In order to have a complete picture of what is going on with a power series (and Taylor series, since Taylor series is a special case of power series), we need to know the radius and interval of convergence. This is analogous to knowing the domain of a function.
Topics You Need To Understand For This Page |
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infinite series (including all the tests) power series taylor series |
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Difference Between Radius and Interval of Convergence
The difference between the radius of convergence and the interval of convergence lies in what information we have about the endpoints.
radius of convergence |
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gives convergence information only between x-values, not AT the x-values |
Example: if \((5,7)\) is the radius of convergence |
the series converges for \(5 < x < 7\) |
says nothing about the endpoints |
interval of convergence |
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gives convergence information inside the interval AND convergence/divergence information at the endpoints |
Example: if \((5,7)\) is the interval of convergence |
the series converges for \(5 < x < 7\) AND we know that the series diverges at \(x = 5\) and \(x = 7\) |
contains the same information as the radius of convergence as well as what is going on at the endpoints |
The interval of convergence is sometimes called the convergence set. |
Once the Taylor series or power series is calculated, we use the ratio test to determine the radius convergence and other tests to determine the interval of convergence. Without knowing the radius and interval of convergence, the series is not considered a complete function (This is similar to not knowing the domain of a function. As we say in the page on domain and range, a fully defined function always contains information about the domain, either implicitly or explicitly stated.) In this case, we can't leave off information about where the series converges because the series may not hold for all values of \(x\).
Using The Ratio Test
The ratio test looks like this. We have a series \( \sum{a_n} \) with non-zero terms. We calculate the limit
\( \displaystyle{\lim_{n \to \infty}{\left| \frac{a_{n+1}}{a_n} \right|} = L} \).
There are three possible cases for the value \(L\).
\( L < 1 \): The series converges absolutely.
\( L = 1 \): The ratio test is inconclusive.
\( L > 1 \): The series diverges.
So we use the first case (\( L < 1 \), since we want convergence) and we set up the inequality \( \displaystyle{\lim_{n \to \infty}{\left| \frac{a_{n+1}}{a_n} \right|} < 1} ~~~~~ [ 1 ]\)
Key: Do not drop the absolute values.
Radius of Convergence
To find the radius of convergence, we need to simplify the inequality [1] to the point that we have \( \left| x-a \right| < R \). This gives the radius of convergence as \(R\). This seems very simple but you need to be careful of the notation and wording your textbooks. Some textbooks use a small \(r\). Some textbooks ask for the ratio of convergence in which case you need to give the answer as \( \rho = 1/R \). This has always seemed kind of strange to me but there must be some reason for it. However, on this site, whenever we talk about the radius of convergence in this context, we will use \(R\) as defined above.
There are three possible cases for the radius of convergence.
\(R = 0\) |
series converges only at the point \(x = a\) |
\( 0 < R < \infty \) |
series converges within the interval |
\(R = \infty \) |
series converges for all \(x\) |
The interesting thing is that we have a strict inequality in \( 0 < R < \infty \) and, because of the definition of the ratio test, we have no idea what happens when \( \left| x-a \right| = R \). The series could converge or diverge. The ratio test doesn't give us a clue on what happens in that case. That's where we need to find the interval of convergence, which we discuss next.
Interval of Convergence
We use the radius of convergence, \(R\), to calculate the interval of convergence as follows
\(\begin{array}{rcccl}
& & \left| x-a \right| & < & R \\
-R & < & x-a & < & R \\
-R + a & < & x & < & R + a
\end{array}\)
So now we have an open interval \( (-R+a, R+a) \) in which the series converges. Now we need to look at the endpoints to determine what goes on. To do that we substitute each endpoint individually for \(x\) into the series and then use the other series test to determine convergence or divergence.
Notice when we substitute \(x=-R+a\) into the \((x-a)^n\) term, we end up with \((-R)^n\) which can be simplified to \((-1)^n R^n\). Now we have an alternating series. So often, the alternating series test can be used to determine convergence or divergence. The point is that, using other tests, we need to definitively determine convergence or divergence at each endpoint. The result will be an open interval, a half-open interval or a closed interval. We call this interval, the interval of convergence.
Notice the difference between the terms radius of convergence and interval of convergence. The radius of convergence gives information about the open interval but says nothing about the endpoints. The interval of convergence includes the radius of convergence AND information about convergence or divergence of the endpoints.
Here is a video clip that explains how to show that a series converges for all x.
video by Dr Chris Tisdell |
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Sometimes you are given the interval of convergence and asked to determine the radius convergence. Here is a short video explaining how to do that.
video by PatrickJMT |
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Okay, let's try some practice problems.
Practice
Unless otherwise instructed, determine the radius and interval of convergence for each series.
Basic
\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{x^n}{n!} } }\)
Problem Statement
Determine the radius and interval of convergence for the series \(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{x^n}{n!} } }\).
Solution
In this video, he writes the geometric series in a little different form than we use. He writes it as \( \sum{c_n(x-a)^n} \).
video by Dr Chris Tisdell |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^n}{n} (x-5)^n } }\)
Problem Statement |
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Determine the radius and interval of convergence for the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^n}{n} (x-5)^n } }\)
Final Answer |
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radius of convergence \(R=1\), interval of convergence \((4,6]\)
Problem Statement
Determine the radius and interval of convergence for the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^n}{n} (x-5)^n } }\)
Solution
Set up the ratio test and require the limit be less than one. |
For convergence \( L < 1 \) |
In this problem \(a_n = \dfrac{(-1)^n}{n}(x-5)^n\) |
and \(a_{n+1} = \dfrac{(-1)^{n+1}}{n+1}(x-5)^{n+1}\) |
Set up the limit |
\(\displaystyle{ \lim_{n \to \infty}{ \left| \frac{(-1)^{n+1}(x-5)^{n+1}/(n+1)}{(-1)^n (x-5)^n/n} \right| } }\) |
\(\displaystyle{ \lim_{n \to \infty}{ \left| \frac{n}{n+1} \frac{(x-5)^{n+1}}{(x-5)^n} \right| } }\) |
\(\displaystyle{ \lim_{n \to \infty}{ \left| \frac{n}{n+1} (x-5) \right| } }\) |
\(\displaystyle{ |x-5| \lim_{n \to \infty}{ \left| \frac{n}{n+1} \frac{1/n}{1/n} \right| } }\) |
\(\displaystyle{ |x-5| \lim_{n \to \infty}{ \left| \frac{1}{1+1/n} \right| } }\) |
\(\displaystyle{ |x-5| \lim_{n \to \infty}{ \left| 1 \right| } }\) |
\(\displaystyle{ |x-5| < 1 }\) |
radius of convergence \(R = 1\) |
Now we need to find the interval of convergence. |
\( |x-5| < 1 \) |
\( -1 < x-5 < 1 \) |
\( 4 < x < 6 \) |
Test the endpoints \(x=4\) and \(x=6\) |
First \(x=4\) |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^n}{n} (4-5)^n } }\) |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^n}{n} (-1)^n } }\) |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n} } }\) |
This is a p-series with \(p=1\), which diverges. So \(x=4\) is not in the interval of convergence. |
Now let's look at \(x=6\). |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^n}{n} (6-5)^n } }\) |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^n}{n} } }\) |
This series converges by the alternating series test. (See practice problem 173.) |
So the interval of convergence is \( (4,6] \), which we can also write as \( 4 \lt x \leq 6 \) |
Final Answer
radius of convergence \(R=1\), interval of convergence \((4,6]\)
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\(\displaystyle{ \sum_{n=1}^{\infty}{ nx^n } }\)
Problem Statement
Determine the radius and interval of convergence for the series \(\displaystyle{ \sum_{n=1}^{\infty}{ nx^n } }\).
Solution
In this video, he writes the geometric series in a little different form than we use. He writes it as \( \sum{ c_n (x-a)^n } \).
video by Dr Chris Tisdell |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{x^n}{\sqrt{n}} } }\)
Problem Statement
Determine the radius and interval of convergence for the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{x^n}{\sqrt{n}} } }\).
Solution
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\(\displaystyle{ e^{x^2} = \sum_{n=0}^{\infty}{ \frac{x^{2n}}{n!} } }\)
Problem Statement
Determine the radius and interval of convergence for the series \(\displaystyle{ e^{x^2} = \sum_{n=0}^{\infty}{ \frac{x^{2n}}{n!} } }\).
Solution
video by PatrickJMT |
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\(\displaystyle{ \sum_{n=0}^{\infty}{ n^3 (x-5)^n } }\)
Problem Statement |
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Determine the radius and interval of convergence for the series \(\displaystyle{ \sum_{n=0}^{\infty}{ n^3 (x-5)^n } }\)
Final Answer |
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radius of convergence \(R=1\), interval of convergence \((4,6)\)
Problem Statement
Determine the radius and interval of convergence for the series \(\displaystyle{ \sum_{n=0}^{\infty}{ n^3 (x-5)^n } }\)
Solution
\(\displaystyle{ \sum_{n=0}^{\infty}{ n^3 (x-5)^n } }\) |
\(\displaystyle{ \lim_{n\to\infty}{ \left| \frac{(n+1)^3 (x-5)^{n+1}}{n^3 (x-5)^n} \right| } }\) |
\(\displaystyle{ \lim_{n\to\infty}{ \left| \frac{(n+1)^3}{n^3} \frac{(x-5)^{n+1}}{(x-5)^n} \right| } }\) |
\(\displaystyle{ \lim_{n\to\infty}{ \left| (1+1/n)^3 (x-5) \right| } }\) |
\(\displaystyle{ \left| x-5 \right| \lt 1 }\) |
\(\displaystyle{ -1 \lt x-5 \lt 1 }\) |
radius of convergence \(R = 1\), \(( 4, 6 )\) |
For the interval of convergence, we need to test the endpoints. |
\(x=4\) |
\(\displaystyle{ \sum_{n=0}^{\infty}{ n^3 (4-5)^n } }\) |
\(\displaystyle{ \sum_{n=0}^{\infty}{ n^3 (-1)^n } }\) |
By the divergence test, the series diverges, since \(\sum a_n \neq 0\) |
\(x=6\) |
\(\displaystyle{ \sum_{n=0}^{\infty}{ n^3 (6-5)^n } }\) |
\(\displaystyle{ \sum_{n=0}^{\infty}{ n^3 } }\) |
By the divergence test, the series diverges, since \(\sum a_n \neq 0\) |
The series diverges at both endpoints, so the interval of convergence is \((4,6)\) |
Final Answer
radius of convergence \(R=1\), interval of convergence \((4,6)\)
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\(\displaystyle{ \sum{ \left(\frac{x}{2}\right)^n } }\)
Problem Statement
Determine the radius and interval of convergence for the series \(\displaystyle{ \sum{ \left(\frac{x}{2}\right)^n } }\)
Solution
video by MIT OCW |
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video by Dr Chris Tisdell |
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\(\displaystyle{ \sum{ \frac{x^n}{n~2^n} } }\)
Problem Statement
Determine the radius and interval of convergence for the series \(\displaystyle{ \sum{ \frac{x^n}{n~2^n} } }\).
Solution
video by MIT OCW |
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\(\displaystyle{ \sum{ \frac{x^{2n}}{(2n)!} } }\)
Problem Statement
Determine the radius and interval of convergence for the series \(\displaystyle{ \sum{ \frac{x^{2n}}{(2n)!} } }\).
Solution
video by MIT OCW |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ (-1)^n n x^n } }\)
Problem Statement
Determine the radius and interval of convergence for the series \(\displaystyle{ \sum_{n=1}^{\infty}{ (-1)^n n x^n } }\).
Solution
video by MIP4U |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^n x^n}{n} } }\)
Problem Statement
Determine the radius and interval of convergence for the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^n x^n}{n} } }\).
Solution
video by MIP4U |
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\( \displaystyle{ \sum_{n=1}^{\infty}{ \frac{(x+1)^{2n}}{n!} } }\)
Problem Statement
Determine the radius and interval of convergence for the series \( \displaystyle{ \sum_{n=1}^{\infty}{ \frac{(x+1)^{2n}}{n!} } }\).
Solution
video by MIP4U |
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\( \displaystyle{ \sum_{n=1}^{\infty}{ \frac{5^n x^n}{n!} } }\)
Problem Statement
Determine the radius and interval of convergence for the series \( \displaystyle{ \sum_{n=1}^{\infty}{ \frac{5^n x^n}{n!} } }\).
Solution
video by Krista King Math |
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\( \displaystyle{ \sum_{n=1}^{\infty}{ \frac{x^n 2^n}{n!} } }\)
Problem Statement
Determine the radius and interval of convergence for the series \( \displaystyle{ \sum_{n=1}^{\infty}{ \frac{x^n 2^n}{n!} } }\).
Solution
video by PatrickJMT |
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\( \cos(x^2) \)
Problem Statement
Determine the radius and interval of convergence of the Maclaurin series for \( \cos(x^2) \).
Solution
video by Krista King Math |
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\(\displaystyle{ \sum_{k=1}^{\infty}{ \frac{x^k}{3^kk^2} } }\)
Problem Statement
Determine the radius and interval of convergence for the series \(\displaystyle{ \sum_{k=1}^{\infty}{ \frac{x^k}{3^kk^2} } }\).
Solution
video by Dr Chris Tisdell |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(x^n) \sqrt{n}}{n!} } }\)
Problem Statement
Determine the radius and interval of convergence for the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(x^n) \sqrt{n}}{n!} } }\)
Solution
video by JDS1854 |
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Intermediate
\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-2)^n (x-2)^n}{3n+1} } }\)
Problem Statement
Determine the radius and interval of convergence for the series \(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-2)^n (x-2)^n}{3n+1} } }\)
Solution
video by JDS1854 |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(x-2)^n}{(n+1)3^n} } }\)
Problem Statement
Determine the radius and interval of convergence for the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(x-2)^n}{(n+1)3^n} } }\).
Solution
video by MIP4U |
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\( \displaystyle{ \sum_{n=1}^{\infty}{ \frac{(x-4)^n}{\sqrt[3]{n}} } }\)
Problem Statement
Determine the radius and interval of convergence for the series \( \displaystyle{ \sum_{n=1}^{\infty}{ \frac{(x-4)^n}{\sqrt[3]{n}} } }\).
Solution
video by Krista King Math |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^3 (x+5)^n}{6^n} } }\)
Problem Statement
Determine the radius and interval of convergence for the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^3 (x+5)^n}{6^n} } }\).
Solution
video by PatrickJMT |
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\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(x+2)^n}{n+3} } }\)
Problem Statement
Determine the radius and interval of convergence for the series \(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(x+2)^n}{n+3} } }\).
Solution
video by PatrickJMT |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{x^n 5^n}{n^n} } }\)
Problem Statement
Determine the radius and interval of convergence for the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{x^n 5^n}{n^n} } }\).
Solution
video by PatrickJMT |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(x-2)^n}{\ln(n+4)} } }\)
Problem Statement
Determine the radius and interval of convergence for the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(x-2)^n}{\ln(n+4)} } }\).
Solution
video by PatrickJMT |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(x+3)^n}{n^2+2n} } }\)
Problem Statement
Determine the radius and interval of convergence for the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(x+3)^n}{n^2+2n} } }\).
Solution
video by PatrickJMT |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{ (-1)^n (x+1)^n }{ 5^n \sqrt{n} } } }\)
Problem Statement
Determine the radius and interval of convergence for the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{ (-1)^n (x+1)^n }{ 5^n \sqrt{n} } } }\).
Solution
video by PatrickJMT |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{ (-1)^n (x+2)^n }{ (n+1)\ln(n+1) } } }\)
Problem Statement
Determine the radius and interval of convergence for the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{ (-1)^n (x+2)^n }{ (n+1)\ln(n+1) } } }\).
Solution
video by PatrickJMT |
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