Sigma notation: sums and standard formulae

Sigma notation writes a sum without listing every term. The capital Greek letter sigma, $\Sigma$, means “add”. For example,

$$\begin{aligned} \sum_{r=1}^{4}(2r+3) &=(2(1)+3)+(2(2)+3)+(2(3)+3)+(2(4)+3)\\ &=5+7+9+11\\ &=32. \end{aligned}$$

The expression below $\Sigma$ says where the index starts, the number above it says where the index stops, and the expression to the right gives each term.

Prerequisites

You should be able to:

• substitute, expand and simplify algebraic expressions;
• use arithmetic sequences and series;
• apply the laws of indices.

Reading a sigma sum

In

$\sum_{r=3}^{7}f(r),$

$r$ is the index of summation, $3$ is the lower limit, $7$ is the upper limit, and $f(r)$ is the summand. Substitute each integer from $3$ to $7$, including both endpoints:

$f(3)+f(4)+f(5)+f(6)+f(7).$

The number of terms is therefore

$\boxed{\text{upper limit}-\text{lower limit}+1}.$

Here there are $7-3+1=5$ terms. The extra $1$ accounts for including both endpoints.

The index is a temporary label. Provided it is changed everywhere inside that sum,

$\sum_{r=1}^{n}r^2=\sum_{k=1}^{n}k^2.$

It does not matter whether the letter is $r$, $k$, $i$ or another symbol.

Worked example 1: expand and evaluate

Evaluate

$\sum_{k=2}^{5}(k^2-1).$

There are $5-2+1=4$ terms. Substitute $k=2,3,4,5$:

$$\begin{aligned} \sum_{k=2}^{5}(k^2-1) &=(2^2-1)+(3^2-1)+(4^2-1)+(5^2-1)\\ &=3+8+15+24\\ &=50. \end{aligned}$$

Brackets prevent the common error of applying the summation only to $k^2$ and forgetting the $-1$ in every term.

Worked example 2: a non-unit lower limit

Write out, but do not evaluate,

$\sum_{j=-2}^{2}3^j.$

The index takes the five integer values $-2,-1,0,1,2$, so

$3^{-2}+3^{-1}+3^0+3^1+3^2.$

The limits need not begin at $1$. They may include $0$ or negative integers.

The upper limit is not usually the number of terms

$\sum_{r=4}^{10}u_r$ contains $10-4+1=7$ terms, not $10$ terms. It is $u_4+u_5+\cdots+u_{10}$.

Self-check 1

1. Expand and evaluate $\displaystyle\sum_{r=0}^{3}(2^r+1)$.
2. How many terms are in $\displaystyle\sum_{k=8}^{25}(3k-2)$?
3. Write the first and last terms of $\displaystyle\sum_{j=2}^{n}(j+4)$.

Answers

1. $(2^0+1)+(2^1+1)+(2^2+1)+(2^3+1)=2+3+5+9=19$.
2. $25-8+1=18$ terms.
3. The first term is $2+4=6$ and the last term is $n+4$.

Writing a series in sigma notation

To compress a listed sum:

1. identify a formula for its general term;
2. choose an index and match its first value to the first term;
3. choose the final index to produce the last term;
4. expand the first two terms and the last term as a check.

There can be several correct forms because the index is only a label.

Worked example 3: arithmetic terms

Write

$5+8+11+\cdots+50$

in sigma notation.

The terms have common difference $3$. Starting with $r=1$, their formula is

$3r+2,$

because $3(1)+2=5$. Find the final index from

$3r+2=50,$

which gives $r=16$. Hence

$\boxed{5+8+11+\cdots+50=\sum_{r=1}^{16}(3r+2)}.$

A form beginning at $r=0$ is equally valid:

$\sum_{r=0}^{15}(3r+5).$

Both forms have $16$ terms and generate the same list.

Worked example 4: geometric terms

Write

$6+18+54+\cdots+1458$

in sigma notation.

The first term is $6$ and the common ratio is $3$, so the $r$th term is

$6(3^{r-1}).$

Since $1458=6(3^5)$, the last term occurs when $r-1=5$, so $r=6$. Therefore

$\boxed{\sum_{r=1}^{6}6(3^{r-1})}.$

The exponent is $r-1$ because the first term contains $3^0$.

Worked example 5: signs and powers

Write

$1-4+9-16+25-36$

in sigma notation.

The magnitudes are $r^2$. The sign is positive for odd $r$ and negative for even $r$, which is produced by $(-1)^{r+1}$. Thus

$\boxed{\sum_{r=1}^{6}(-1)^{r+1}r^2}.$

Check $r=1$: $(-1)^2(1^2)=1$. Checking the first term is a quick way to detect a sign shift.

Rules for finite sums

Sigma notation obeys the usual laws of addition and multiplication. For constants $a$ and $b$,

$$\boxed{ \sum_{r=m}^{n}\bigl(af(r)+bg(r)\bigr) =a\sum_{r=m}^{n}f(r)+b\sum_{r=m}^{n}g(r) }.$$

In particular, a constant factor can be taken outside:

$\sum_{r=1}^{n}5r^2=5\sum_{r=1}^{n}r^2.$

However, a constant term is repeated once for every index value:

$$\boxed{\sum_{r=m}^{n}c=c(n-m+1)}.$$

For example,

$\sum_{r=3}^{8}7=7(8-3+1)=42.$

Worked example 6: split a sum correctly

Simplify

$\sum_{r=1}^{n}(4r^2-3r+5).$

Apply linearity term by term:

$$\begin{aligned} \sum_{r=1}^{n}(4r^2-3r+5) &=4\sum_{r=1}^{n}r^2-3\sum_{r=1}^{n}r+\sum_{r=1}^{n}5\\ &=4\sum_{r=1}^{n}r^2-3\sum_{r=1}^{n}r+5n. \end{aligned}$$

The final term is $5n$, not $5$. The number $5$ appears in every one of the $n$ summands.

Combining and separating ranges

Adjacent parts of the same sum may be joined:

$$\sum_{r=1}^{p}f(r)+\sum_{r=p+1}^{n}f(r) =\sum_{r=1}^{n}f(r).$$

Consequently,

$$\boxed{ \sum_{r=m}^{n}f(r) =\sum_{r=1}^{n}f(r)-\sum_{r=1}^{m-1}f(r) }.$$

Subtract up to $m-1$, not up to $m$, because the term $f(m)$ must remain.

Worked example 7: sum over a restricted range

Given

$\sum_{r=1}^{20}u_r=148 \qquad \text{and} \qquad \sum_{r=1}^{7}u_r=31,$

find $\displaystyle\sum_{r=8}^{20}u_r$.

The first seven terms must be removed from the first twenty:

$$\sum_{r=8}^{20}u_r =\sum_{r=1}^{20}u_r-\sum_{r=1}^{7}u_r =148-31 =117.$$

Standard sums

The following formulae are supplied in many A-level formula booklets, but you must know how to choose and use them:

$\boxed{\sum_{r=1}^{n}r=\frac{n(n+1)}{2}},$

$\boxed{\sum_{r=1}^{n}r^2=\frac{n(n+1)(2n+1)}{6}},$

$\boxed{\sum_{r=1}^{n}r^3=\left(\frac{n(n+1)}{2}\right)^2}.$

The third identity has the memorable form

$1^3+2^3+\cdots+n^3=(1+2+\cdots+n)^2.$

These formulae begin at $1$. If a sum begins elsewhere, subtract the unwanted initial terms.

Worked example 8: evaluate a quadratic sum

Evaluate

$\sum_{r=1}^{20}(3r^2-2r+4).$

Split the sum and remember that $4$ occurs $20$ times:

$$\begin{aligned} \sum_{r=1}^{20}(3r^2-2r+4) &=3\sum_{r=1}^{20}r^2-2\sum_{r=1}^{20}r+4(20)\\ &=3\left(\frac{20(21)(41)}6\right) -2\left(\frac{20(21)}2\right)+80\\ &=8610-420+80\\ &=8270. \end{aligned}$$

A rough check helps: the largest summand is $3(20)^2-40+4=1164$, so a total of several thousand is plausible.

Worked example 9: lower limit greater than 1

Evaluate

$\sum_{r=6}^{15}r^2.$

The standard formula gives a sum from $1$, so subtract the squares from $1$ to $5$:

$$\begin{aligned} \sum_{r=6}^{15}r^2 &=\sum_{r=1}^{15}r^2-\sum_{r=1}^{5}r^2\\ &=\frac{15(16)(31)}6-\frac{5(6)(11)}6\\ &=1240-55\\ &=1185. \end{aligned}$$

There are $15-6+1=10$ terms. This is a useful check on the range.

Worked example 10: sum an algebraic sequence

Find a formula for

$S_n=\sum_{r=1}^{n}(2r-1)^2.$

Expand before applying the standard sums:

$$(2r-1)^2=4r^2-4r+1.$$

Therefore

$$\begin{aligned} S_n &=4\sum_{r=1}^{n}r^2-4\sum_{r=1}^{n}r+\sum_{r=1}^{n}1\\ &=4\left(\frac{n(n+1)(2n+1)}6\right) -4\left(\frac{n(n+1)}2\right)+n\\ &=\frac{2n(n+1)(2n+1)}3-2n(n+1)+n\\ &=\frac{n(4n^2-1)}3. \end{aligned}$$

Check $n=1$: the original sum is $1$, and the formula gives $1(4-1)/3=1$.

Self-check 2

1. Evaluate $\displaystyle\sum_{r=1}^{12}(5r-2)$.
2. Evaluate $\displaystyle\sum_{r=4}^{10}r^2$.
3. Simplify $\displaystyle\sum_{r=1}^{n}(r^3+2r)$.

Answers

1. $5\left(\frac{12(13)}2\right)-2(12)=390-24=366.$
2. $\frac{10(11)(21)}6-\frac{3(4)(7)}6=385-14=371.$
3. $$\left(\frac{n(n+1)}2\right)^2+n(n+1) =\frac{n(n+1)(n^2+n+4)}4.$$

Changing the index

Sometimes shifting the index makes a sum match a familiar pattern. The values of the terms must stay unchanged, so change the limits and the summand together.

Worked example 11: shift an index

Rewrite

$\sum_{r=3}^{8}(r-2)^2$

using an index that begins at $1$.

Let

$k=r-2.$

When $r=3$, $k=1$. When $r=8$, $k=6$. Also $(r-2)^2=k^2$. Hence

$\boxed{\sum_{r=3}^{8}(r-2)^2=\sum_{k=1}^{6}k^2}.$

Both sides expand to $1^2+2^2+\cdots+6^2$.

Do not change only the limits

Replacing $r$ by $k+2$ changes every occurrence of $r$. A valid index shift preserves the actual list of terms. Expand the first and last terms before and after the shift to verify it.

Telescoping sums: a useful extension

In a telescoping sum, most terms cancel after the summand is rewritten as a difference. This is useful preparation for later work with series.

Worked example 12: cancellation

Evaluate

$\sum_{r=1}^{n}\frac{1}{r(r+1)}.$

First use partial fractions:

$\frac{1}{r(r+1)}=\frac1r-\frac1{r+1}.$

Now expand several terms:

$$\begin{aligned} \sum_{r=1}^{n}\left(\frac1r-\frac1{r+1}\right) &=\left(1-\frac12\right) +\left(\frac12-\frac13\right) +\left(\frac13-\frac14\right) +\cdots +\left(\frac1n-\frac1{n+1}\right)\\ &=1-\frac1{n+1}\\ &=\frac{n}{n+1}. \end{aligned}$$

Everything except the first positive term and the final negative term cancels. Writing at least three expanded terms makes the pattern visible.

Common misconceptions

• Forgetting inclusive limits: from $m$ to $n$ there are $n-m+1$ terms.
• Treating the index as a fixed number: it takes every integer value between the limits.
• Summing a constant only once: $\sum_{r=1}^{n}c=cn$.
• Using standard sums from the wrong lower limit: $\sum_{r=5}^{n}r^2$ requires subtracting $\sum_{r=1}^{4}r^2$.
• Splitting products: in general, $\sum a_rb_r\ne(\sum a_r)(\sum b_r)$.
• Squaring a sum term by term: in general, $\sum r^2\ne(\sum r)^2$. The cube formula is a special identity, not a general rule.
• Losing brackets with negative terms: $(-1)^r$ alternates sign, whereas $-1^r=-1$ under the usual order of operations.

Mixed self-check

1. Write $7+12+17+\cdots+102$ in sigma notation.
2. Evaluate $\displaystyle\sum_{r=5}^{12}(2r+1)$.
3. Evaluate $\displaystyle\sum_{r=1}^{8}(r^3-r)$.
4. Rewrite $\displaystyle\sum_{r=0}^{n-1}(r+1)^2$ with an index beginning at $1$.
5. Find $\displaystyle\sum_{r=1}^{n}\frac{2}{r(r+1)}$.

Answers

1. Since $7+5(r-1)=5r+2$ and $5r+2=102$ gives $r=20$, $\sum_{r=1}^{20}(5r+2).$
2. There are $8$ terms. Using standard sums, $$2\left(\sum_{r=1}^{12}r-\sum_{r=1}^{4}r\right)+8 =2(78-10)+8 =144.$$
3. $$\left(\frac{8(9)}2\right)^2-\frac{8(9)}2 =1296-36 =1260.$$
4. Let $k=r+1$. The limits become $1$ and $n$, so $\sum_{k=1}^{n}k^2.$
5. $$2\sum_{r=1}^{n}\left(\frac1r-\frac1{r+1}\right) =\frac{2n}{n+1}.$$

Next steps

Use sigma notation to describe the sums in arithmetic sequences and series and geometric sequences and series. It also appears naturally in the binomial expansion, where the index selects each term of the expansion. For applications involving repeated change, continue to sequences and series in modelling.