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# Geometric Series – Definition, Formula, and Examples

The**geometric series**plays an important part in the early stages of calculus and contributes to our understanding of the convergence series. We can also use the geometric series in physics, engineering, finance, and finance. This shows that is essential that we know how to identify and find the sum of geometric series.

**In this article, we’ll understand how closely related the geometric sequence and series are. We’ll also show you how the infinite and finite sums are calculated. You’ll also get the chance to try out word problems that make use of geometric series.**

*The geometric series represents the sum of the terms in a finite or infinite geometric sequence. The consecutive terms in this series share a common ratio.***What is a geometric series?**

**The geometric series**represents the

**sum of the geometric sequence’s terms**. This means that the

**terms of a geometric series**will also

**share a common ratio**, $r$.Since the geometric series is closely related to the geometric sequence, we’ll do a quick refresher on the geometric sequence’s definition to understand the geometric series’ components.Does this image look familiar? That’s because this is one known way for us to visualize what happens with a geometric sequence with the following terms: $\left\{1, \dfrac{1}{2}, \dfrac{1}{4},\dfrac{1}{8},\dfrac{1}{16}\right\}$. This is an example of a geometric sequence that has a common ratio that is less than $1$.

**This sequence becomes a series if we express the terms’ sum**as shown below:\begin{aligned}1 \underbrace{+}_{\color{Orchid} \times \frac{1}{2}} \dfrac{1}{2} \underbrace{+}_{\color{Orchid} \times \frac{1}{2}} \dfrac{1}{4} \underbrace{+}_{\color{Orchid} \times \frac{1}{2}} \dfrac{1}{16} &= \sum_{i=1}^{n} 1\cdot \dfrac{1}{2}^{n -1}\end{aligned}Here are a few more examples highlighting the difference between the arithmetic sequence and series.

Geometric Sequence | Geometric Series | Common Ratio |

\begin{aligned}4,8,16,32, …, 1024\end{aligned} | \begin{aligned}4+8 +16+32+ …+ 1024\end{aligned} | \begin{aligned}r =2\end{aligned} |

\begin{aligned} 1,-3,9, -27,…,-2187 \end{aligned} | \begin{aligned} 1+-3 +9 -27+…-2187\end{aligned} | \begin{aligned}r = -3\end{aligned} |

\begin{aligned} 625,125, 25, …\end{aligned} | \begin{aligned} 625+125+25+ …\end{aligned} | \begin{aligned}r = \dfrac{1}{2}\end{aligned} |

**because it’s possible for a geometric series to either be a finite or infinite series**:

- A
**finite geometric series contains a finite number of terms.**This means that the series will**have both first and last terms**. Finite geometric series are also convergent. - The
**infinite geometric series**, on the other hand,**goes on and approaches infinity**. This means that the geometric series that is infinite**does not have the last term**.

Geometric Sequence | \begin{aligned} a\underbrace{\phantom{x}} _{\color{Teal} \times r}ar\underbrace{\phantom{x}} _{\color{Teal} \times r}ar^2,…,ar^{n -2}\underbrace{\phantom{x}} _{\color{Teal}\times r}ar^{n -1}\end{aligned} |

Geometric Series (Finite) | \begin{aligned} a\underbrace{+} _{\color{Teal}\times r}ar\underbrace{+} _{\color{Teal}\times r}ar^2+…+ar^{n – 2}\underbrace{+}_{\color{Teal}\times r}ar^{n – 1}\end{aligned} |

Geometric Series (Infinite) | \begin{aligned} a\underbrace{+} _{\color{Teal}\times r}ar\underbrace{+} _{\color{Teal}\times r}ar^2+…\end{aligned} |

**Derivation of geometric series formulas **

We can express the $\boldsymbol{n}$**th term of any geometric series as**$\boldsymbol{a_n = ar^{n -1}}$. This means that we can use this formula to express the sum of the series, $S_n$, as shown below.\begin{aligned} S_n &= a_1 + a_2 + a_3 + …+ a_{n- 1}+ a_n\\&=ar^{1 -1} + ar^{2 -1} + ar^{3 -1} + …+ar^{n – 1 -1} + ar^{n -1}\\&=a + ar + ar^2 +…+ar^{n-2} + ar^{n-1},\phantom{xxxxx}\text{Equation 1}\end{aligned}Multiply both sides of the first equation by $r$.\begin{aligned} S_nr &=ar + ar^2 + ar^3 +…+ar^{n-1} + ar^{n},\phantom{xxxxx}\text{Equation 2}\end{aligned}Subtract the second equation from the first one then isolate $S_n$ on the left-hand side of the equation.\begin{aligned} S_n &= a + ar + ar^2 +…+ar^{n -2} + ar^{n- 1},\phantom{x}(1)\\-\underline{\phantom{xxxx}S_nr }&= \underline{ar + ar^2 + ar^3 +…+ar^{n -1} + ar^{n}},\phantom{x}(2)\\S_n – S_nr &= a-ar^n\end{aligned}\begin{aligned} S_n(1 – r) &= a(1 – r^n)\\S_n&=\dfrac{a(1 – r^n)}{1 – r}\end{aligned}Hence, we have the formula for the finite geometric series’ sum as shown below.

\begin{aligned} S_n&=\dfrac{a(1 – r^n)}{1 – r}\\\\S_n&:\text{Geometric series’s sum}\\a &: \text{First term}\\r &: \text{Common ratio}\end{aligned} |

\begin{aligned} S_\infty&=\dfrac{a}{1 – r}\\\\S_\infty&:\text{Infinite series’ sum}\\a &: \text{First term}\\r &: \text{Common ratio}\\ &:-1< r<1\end{aligned} |

**How to solve geometric series?**

Here are some important pointers to remember when solving a geometric series problem:- Confirm whether the given series is a finite or an infinite geometric series.
- Identify the values that are available: $a$, $r$, and $n$. Find $n$ using the equation, $a_n=ar^{n –1}$.
- Apply the appropriate formula depending on whether the series is finite or infinite.

**How to tell if a series is geometric?**

It’s easy to determine whether a given series is geometric or not – we simply find the ratio shared between two consecutive terms. **Compare this ratio with the rest and see if they’re equal**. When they are, the series is said to be geometric.\begin{aligned} a_1, a_2, a_3, a_4, &…,a_{n-1}, a_n\\\dfrac{a_2}{a_1} &= r\\\dfrac{a_3}{a_2}&= r\\\dfrac{a_4}{a_3}&= r\\&.\\&.\\&.\\\dfrac{a_n}{a_{n -1}}&= r\end{aligned}When you can

**confirm the existence of the common ratio**, $\boldsymbol{r}$, from the series you’re observing,

**then the series is geometric**.

**How to apply the geometric series’ formula?**

Now that we have confirmed that the series is indeed geometric, your next goal is to check if the geometric series is finite or infinite. Apply the appropriate formula as shown below.Finite Series | Infinite Series |

\begin{aligned} S_n&=\dfrac{a(1 – r^n)}{1 – r} \end{aligned} | \begin{aligned} S_\infty&=\dfrac{a}{1 – r}\\-1&< r<1\end{aligned} |

$1)$ | \begin{aligned} \dfrac{4}{2} = 2,\phantom{x}\dfrac{8}{4} = 2, &…,\phantom{x}\dfrac{1024}{512} =2 \\\\2 \underbrace{+}_{\color{DarkOrange}\times 2}4\underbrace{+}_{\color{DarkOrange}\times 2} 8+&…+512 \underbrace{+}_{\color{DarkOrange}\times 2} 1024 \\\end{aligned} |

$2)$ | \begin{aligned} \dfrac{27}{81} = \dfrac{1}{3},\phantom{x}\dfrac{9}{27} &= \dfrac{1}{3}, \phantom{x}\dfrac{3}{9} = \dfrac{1}{3}…\\\\81 \underbrace{+}_{\color{Purple}\times \frac{1}{3}}27 &\underbrace{+}_{\color{Purple}\times \frac{1}{3}} 9 \underbrace{+}_{\color{Purple}\times \frac{1}{3}} 3+…\\\end{aligned} |

**1) is a finite geometric series**while

**2) is an infinite geometric series**.Focusing on the first series, $2+ 4+ 8+…+512+ 1024$, we know that $a = 2$ and $r = 2$. Now,

**to find the number of terms, let’s use the fact that**$\boldsymbol{a_n = ar^{n -1}}$ to solve for $n$.\begin{aligned}a_n &= ar^{n – 1}\\1024 &= 2\cdot 2^{n -1}\\512 &= 2^{n – 1}\\2^9 &= 2^{n -1}\\9&= n -1\\n&= 10\end{aligned}Since we now have all the values we need, let’s apply the sum formula for the finite geometric series.\begin{aligned} S_n&=\dfrac{a(1 – r^n)}{1 – r}\\S_{10} &=\dfrac{2(1 – 2^{10})}{1 – 2}\\&=\dfrac{2(-1023)}{-1}\\ &= 2046 \end{aligned}Working on the second series, it’s important to double check if $-1 the infinite series’ sum will converge and we can apply the sum formula.\begin{aligned} S_\infty&=\dfrac{a}{1 – r}\\&= \dfrac{81}{1 – \dfrac{1}{3}}\\&=\dfrac{243}{2}\end{aligned}These two examples clearly show how we can apply the two formulas to simplify the sum of infinite and finite geometric series. Don’t worry, we’ve prepared more problems for you to work on as well!

**Find the sum of the series, $-3 – 6 – 12 -… – 768-1536$.**

*Example 1*__Solution__Factor out $-1$ from each term then check the common ratio shared by each pair of consecutive terms.\begin{aligned}-3 – 6 -12 – …- 768-1536&= -(3 + 6 +12 + …+ 768+ 1536)\\&= -(3 \underbrace{+}_{\color{Teal}\times 2} 6 \underbrace{+}_{\color{Teal}\times 2}12 + …+ 768\underbrace{+}_{\color{Teal}\times 2} 1536)\end{aligned}From this, we can see that the common ratio is $r = 2$. Since the series has a first and last term, we’ll need the number of terms in the given series before we can apply the sum formula for the finite geometric series.\begin{aligned}a_n &= ar^{n – 1}\\1536 &= 3\cdot 2^{n -1}\\512 &= 2^{n – 1}\\2^9 &= 2^{n -1}\\9&= n -1\\n&= 10\end{aligned}Apply the sum formula to find the sum of the finite geometric series. Don’t forget to account for the $-1$ factored out of the series.\begin{aligned} S_n&=\dfrac{a(1 – r^n)}{1 – r}\\-S_{10} &=-\dfrac{3(1 – 2^{10})}{1 – 2}\\&=-\dfrac{3(-1023)}{-1}\\ &= -3069 \end{aligned}This means that the finite geometric series’ sum is equal to $-3069$.

**Find the sum of the series, $1024 – 512 + 256 – 128 + …$.**

*Example 2*__Solution__By inspection alone, we can see that we’re working with an infinite series. Let’s now divide each pair of consecutive terms to confirm if the series has a common ratio, $r$.\begin{aligned}1024 \underbrace{-}_{\color{DarkOrange} \times -\frac{1}{2}} 512 \underbrace{+}_{\color{DarkOrange} \times -\frac{1}{2}} 256 \underbrace{-}_{\color{DarkOrange} \times -\frac{1}{2}} 128 +…\end{aligned}From this, we can see that we have an infinite geometric series, where $r = -\dfrac{1}{2}$. Since $-1 < -\dfrac{1}{2}< 1$, we can apply the sum formula for the infinite geometric series. Use the first term of the series, $a = 1024$, and the common ratio, $r = -\dfrac{1}{2}$.\begin{aligned} S_\infty&=\dfrac{a}{1 -r}\\&= \dfrac{1024}{1- \left(-\dfrac{1}{2} \right )}\\&= \dfrac{1024}{\dfrac{3}{2}}\\&= \dfrac{2048}{3}\end{aligned}Hence, the sum of the infinite geometric series is $\dfrac{2048}{3}$.

**Marie is observing a certain ball that bounces back to three-fourths of the height it fell from. She initially dropped the ball from $16$ feet. What is the approximate total distance traveled by the ball?**

*Example 3*__Solution__Observe the height reached by the ball after each bounce. Keep in mind that we’ll have to account for the distance it reaches going up and down.

Number of Bounce | Distance Reached Going Up | Distance Reached Going Down |

Initial Position | \begin{aligned}16\end{aligned} | |

First Bounce | \begin{aligned}16 \cdot \dfrac{3}{4} = 12\end{aligned} | \begin{aligned}16 \cdot \dfrac{3}{4} = 12\end{aligned} |

Second Bounce | \begin{aligned}12 \cdot \dfrac{3}{4} = 9\end{aligned} | \begin{aligned}12 \cdot \dfrac{3}{4} = 9\end{aligned} |

Third Bounce | \begin{aligned}9 \cdot \dfrac{3}{4} = \dfrac{27}{4}\end{aligned} | \begin{aligned}9 \cdot \dfrac{3}{4} = \dfrac{27}{4}\end{aligned} |

…and so on |

\begin{aligned}S_{\infty_1} &= 16 + 12 + 9 + \dfrac{27}{4} +…\\&= \dfrac{16}{\dfrac{3}{4}}\\&=\dfrac{64}{3}\end{aligned} | \begin{aligned} S_{\infty_2} &= 12 + 9 + \dfrac{27}{4} +…\\&= \dfrac{12}{\dfrac{3}{4}}\\&=16\end{aligned} |

\begin{aligned}S_{\infty_1} + S_{\infty_2} &= \dfrac{64}{3} + 16\\&= \dfrac{112}{3}\\&\approx 37.33\end{aligned} |