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Geometric Series 
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► geometric series convergence ► how to use geometric series ► finite geometric series ► additional topics

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 

If the series converges, it converges to \(\displaystyle{\frac{1}{1r}} \). 
This is usually written \(\displaystyle{ \sum_{n=0}^{\infty}{r^n} = \frac{1}{1r}, ~~ 0 \lt \abs{r} \lt 1 }\). 
Quick Summary
used to prove convergence  yes 
used to prove divergence  yes 
can be inconclusive  no 
can find convergence value  yes 
useful for finding power series 

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 \( \displaystyle{\frac{1}{1r}} \)
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.
In the following discussion, we are assuming that \( 0 < r < 1 \).
The equation for the value of a finite geometric series is
\(\displaystyle{ \sum_{n=0}^{k}{r^n} = \frac{1r^{k+1}}{1r} ~~~~~ (1) }\)
where \(k\) is a finite positive integer.
Practice 1 
Show that the above formula holds for \(k=1, k=2\) and \(k=3\). 

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
\(\displaystyle{ \frac{1r^{1+1}}{1r} = \frac{1r^2}{1r} = \frac{(1+r)(1r)}{1r} = 1+r }\)
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
\(\displaystyle{ \frac{1r^{1+2}}{1r} = \frac{1r^3}{1r} = \frac{(1+r+r^2)(1r)}{1r} = 1+r+r^2 }\)
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
\( (1r)/(1r)\).
\( \begin{array}{rcl}
1+r+r^2+r^3 & = & \displaystyle{(1+r+r^2+r^3)\frac{1r}{1r}} \
& = & \displaystyle{\frac{(1+r+r^2+r^3)r(1+r+r^2+r^3)}{1r}} \
& = & \displaystyle{\frac{1+r+r^2+r^3 r  r^2  r^3  r^4}{1r}} \
& = & \displaystyle{\frac{1r^4}{1r}}
\end{array} \)
Notice that all the inside terms cancel and we are left with \(1r^4\) in the numerator.
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 \(1r^{n+1}\). So now we have \(S_nrS_n = 1r^{n+1}\). Now we solve for \(S_n\).
\(\begin{array}{rcl}
S_nrS_n & = & 1r^{n+1} \\
S_n(1r) & = & 1r^{n+1} \\
S_n & = & \displaystyle{ \frac{1r^{n+1}}{1r} }
\end{array}\)
\(\displaystyle{ \sum_{k=0}^{n}{r^k} = \frac{1r^{k+1}}{1r} }\) [qed]
Additional Geometric Series Topics 
video: Financial Application of Geometric Series
Financial Application of Geometric Series
 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. 

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.
Practice A01 
\(\displaystyle{\sum_{n=0}^{\infty}{\frac{3}{4^n}}}\) 


The series \( \displaystyle{ \sum_{n=0}^{\infty}{ \frac{3}{4^n} } }\) converges to 4 by the Geometric Series Test. 
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 \(\leftr\right=1/4 < 1\), which converges by the geometric series test.
Additionally, we can determine that it converges to
\(\displaystyle{ \frac{3}{1(1/4)} = \frac{3}{3/4} = 4 }\)
Practice A01 Final Answer 
The series \( \displaystyle{ \sum_{n=0}^{\infty}{ \frac{3}{4^n} } }\) converges to 4 by the Geometric Series Test. 
Practice A02 
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^n}{e^n} } }\) 


The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^n}{e^n} } }\) converges to \(\displaystyle{\frac{2}{e2}}\) by the Geometric Series Test. 
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
\(\leftr\right=2/e < 1 \) so the series converges.
Convergence Value
When determining only convergence or divergence, you don't need to worry about the starting nvalue. 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).
\(\displaystyle{ \frac{1}{1(2/e)}  1 = \frac{e}{e2}  \frac{e2}{e2} = \frac{2}{e2} }\)
Practice A02 Final Answer 
The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^n}{e^n} } }\) converges to \(\displaystyle{\frac{2}{e2}}\) by the Geometric Series Test. 
Practice A03 
\(\displaystyle{\sum_{n=1}^{\infty}{\left[\frac{1+\sqrt{5}}{2}\right]^n}}\) 


The geometric series \(\displaystyle{\sum_{n=1}^{\infty}{\left[\frac{1+\sqrt{5}}{2}\right]^n}}\) diverges. 
Practice A03 Final Answer 
The geometric series \(\displaystyle{\sum_{n=1}^{\infty}{\left[\frac{1+\sqrt{5}}{2}\right]^n}}\) diverges. 
Practice A04 
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 \(5/3\). 
Practice A04 Final Answer 
The geometric series \(\displaystyle{ 1+0.4+0.16+0.064+ . . . }\) converges to \(5/3\). 
Practice A05 
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\pi^n }{ 3^{n+2}} } }\) 


The geometric series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\pi^n }{ 3^{n+2}} } }\) diverges. 
Practice A05 Final Answer 
The geometric series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\pi^n }{ 3^{n+2}} } }\) diverges. 
Practice A06 
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{ 3^n+2^n }{ 6^n } } }\) 


The geometric series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{ 3^n+2^n }{ 6^n } } }\) converges. 
Practice A06 Final Answer 
The geometric series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{ 3^n+2^n }{ 6^n } } }\) converges. 
Practice A07 
Express this sum using sigma notation \(\displaystyle{ 1+0.1+0.01+0.001+0.0001+ . . . }\) 


\(\displaystyle{ 1+0.1+0.01+0.001+0.0001+ . . . = \sum_{n=0}^{\infty}{\frac{1}{10^n}} }\) 
Practice A07 Final Answer 
\(\displaystyle{ 1+0.1+0.01+0.001+0.0001+ . . . = \sum_{n=0}^{\infty}{\frac{1}{10^n}} }\) 
Practice A08 
\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{ 1 }{ 5^n } } }\) 


The geometric series \(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{ 1 }{ 5^n } } }\) converges to \( 5/4 \). 
Practice A08 Final Answer 
The geometric series \(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{ 1 }{ 5^n } } }\) converges to \( 5/4 \). 
Practice A09 
Express this sum using sigma notation \(\displaystyle{ 2/13  4/13^2 + 8/13^3  16/13^4 + . . . }\) 


\(\displaystyle{ 2/13  4/13^2 + 8/13^3  16/13^4 + . . . = \sum_{n=1}^{\infty}{ \left[ \frac{(1)^{n+1} (2)^n}{(13)^n} \right] } }\) 
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 \((1)^{n+1} (2)^n\).
Practice A09 Final Answer 
\(\displaystyle{ 2/13  4/13^2 + 8/13^3  16/13^4 + . . . = \sum_{n=1}^{\infty}{ \left[ \frac{(1)^{n+1} (2)^n}{(13)^n} \right] } }\) 
Practice A10 
\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{3^n}{4^n} + \frac{2}{7^n} \right] } }\) 


The geometric series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{3^n}{4^n} + \frac{2}{7^n} \right] } }\) converges. 
Practice A10 Final Answer 
The geometric series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{3^n}{4^n} + \frac{2}{7^n} \right] } }\) converges. 
Practice A11 
\(\displaystyle{ \sum_{k=0}^{\infty}{ \left( \frac{2}{3} \right)^k } }\) 


The geometric series \(\displaystyle{ \sum_{k=0}^{\infty}{ \left( \frac{2}{3} \right)^k } }\) converges to \(3\). 
Practice A11 Final Answer 
The geometric series \(\displaystyle{ \sum_{k=0}^{\infty}{ \left( \frac{2}{3} \right)^k } }\) converges to \(3\). 
Practice A12 
\(\displaystyle{ 1 + e^{1} + }\) \(\displaystyle{ e^{2} + e^{3} + . . . }\) 


The geometric series \(\displaystyle{1 + e^{1} + e^{2} + e^{3} + . . .}\) converges to \(\displaystyle{\frac{e}{e1}}\). 
Practice A12 Final Answer 
The geometric series \(\displaystyle{1 + e^{1} + e^{2} + e^{3} + . . .}\) converges to \(\displaystyle{\frac{e}{e1}}\). 
Practice A13 
\(\displaystyle{ \frac{3}{2}  \frac{3}{4} + \frac{3}{8}  \frac{3}{16} + . . . }\) 


The geometric series \(\displaystyle{ \frac{3}{2}  \frac{3}{4} + \frac{3}{8}  \frac{3}{16} + . . . }\) converges to \( 1 \). 
Practice A13 Final Answer 
The geometric series \(\displaystyle{ \frac{3}{2}  \frac{3}{4} + \frac{3}{8}  \frac{3}{16} + . . . }\) converges to \( 1 \). 
Practice A14 
\(\displaystyle{ 1  \frac{1}{5} + \left( \frac{1}{5} \right)^2  \left( \frac{1}{5} \right)^3 + . . . }\) 


The geometric series \(\displaystyle{ 1  \frac{1}{5} + \left( \frac{1}{5} \right)^2  \left( \frac{1}{5} \right)^3 + . . . }\) converges to \(\displaystyle{ \frac{5}{6} }\). 
Practice A14 Final Answer 
The geometric series \(\displaystyle{ 1  \frac{1}{5} + \left( \frac{1}{5} \right)^2  \left( \frac{1}{5} \right)^3 + . . . }\) converges to \(\displaystyle{ \frac{5}{6} }\). 
Practice B01 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{(1)^n2^{n1}}{3^n}}}\) 


The geometric series \(\displaystyle{\sum_{n=1}^{\infty}{\frac{(1)^n2^{n1}}{3^n}}}\) converges absolutely to \(1/5\). 
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 \(2^{n1}\). This can be written as \(\displaystyle{ 2^{n1} = (2^n)(2^{1}) = \frac{2^n}{2}}\). So, now we can move things around so that we can easily see the geometric series.
\(\displaystyle{ \frac{(1)^n2^{n1}}{3^n} = \frac{(1)^n2^n}{3^n \cdot 2} = \frac{1}{2} \left( \frac{2}{3} \right)^n }\)
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 \(\leftr\right=2/3<1\), 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 \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^{n1}}{3^n} } }\) converges. Using similar algebra as we did above, this series reduces to a geometric series with \( r = 2/3 \). Since \(\leftr=2/3\right < 1 \), this series converges. Therefore, the original series converges absolutely.
Practice B01 Final Answer 
The geometric series \(\displaystyle{\sum_{n=1}^{\infty}{\frac{(1)^n2^{n1}}{3^n}}}\) converges absolutely to \(1/5\). 
Practice B02 
\(\displaystyle{ \sum_{n=3}^{\infty}{ 5 (2/3)^{n1} } }\) 


The geometric series \(\displaystyle{ \sum_{n=3}^{\infty}{ 5 (2/3)^{n1} } }\) converges to \(20/3\). 
Practice B02 Final Answer 
The geometric series \(\displaystyle{ \sum_{n=3}^{\infty}{ 5 (2/3)^{n1} } }\) converges to \(20/3\). 
Practice B03 
\(\displaystyle{ \sum_{n=3}^{\infty}{ \frac{ \pi^{n+1} }{ 6^n } } }\) 


The geometric series \(\displaystyle{ \sum_{n=3}^{\infty}{ \frac{ \pi^{n+1} }{ 6^n } } }\) converges to \(\displaystyle{ \frac{ \pi^4 }{ 36(6\pi) } }\). 
Practice B03 Final Answer 
The geometric series \(\displaystyle{ \sum_{n=3}^{\infty}{ \frac{ \pi^{n+1} }{ 6^n } } }\) converges to \(\displaystyle{ \frac{ \pi^4 }{ 36(6\pi) } }\). 
Practice B04 
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 \( 7/33 \) 
Practice B04 Final Answer 
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 \( 7/33 \) 