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 infinite limits infinite series basics

related topics

For a discussion of geometric sequences, see the page on sequences

A main application of a geometric series is to build power series

WikiBooks - Geometric Series

### Geometric Series Quick Breakdown

 used to prove convergence yes used to prove divergence yes can be inconclusive no can find convergence value yes useful for finding power series

### 17Calculus Subjects Listed Alphabetically

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### Search Practice Problems

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17calculus > infinite series > geometric series

 Geometric Series How To Use The Geometric Series Finite Geometric Series Additional Geometric Series Topics Practice

Geometric Series are an important type of series that you will come across while studying infinite series. This series type is unusual because not only can you easily tell whether a geometric series converges or diverges but, if it converges, you can calculate exactly what it converges to. This is extremely unusual for an infinite series. Let's break down what this theorem is saying and how we can use it.

Geometric Series Convergence

A series in the form $$\displaystyle{ \sum_{n=0}^{\infty}{r^n}}$$ is called a Geometric Series with ratio $$r$$.

$$0 \lt \abs{r} \lt 1$$

$$\abs{r} \geq 1$$

converges

diverges

Series Convergence Value

If the geometric series converges, it converges to $$\displaystyle{\frac{1}{1-r}}$$.

This is usually written $$\displaystyle{ \sum_{n=0}^{\infty}{r^n} = \frac{1}{1-r}, ~~ 0 \lt \abs{r} \lt 1 }$$.

How To Use The Geometric Series

A geometric series looks like this $$\displaystyle{ \sum_{n=0}^{\infty}{ r^n } }$$ where r is an expression of some sort, not containing n.

If you can get your series into this form using algebra, then $$r$$ will tell you whether the series converges or diverges.
If $$\abs{r} \geq 1$$, then the series diverges.
If $$\abs{r} < 1,$$ then the series converges and it converges to $$\displaystyle{\frac{1}{1-r}}$$

Notes

1. r is called the ratio.
2. The absolute values on r to determine convergence or divergence are absolutely critical. Do not drop them unless you are sure that r is positive all the time. If you are not sure, keeping them is always correct.
3. Watch out! - - Be careful to notice that the series given above starts with index $$n=0$$. When determining the convergence value, make sure you take that into account and adjust your series to exactly match this one, including the starting index value of zero.

Finite Geometric Series

In the following discussion, we are assuming that $$0 < r < 1$$.
The equation for the value of a finite geometric series is
$$\displaystyle{ \sum_{n=0}^{k}{r^n} = \frac{1-r^{k+1}}{1-r} ~~~~~ (1) }$$
where $$k$$ is a finite positive integer.

Show that the above formula holds for $$k=1, k=2$$ and $$k=3$$.

Problem Statement

Show that the above formula holds for $$k=1, k=2$$ and $$k=3$$.

Solution

Let $$k=1$$.
$$\displaystyle{ \sum_{n=0}^{1}{r^n} = r^0 + r^1 = 1+r }$$
If we substitute $$k=1$$ into the right side of equation $$(1)$$, factor and simplify, we get
$$\displaystyle{ \frac{1-r^{1+1}}{1-r} = \frac{1-r^2}{1-r} = \frac{(1+r)(1-r)}{1-r} = 1+r }$$

So we have shown that the equation holds for $$k=1$$. Let's try $$k=2$$.

Let $$k=2$$ and go through the same steps.
$$\displaystyle{ \sum_{n=0}^{2}{r^n} = r^0 + r^1 + r^2 = 1+r+r^2 }$$
If we substitute $$k=2$$ into the right side of equation $$(1)$$, factor and simplify, we get
$$\displaystyle{ \frac{1-r^{1+2}}{1-r} = \frac{1-r^3}{1-r} = \frac{(1+r+r^2)(1-r)}{1-r} = 1+r+r^2 }$$

It works again. How about for $$k=3$$?

Let $$k=3$$.
$$\displaystyle{ \sum_{n=0}^{3}{r^n} = r^0 + r^1 + r^2 + r^3 = 1+r+r^2+r^3 }$$
This time, if we set up the right side of equation $$(1)$$, the factoring is getting a little bit more complicated. So let's work backwards and start with $$1+r+r^2+r^3$$ and try to get the right side of equation $$(1)$$. To do this we multiple by $$(1-r)/(1-r)$$.

 $$1+r+r^2+r^3$$ $$\displaystyle{ \displaystyle{(1+r+r^2+r^3)\frac{1-r}{1-r}} }$$ $$\displaystyle{ \displaystyle{\frac{(1+r+r^2+r^3)-r(1+r+r^2+r^3)}{1-r}} }$$ $$\displaystyle{ \displaystyle{\frac{1+r+r^2+r^3 -r - r^2 - r^3 - r^4}{1-r}} }$$ $$\displaystyle{ \displaystyle{\frac{1-r^4}{1-r}} }$$

Notice that all the inside terms cancel and we are left with $$1-r^4$$ in the numerator.

Derivation

In this section, we will derive equation (1).
Let $$\displaystyle{ S_n = \sum_{k=0}^{n}{r^k} = }$$ $$1+r+r^2+r^3+ . . . + r^n$$
Multiplying this by $$r$$ gives us $$rS_n = r + r^2 + r^3 + . . . +r^n + r^{n+1}$$
Now we subtract $$rS_n$$ from $$S_n$$ to get $$S_n - rS_n = (1+ r + r^2 + r^3 + . . . +r^n)$$ $$-$$ $$(r+r^2+r^3+ . . . + r^n + r^{n+1} )$$
Looking closely at the terms on the right side of the equal sign, we can see that all the terms cancel except for $$1-r^{n+1}$$. So now we have $$S_n-rS_n = 1-r^{n+1}$$. Now we solve for $$S_n$$.
$$\begin{array}{rcl} S_n-rS_n & = & 1-r^{n+1} \\ S_n(1-r) & = & 1-r^{n+1} \\ S_n & = & \displaystyle{ \frac{1-r^{n+1}}{1-r} } \end{array}$$

$$\displaystyle{ \sum_{k=0}^{n}{r^k} = \frac{1-r^{k+1}}{1-r} }$$ [qed]

This video shows a geometric series applied to a financial equation. The application of calculus to finances is not covered on this site. But you may find this video helpful if you are in business calculus.

### Krista King Math - Financial Application of Geometric Series [9min-49secs]

video by Krista King Math

### Practice

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on.

GOT IT. THANKS!

Instructions - - Unless otherwise instructed, for each of the following series
1. determine whether it converges or diverges, using the geometric series test, if possible.
2. If it converges, determine what it converges to (if possible). Give your answers in exact form.

Basic Problems

$$\displaystyle{ \sum_{n=0}^{\infty}{ \frac{3}{4^n} } }$$

Problem Statement

$$\displaystyle{ \sum_{n=0}^{\infty}{ \frac{3}{4^n} } }$$

The series $$\displaystyle{ \sum_{n=0}^{\infty}{ \frac{3}{4^n} } }$$ converges to 4 by the Geometric Series Test.

Problem Statement

$$\displaystyle{ \sum_{n=0}^{\infty}{ \frac{3}{4^n} } }$$

Solution

There is a geometric series hidden here. First key: If there is a constant in a series, pull it outside the sum. This will often simplify the problem so that you can see it more clearly.
After you do that, you get $$\displaystyle{ a_n = \frac{1}{4^n} }$$
Now, since there is a one in numerator, you can rewrite this as:
$$\displaystyle{ a_n = \frac{1^n}{4^n} = \left( \frac{1}{4} \right)^n }$$
You can now see that you have $$\left|r\right|=1/4 < 1$$, which converges by the geometric series test.
Additionally, we can determine that it converges to
$$\displaystyle{ \frac{3}{1-(1/4)} = \frac{3}{3/4} = 4 }$$

The series $$\displaystyle{ \sum_{n=0}^{\infty}{ \frac{3}{4^n} } }$$ converges to 4 by the Geometric Series Test.

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^n}{e^n} } }$$

Problem Statement

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^n}{e^n} } }$$

The series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^n}{e^n} } }$$ converges to $$\displaystyle{\frac{2}{e-2}}$$ by the Geometric Series Test.

Problem Statement

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^n}{e^n} } }$$

Solution

Convergence Or Divergence
Since we have the same power in the numerator and denominator, we can combine them to get $$\displaystyle{ \frac{2^n}{e^n} = \left( \frac{2}{e} \right)^n }$$
Now we can see that we have a geometric series with
$$\left|r\right|=2/e < 1$$ so the series converges.
Convergence Value
When determining only convergence or divergence, you don't need to worry about the starting n-value. But you do need to consider it if you are trying to determine what a series converges to.
Notice in the definition of a geometric series, n starts at zero. In this series, n starts at one. So we need to take that into account when determining what it converges to. This means, for this problem, we need to subtract 1 (because when $$n=0$$, the term is 1; this is not always the case, so you need to determine this in each problem).
$$\displaystyle{ \frac{1}{1-(2/e)} - 1 = \frac{e}{e-2} - \frac{e-2}{e-2} = \frac{2}{e-2} }$$

The series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^n}{e^n} } }$$ converges to $$\displaystyle{\frac{2}{e-2}}$$ by the Geometric Series Test.

$$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{1+\sqrt{5}}{2} \right]^n } }$$

Problem Statement

$$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{1+\sqrt{5}}{2} \right]^n } }$$

The geometric series diverges.

Problem Statement

$$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{1+\sqrt{5}}{2} \right]^n } }$$

Solution

### 274 solution video

video by MIT OCW

The geometric series diverges.

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\pi^n }{ 3^{n+2}} } }$$

Problem Statement

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\pi^n }{ 3^{n+2}} } }$$

The geometric series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\pi^n }{ 3^{n+2}} } }$$ diverges.

Problem Statement

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\pi^n }{ 3^{n+2}} } }$$

Solution

### 277 solution video

video by PatrickJMT

The geometric series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\pi^n }{ 3^{n+2}} } }$$ diverges.

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{ 3^n+2^n }{ 6^n } } }$$

Problem Statement

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{ 3^n+2^n }{ 6^n } } }$$

The geometric series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{ 3^n+2^n }{ 6^n } } }$$ converges.

Problem Statement

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{ 3^n+2^n }{ 6^n } } }$$

Solution

### 278 solution video

video by PatrickJMT

The geometric series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{ 3^n+2^n }{ 6^n } } }$$ converges.

$$\displaystyle{ \sum_{n=0}^{\infty}{ \frac{ 1 }{ 5^n } } }$$

Problem Statement

$$\displaystyle{ \sum_{n=0}^{\infty}{ \frac{ 1 }{ 5^n } } }$$

The geometric series $$\displaystyle{ \sum_{n=0}^{\infty}{ \frac{ 1 }{ 5^n } } }$$ converges to $$5/4$$.

Problem Statement

$$\displaystyle{ \sum_{n=0}^{\infty}{ \frac{ 1 }{ 5^n } } }$$

Solution

### 280 solution video

video by PatrickJMT

The geometric series $$\displaystyle{ \sum_{n=0}^{\infty}{ \frac{ 1 }{ 5^n } } }$$ converges to $$5/4$$.

$$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{3^n}{4^n} + \frac{2}{7^n} \right] } }$$

Problem Statement

$$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{3^n}{4^n} + \frac{2}{7^n} \right] } }$$

The geometric series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{3^n}{4^n} + \frac{2}{7^n} \right] } }$$ converges.

Problem Statement

$$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{3^n}{4^n} + \frac{2}{7^n} \right] } }$$

Solution

### 282 solution video

video by PatrickJMT

The geometric series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{3^n}{4^n} + \frac{2}{7^n} \right] } }$$ converges.

$$\displaystyle{ \sum_{k=0}^{\infty}{ \left( \frac{2}{3} \right)^k } }$$

Problem Statement

$$\displaystyle{ \sum_{k=0}^{\infty}{ \left( \frac{2}{3} \right)^k } }$$

The geometric series $$\displaystyle{ \sum_{k=0}^{\infty}{ \left( \frac{2}{3} \right)^k } }$$ converges to $$3$$.

Problem Statement

$$\displaystyle{ \sum_{k=0}^{\infty}{ \left( \frac{2}{3} \right)^k } }$$

Solution

### 285 solution video

video by Krista King Math

The geometric series $$\displaystyle{ \sum_{k=0}^{\infty}{ \left( \frac{2}{3} \right)^k } }$$ converges to $$3$$.

Express $$\displaystyle{ 2/13 - 4/13^2 + 8/13^3 - 16/13^4 + . . . }$$ using sigma notation.

Problem Statement

Express $$\displaystyle{ 2/13 - 4/13^2 + 8/13^3 - 16/13^4 + . . . }$$ using sigma notation.

$$\displaystyle{ 2/13 - 4/13^2 + 8/13^3 - 16/13^4 + . . . = \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^{n+1} (2)^n}{(13)^n} \right] } }$$

Problem Statement

Express $$\displaystyle{ 2/13 - 4/13^2 + 8/13^3 - 16/13^4 + . . . }$$ using sigma notation.

Solution

In this video, his answer is incorrect. He writes $$(-2)^n$$ but the signs do not match the ones given in the problem statement. We could have written this as $$-(-2)^n$$ but we chose the more standard $$(-1)^{n+1} (2)^n$$.

### 281 solution video

video by PatrickJMT

$$\displaystyle{ 2/13 - 4/13^2 + 8/13^3 - 16/13^4 + . . . = \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^{n+1} (2)^n}{(13)^n} \right] } }$$

Express $$\displaystyle{ 1+0.1+0.01+0.001+0.0001+ . . . }$$ using sigma notation.

Problem Statement

Express $$\displaystyle{ 1+0.1+0.01+0.001+0.0001+ . . . }$$ using sigma notation.

$$\displaystyle{ 1+0.1+0.01+0.001+0.0001+ . . . = \sum_{n=0}^{\infty}{ \frac{1}{10^n} } }$$

Problem Statement

Express $$\displaystyle{ 1+0.1+0.01+0.001+0.0001+ . . . }$$ using sigma notation.

Solution

### 279 solution video

video by PatrickJMT

$$\displaystyle{ 1+0.1+0.01+0.001+0.0001+ . . . = \sum_{n=0}^{\infty}{ \frac{1}{10^n} } }$$

Express $$\displaystyle{ 1+0.4+0.16+0.064+ . . . }$$ using sigma notation and determine convergence or divergence.

Problem Statement

Express $$\displaystyle{ 1+0.4+0.16+0.064+ . . . }$$ using sigma notation and determine convergence or divergence.

The geometric series $$\displaystyle{ 1+0.4+0.16+0.064+ . . . }$$ converges to $$5/3$$.

Problem Statement

Express $$\displaystyle{ 1+0.4+0.16+0.064+ . . . }$$ using sigma notation and determine convergence or divergence.

Solution

### 275 solution video

video by PatrickJMT

The geometric series $$\displaystyle{ 1+0.4+0.16+0.064+ . . . }$$ converges to $$5/3$$.

$$\displaystyle{ 1 + e^{-1} + e^{-2} + e^{-3} + . . . }$$

Problem Statement

$$\displaystyle{ 1 + e^{-1} + e^{-2} + e^{-3} + . . . }$$

The geometric series $$\displaystyle{1 + e^{-1} + e^{-2} + e^{-3} + . . . }$$ converges to $$\displaystyle{\frac{e}{e-1}}$$.

Problem Statement

$$\displaystyle{ 1 + e^{-1} + e^{-2} + e^{-3} + . . . }$$

Solution

### 286 solution video

video by Krista King Math

The geometric series $$\displaystyle{1 + e^{-1} + e^{-2} + e^{-3} + . . . }$$ converges to $$\displaystyle{\frac{e}{e-1}}$$.

$$\displaystyle{ \frac{3}{2} - \frac{3}{4} + \frac{3}{8} - \frac{3}{16} + . . . }$$

Problem Statement

$$\displaystyle{ \frac{3}{2} - \frac{3}{4} + \frac{3}{8} - \frac{3}{16} + . . . }$$

The geometric series $$\displaystyle{ \frac{3}{2} - \frac{3}{4} + \frac{3}{8} - \frac{3}{16} + . . . }$$ converges to $$1$$.

Problem Statement

$$\displaystyle{ \frac{3}{2} - \frac{3}{4} + \frac{3}{8} - \frac{3}{16} + . . . }$$

Solution

### 287 solution video

video by Krista King Math

The geometric series $$\displaystyle{ \frac{3}{2} - \frac{3}{4} + \frac{3}{8} - \frac{3}{16} + . . . }$$ converges to $$1$$.

$$\displaystyle{ 1 - \frac{1}{5} + \left( \frac{1}{5} \right)^2 - \left( \frac{1}{5} \right)^3 + \ldots }$$

Problem Statement

$$\displaystyle{ 1 - \frac{1}{5} + \left( \frac{1}{5} \right)^2 - \left( \frac{1}{5} \right)^3 + \ldots }$$

The geometric series $$\displaystyle{ 1 - \frac{1}{5} + \left( \frac{1}{5} \right)^2 - \left( \frac{1}{5} \right)^3 + \ldots }$$ converges to $$\displaystyle{ \frac{5}{6} }$$.

Problem Statement

$$\displaystyle{ 1 - \frac{1}{5} + \left( \frac{1}{5} \right)^2 - \left( \frac{1}{5} \right)^3 + \ldots }$$

Solution

### 288 solution video

video by Krista King Math

The geometric series $$\displaystyle{ 1 - \frac{1}{5} + \left( \frac{1}{5} \right)^2 - \left( \frac{1}{5} \right)^3 + \ldots }$$ converges to $$\displaystyle{ \frac{5}{6} }$$.

Intermediate Problems

$$\displaystyle{\sum_{n=1}^{\infty}{\frac{(-1)^n2^{n-1}}{3^n}}}$$

Problem Statement

$$\displaystyle{\sum_{n=1}^{\infty}{\frac{(-1)^n2^{n-1}}{3^n}}}$$

The geometric series $$\displaystyle{\sum_{n=1}^{\infty}{\frac{(-1)^n2^{n-1}}{3^n}}}$$ converges absolutely to $$-1/5$$.

Problem Statement

$$\displaystyle{\sum_{n=1}^{\infty}{\frac{(-1)^n2^{n-1}}{3^n}}}$$

Solution

Converges Or Diverges?
With this series, you may have immediately recognized that it is an alternating series. Using the Alternating Series Test does tell you correctly that this series converges. However, if you look more closely, you might see there is a geometric series hidden here. Let's do some algebra to make it more obvious.
What we want to do is pull out all constants (not dependent on n), then try to get all the exponents to be the same, so that we can combine them. The tricky term is the $$2^{n-1}$$. This can be written as $$\displaystyle{ 2^{n-1} = (2^n)(2^{-1}) = \frac{2^n}{2}}$$. So, now we can move things around so that we can easily see the geometric series.

$$\displaystyle{ \frac{(-1)^n2^{n-1}}{3^n} = \frac{(-1)^n2^n}{3^n \cdot 2} = \frac{1}{2} \left( \frac{-2}{3} \right)^n }$$
First, we pulled out any constants (not dependent on n). Then we combined all the terms with the same power. Can you now see that we have a geometric series where $$a=1/2$$ and $$r=-2/3$$?
In this case $$\left|r\right|=2/3<1$$, so the series converges.

Convergence Value
We can also determine what it converges to, but we need to be careful here. Notice in the Geometric Series theorem, the index n starts with zero. But our series starts with one. We need to take that into account here. In our case, when $$n=0$$, we have $$1/2$$. So we need to subtract $$1/2$$ from the formula to get

$$\displaystyle{\frac{1/2}{1-(-2/3)} - \frac{1}{2} = \frac{3/2}{3+2} - \frac{1}{2} = \frac{3}{10} - \frac{5}{10} = \frac{-2}{10} = \frac{-1}{5} }$$

Absolute Or Conditional Convergence
Now we need to determine whether the series converges absolutely or conditionally. To do that we need to determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^{n-1}}{3^n} } }$$ converges. Using similar algebra as we did above, this series reduces to a geometric series with $$r = 2/3$$. Since $$\left|r=2/3\right| < 1$$, this series converges. Therefore, the original series converges absolutely.

The geometric series $$\displaystyle{\sum_{n=1}^{\infty}{\frac{(-1)^n2^{n-1}}{3^n}}}$$ converges absolutely to $$-1/5$$.

$$\displaystyle{ \sum_{n=3}^{\infty}{ 5 (2/3)^{n-1} } }$$

Problem Statement

$$\displaystyle{ \sum_{n=3}^{\infty}{ 5 (2/3)^{n-1} } }$$

The geometric series $$\displaystyle{ \sum_{n=3}^{\infty}{ 5 (2/3)^{n-1} } }$$ converges to $$20/3$$.

Problem Statement

$$\displaystyle{ \sum_{n=3}^{\infty}{ 5 (2/3)^{n-1} } }$$

Solution

### 276 solution video

video by PatrickJMT

The geometric series $$\displaystyle{ \sum_{n=3}^{\infty}{ 5 (2/3)^{n-1} } }$$ converges to $$20/3$$.

$$\displaystyle{ \sum_{n=3}^{\infty}{ \frac{ \pi^{n+1} }{ 6^n } } }$$

Problem Statement

$$\displaystyle{ \sum_{n=3}^{\infty}{ \frac{ \pi^{n+1} }{ 6^n } } }$$

The geometric series $$\displaystyle{ \sum_{n=3}^{\infty}{ \frac{ \pi^{n+1} }{ 6^n } } }$$ converges to $$\displaystyle{ \frac{ \pi^4 }{ 36(6-\pi) } }$$.

Problem Statement

$$\displaystyle{ \sum_{n=3}^{\infty}{ \frac{ \pi^{n+1} }{ 6^n } } }$$

Solution

### 283 solution video

video by PatrickJMT

The geometric series $$\displaystyle{ \sum_{n=3}^{\infty}{ \frac{ \pi^{n+1} }{ 6^n } } }$$ converges to $$\displaystyle{ \frac{ \pi^4 }{ 36(6-\pi) } }$$.

Express $$\displaystyle{ 0.\overline{21} }$$ as a fraction.

Problem Statement

Express $$\displaystyle{ 0.\overline{21} }$$ as a fraction.

The decimal $$\displaystyle{ 0.2121 \overline{21} }$$ can be written as a geometric series $$\displaystyle{ \sum_{k=0}^{\infty}{ \frac{21}{100} \left[ \frac{1}{100} \right]^k } }$$ which converges to $$7/33$$.

Problem Statement

Express $$\displaystyle{ 0.\overline{21} }$$ as a fraction.

Solution

### 284 solution video

video by Krista King Math

The decimal $$\displaystyle{ 0.2121 \overline{21} }$$ can be written as a geometric series $$\displaystyle{ \sum_{k=0}^{\infty}{ \frac{21}{100} \left[ \frac{1}{100} \right]^k } }$$ which converges to $$7/33$$.