formula to sum the first n terms of a geometric sequence is S_n=\dfrac{a(r^n-1)}{r-1}


In this case, the number of terms is 10; a=3 and r=2.

When written out, the sequence reads


Using the formula, we have


Example 4 Square Numbers


In this course, we don’t derive the formula to add together the square numbers. It takes a little more time to derive this formula – here is one youtube that explains where the formula comes from.

Here it is:

    \[1 + 4+ 9 + \dots n^2 = \dfrac{n(n+1)(2n+1)}{6}\]



Example 5 Summing n terms

When the number of terms to add has not been specified, we generally use n to describe the number of terms to be added. This means that we need an additional variable to use to describe the sequence. Let’s add together n terms of the sequence in example 1:


This reads ‘sum the terms of the sequence 4k+1 starting with the first term and ending with the n^{\text{th}} term.

Writing the sequence out we have:

    \[\sum_{k=1}^{n}(4k+1)=5 + 9 + 13 + \dots +(4n+1)\]

This sequence is arithmetic, with n terms, t_1=5 and t_n=4n+1.


Before evaluating a sum presented with sigma notation, it is good to ask whether or not a formula would be necessary. If only a few terms are to be added, it might be more efficient to write out the terms and add them. If there are many terms and you are not permitted to use technology, then you then need to determine the kind of sequence in order to select the correct formula for adding the terms.

Simplifying Sigma Notation

Consider the sum


Written out in full we have


Rearranging the terms, taking all the first terms in each bracket followed by all the second terms, followed by all the third we have


We can identify three separate, simpler sums:

    \[\sum_{n=1}^{5}(n^2+4n+1)=\sum_{n=1}^5 n^2+\sum_{n=1}^5 4n + \sum_{n=1}^5 1\]

Taking out the common factor 4 in the second term, we can simplify once more:

    \begin{align*}\sum_{n=1}^{5}(n^2+4n+1)&=(1+4+9+16+25)+4(1+2+3+4+5)+(1+1+1+1+1)\\[10 pt] &=\sum_{n=1}^5 n^2+4\sum_{n=1}^5 n +\sum_{n=1}^5 1\end{align}

Finally, notice that the constant term is simply a multiplication:

    \begin{align*}\sum_{n=1}^{5}(n^2+4n+1)&=(1+4+9+16+25)+4(1+2+3+4+5)+(1+1+1+1+1)\\[10 pt] &=\sum_{n=1}^5 n^2+4\sum_{n=1}^5 n +5\times 1\end{align}

We can make the following generalizations:

1. Let t_n and u_n represent two sequences in terms of n. Then:

    \[\sum_{n=1}^{5}(t_n+u_n)=\sum_{n=1}^5 t_n+\sum_{n=1}^5 u_n\]

2. Let at_n be a sequence t_n multiplied by a constant a.

    \[\sum_{n=1}^{5}at_n=a\sum_{n=1}^5 t_n\]

3. Let c be a constant.

    \[\sum_{n=1}^{5}c=5\times c\]


Try the practice test offered by CK12.

Next: Sum to Infinity