Series are the sums of the terms of sequences. Similar to sequences, we can have finite or infinite series and convergent or divergent series. The sigma notation or summations are used to express series in a more concise form.

In this article, we will look at some exercises in which we will practice the relationship between series and summations. Mainly, we will learn how to write a series using sigma notation.

## Writing series in sigma notation or summations

Since the series are the sum of the terms of a sequence, we can use sigma notation (Σ) to write them.

We write the sum of the first $latex n $ terms of a sequence as $latex S_{n}$, where

$$S_{n}=u_{1}+u_{2}+u_{3}+…+u_{n}$$

This is an example of a finite series since there are a finite number of terms. We can express this more concisely as follows:

$$u_{1}+u_{2}+u_{3}+…+u_{n}=\sum_{r=1}^{n} u_{r}$$

To write the series in sigma notation, we have to identify the rule by which each term of the sequence is formed. For example, the finite series $latex 7+11+15+19$ can be written as

$$\sum_{r=1}^{4} (4r+3)$$

We can see that each term is formed by adding 4 to the previous term, so we are adding multiples of 4 to the consecutive terms.

The infinite series $latex 1+4+9+16+…$ can be written as

$$\sum_{r=1}^{\infty} r^2$$

## Series and sigma notation – Examples with answers

**EXAMPLE 1**

Find all terms in the series $latex \sum_{r=1}^{5} r(r+1)$.

##### Solution

We can find the terms of the series by using the value of *r* in the expression $latex r(r+1)$ starting from $latex r=1$ to $latex r=5$:

- When $latex r=1$, we have $latex (1)(1+1)=2$
- When $latex r=2$, we have $latex (2)(2+1)=6$
- When $latex r=3$, we have $latex (3)(3+1)=12$
- When $latex r=4$, we have $latex (4)(4+1)=20$
- When $latex r=4$, we have $latex (5)(5+1)=30$

Then, the terms of the series are $latex 2+6+12+20+30$.

**EXAMPLE **2

**EXAMPLE**

Write all the terms of the series $latex \sum_{r=1}^{7} (\frac{(-1)^{r-1}}{r})$.

##### Solution

In this case, we substitute the value of $latex r=1$ through $latex r=7$ in the expression $latex (\frac{(-1)^{r-1}}{r})$:

- When $latex r=1$, we have $latex (\frac{(-1)^{1-1}}{1})=1$
- When $latex r=2$, we have $latex (\frac{(-1)^{2-1}}{2})=-\frac{1}{2}$
- When $latex r=3$, we have $latex (\frac{(-1)^{3-1}}{3})=\frac{1}{3}$
- When $latex r=4$, we have $latex (\frac{(-1)^{4-1}}{4})=-\frac{1}{4}$
- When $latex r=5$, we have $latex (\frac{(-1)^{5-1}}{5})=\frac{1}{5}$
- When $latex r=6$, we have $latex (\frac{(-1)^{6-1}}{6})=-\frac{1}{6}$
- When $latex r=7$, we have $latex (\frac{(-1)^{7-1}}{7})=\frac{1}{7}$

Then, the terms of the series are

$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}$$

**EXAMPLE **3

**EXAMPLE**

What is the sum of the first four terms of the following sequence?

$latex u_{r}=(-1)^r~3^{r+1}$ for $latex r\geq 1$

##### Solution

This is a series written as $latex S_{4}$, which is the sum of the first four terms. Using sigma notation, we have:

$$S_{4}=\sum_{r=1}^{4}(-1)^r~3^{r+1}$$

$$=(-1)3^2+(-1)^23^3+(-1)^33^4+(-1)^43^5$$

$latex =-3^2+3^3-3^4+3^5$

$latex S_{4}=180$

The sum of the first four terms equals 180.

**EXAMPLE **4

**EXAMPLE**

Represent the following series using sigma notation:

$$\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+…+\frac{1}{20}$$

##### Solution

This series can be formed using the expression $latex \frac{1}{r}$, where *r* increments by 1 in each consecutive term.

Moreover, we can observe that the series starts at $latex r=3$ and ends at $latex r=20$. Then, we write it as follows:

$$\sum_{r=3}^{20}\frac{1}{r}$$

**EXAMPLE **5

**EXAMPLE**

Represent the following series using sigma notation:

$$-1+2+7+14+23$$

##### Solution

We may not be able to identify the pattern of the series at first glance. However, we can consider a series in which its terms are defined by $latex u_{r}=r^2$:

$latex 1+4+9+16+25$

Comparing this series with the given series, we observe that if we subtract 2 from each term of the defined series, we obtain the required series. Then, we have the expression $latex u_{r}=r^2-2$.

Using sigma notation, we have:

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

**EXAMPLE **6

**EXAMPLE**

Use sigma notation to write the following series:

$$6-7+8-9+…$$

##### Solution

This is an infinite series since it is indicated that the series continues indefinitely.

If we ignore the negative sign, we see that each term increments by 1. Therefore, the *r* term of the series $latex 6+7+8+9+…$ is given by $latex u_{r}=r+5$.

Now, we can multiply by $latex (-1)^{r-1}$, since this expression gives us the alternating sign. Then, we have:

$latex u_{r}=(-1)^{r-1}(r+5)$

When $latex (r+1)$ is even, the sign is positive and when $latex (r+1)$ is odd, the sign is negative.

Therefore, in sigma notation, we have:

$$\sum_{r=1}^{\infty}(-1)^{r+1}(r+5)$$

In this case, it is also possible to write the series as follows:

$$\sum_{r=5}^{\infty}(-1)^{r+1}(r+1)$$

**EXAMPLE **7

**EXAMPLE**

Write the following series in sigma notation:

$$4-8+16-32+64-128+256-512+1024$$

##### Solution

Similar to the previous example, here we also have an alternating negative sign. Therefore, we can implement this using the expression $latex (-1)^{r+1}$.

If we ignore the negative sign, we see that the terms are formed by powers of 2. That is, we have $latex 2^2+2^3+2^4+…$.

Then, the terms are found with the expression $latex (2)^{r+1}$ starting from $latex r=1$ to $latex r=9$. Therefore, in sigma notation, we have:

$$\sum_{r=1}^{9}(-2){r+1}$$

**EXAMPLE **8

**EXAMPLE**

Write the following series in sigma notation:

$$\frac{5}{5^2-1}+\frac{6}{6^2-1}+\frac{7}{7^2-1}+…+\frac{n}{n^2-1}$$

##### Solution

The terms of this series are formed using the expression $latex \frac{r}{r^2-1}$, where *r* increments by 1 in each consecutive term.

In this case, the limits range from $latex r=5$ to $latex r=n$. Then, using sigma notation, we have:

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

**EXAMPLE **9

**EXAMPLE**

Write the following series in sigma notation:

$$\frac{1}{2\times 3}+\frac{2}{3\times 4}+\frac{3}{4\times 5}+…+\frac{n}{(n+1)(n+2)}$$

##### Solution

This example is similar to the previous one since we can use the last given term to obtain the expression in sigma notation easily. That is, we have $latex \frac{r}{(r+1)(r+2)}$$.

Moreover, we observe that the series goes from $latex r=1$ to $latex r=n$. Then, we write it as follows:

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

**EXAMPLE **10

**EXAMPLE**

Write the following series in sigma notation:

$$1\times 4-3\times 7 +5 \times 10-…+29\times 46$$

##### Solution

First, we note that the series has alternating negative signs, so we know that we are going to use the expression $latex (-1)^{r+1}$ to implement this.

Then, we see that each term is a multiplication of two numbers. Therefore, we find an expression for each part in terms of *r*.

The first part can be obtained using $latex (2r-1)$ and the second part can be obtained with $latex (3r+1)$.

Finally, we note that the series goes from $latex r=1$ to $latex r=15$. Then, combining all this, we have:

$$\sum_{r=1}^{15}(-1)^{r+1}(2r-1)(3r+1)$$

## Series and sigma notation – Practice problems

## See also

Interested in learning more about series and sequences? You can look at these pages:

### Learn mathematics with our additional resources in different topics

**LEARN MORE**