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A popular types of series are the power series, which are series of the form \(\sum_{n}^\infty c_nx^n\), whose convergence will depend on the value of \(x\). In this article, you will see how to find the values of \(x\), for which the power series converges. In other words, you will learn how to calculate its radius of convergence and its interval of convergence.
Definition of Radius of Convergence
As mentioned above, the convergence of the power series depends on the values of \(x\).
Given a power series
\[\sum\limits_{n=0}^\infty c_n(x-x_0)^n,\]
only one of the following three statements is true:
(a) The series converges only for \(x=x_0\).
(b) The series converges for all \(x\).
(c) There exists a number \(R>0\) such that the series converges for \(|x-x_0| < R\) and diverges for \(|x-x_0| > R\) .
The number \(R\) in case (c) is known as the radius of convergence of the power series, and the interval of convergence is the interval composed by all the points \(x\) where the series converges.
For a power series
\[\sum\limits_{n=1}^\infty c_n(x-x_0)^n ,\]
the relationship between the radius of convergence with the interval of convergence is shown in the table below.
Table 1. Relationship between radius of convergence and interval of convergence.
| Radius of convergence | Interval of convergence |
\(R=0\) | \(\{x_0\}\) |
\(R=\infty\) | \((-\infty,\infty)\) |
\(R>0\) | \((x_0-R,x_0+R)\),\( (x_0-R, x_0+R]\),\( [x_0-R, x_0+R)\), or \( [x_0-R, x_0+R]\) |
Radius of Convergence of Power Series
A power series centered at \(x_0\) (or a power series about \(x_0\)) is a series of the form
\[\sum_{n=0}^\infty c_n(x-x_0)^n=c_0+c_1(x-x_0)+c_2(x-x_0)^2+\dots,\]
where \(x\) is a variable, and \(x_0\) and \(c_n\) are real numbers. The \(c_n\)'s are called the coefficients of the series.
For each value of \(x\), the series can converge or diverge. The radius of convergence of a series is what will tell you for which values of \(x\) it converges.
Ratio Test and Radius of Convergence
To calculate the radius of convergence of a power series, you can use the Ratio Test (or sometimes the Root Test).
Given a series \(\sum\limits_n^\infty a_n\). Let
\[L=\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|,\]
1. if \(L<1\), then the series is absolutely convergence (and hence convergent).
2. if \(L>1\), then the series is divergent.
3. if \(L=1\), then the series may be divergent, conditionally convergent, or absolutely convergent.
Visit the Ratio Test and Root Test articles for more information!
Let's see an example of how to use the Ratio Test to obtain the radius of convergence.
Find the radius of convergence of the series
\[\sum\limits_{n=0}^\infty (3x)^n.\]
Answer:
Note that in this case, \(a_n=(3x)^n\), so
\[\begin{align}\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| &= \lim_{n\to\infty}\left|\frac{(3x)^{n+1}}{(3x)^n}\right| \\ &=\left|3x\right|. \end{align}\]
The Ratio Test states that it converges if \(|3x|<1\), that is, \(|x|<\dfrac{1}{3}\). Therefore, the radius of convergence is \(R=\dfrac{1}{3}\).
Radius of Convergence of Geometric Series
A special case of power series is the geometric series given by
\[\sum\limits_{n=0}^\infty ax^n,\]
where \(a\) is a constant.
You can calculate its radius of convergence using the Ratio Test just like for other power series. In this case, the terms of the series are given by \(a_n=ax^n\), so
\[\begin{align} \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| &=\lim_{n\to\infty}\left|\frac{ax^{n+1}}{ax^n}\right| \\ &=|x|. \end{align}\]
The Ratio Test says that it converges if \(|x|<1\), and therefore the radius of convergence is \(R=1\).
To learn more about this kind of series, visit the article Geometric Series.
Radius of Convergence of \(\sin (x)\)
To calculate the radius of convergence of the function \(\sin x\), remember that you can rewrite the function using its Taylor or Maclaurin series. The Maclaurin series of the function \(\sin x\) is given by
\[\begin{align} \sin x&=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}x^{2n+1} \\ &=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots \end{align}\]
To learn more about series of this kind, see Taylor series and Maclaurin series.
Using the Ratio Test
\[\begin{align}\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| & =\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}}{(2(n+1)+1)!}x^{2(n+1)+1}}{\frac{(-1)^n}{(2n+1)!}x^{2n+1}}\right| \\ &=\lim_{n\to\infty}\frac{(2n+1)!}{(2n+3)!}|x|^2 \\ &= \lim_{n\to\infty}\frac{1}{(2n+3)(2n+2)}|x|^2 \\ &=0. \end{align}\]
Since the limit is zero independently of \(x\), it means that the series converges for any value of \(x\), and therefore the series of \(\sin x\) has radius of convergence \(R=\infty\).
Examples of Radius of Convergence
Here are some examples.
For what values of \(x\) does the series
\[\sum\limits_{n=1}^\infty \frac{5^n(x-2)^n}{n}\]
converge?
Answer:
Note that the terms of the series are given by \(a_n=\dfrac{5^n(x-2)^n}{n}\). Calculating the limit
\[\begin{align} \lim_{n\to\infty} \left| \dfrac{\dfrac{5^{n+1}(x-2)^{n+1}}{n+1}}{\dfrac{5^n(x-2)^n}{n}}\right| &= 5|x-2| \lim_{n\to\infty}\frac{n}{n+1} \\ &= 5|x-2|.\end{align}\]
You can factor the \(5|x-2| \) out in front of the limit because it doesn't depend on \(n\). For a reminder of why the limit of \(\frac{n}{n+1}\) is equal to \(1\) see the article Limit of a Sequence.
By the Ratio Test, you have that the series converges when \(5|x-2|<1\), that is, when
\[\frac{9}{5} < x < \frac{11}{5} .\]
Finally, it remains to see what happens at the endpoints \(x=\dfrac{9}{5}\) and \(x= \dfrac{11}{5}\). You need to check those separately!
For \(x=\dfrac{9}{5}\) you get the alternating series
\[\begin{align}\sum\limits_{n=1}^\infty \frac{5^n(x-2)^n}{n} &= \sum\limits_{n=1}^\infty \frac{5^n\left(\dfrac{9}{5} -2\right)^n}{n} \\ &= \sum\limits_{n=1}^\infty \frac{5^n\left(-\dfrac{1}{5} \right)^n}{n} \\ &= \sum\limits_{n=1}^\infty \frac{(-1)^n}{n} , \end{align}\]
which is convergent. That means \(x=\dfrac{9}{5}\) is in the interval of convergence.
On the other hand, for \(x=\dfrac{11}{5}\) you get the harmonic series
\[\begin{align} \sum\limits_{n=1}^\infty \frac{5^n(x-2)^n}{n} &= \sum\limits_{n=1}^\infty \frac{5^n\left(\dfrac{11}{5} -2\right)^n}{n} \\ &= \sum\limits_{n=1}^\infty \frac{5^n\left(\dfrac{1}{5}\right)^n}{n} \\ &= \sum\limits_{n=1}^\infty\frac{1}{n}, \end{align}\]
which diverges. So \(x=\dfrac{11}{5}\) is not in the interval of convergence.
Therefore, the values of \(x\) for which the series converges are
\[\frac{9}{5} \leq x < \frac{11}{5},\]
and the interval of convergence is
\[ \left[ \frac{9}{5} ,\frac{11}{5} \right). \]
One more example.
Find the radius of convergence and interval of convergence of the series
\[\sum\limits_{n=0}^\infty \left(\frac{x+3}{7}\right)^n .\]
Answer:
Let's calculate the limit first:
\[\lim_{n\to\infty} \left| \frac{\dfrac{(x+3)^{n+1}}{7^{n+1}}}{\dfrac{(x+3)^{n}}{7^{n}}}\right|=\frac{|x+3|}{7}.\]
By the Ratio Test, the series converges for \(x\) where \(\dfrac{|x+3|}{7} < 1\), which is the same as \(|x+3|<7\). This can be written as \(-7<x+3<7\), which means you need \(-10<x<4\). Therefore, the radius of convergence is \(R=7\).
When you find the interval of convergence, don't forget to check the endpoints! For \(x=-10\), you have
\[\begin{align} \sum\limits_{n=0}^\infty \left(\frac{x+3}{7}\right)^n &= \sum\limits_{n=0}^\infty \left(\frac{-10+3}{7}\right)^n \\ &= \sum\limits_{n=1}^\infty (-1)^n , \end{align}\]
which is an alternating series that diverges. So \(x=-10\) is not in the interval of convergence.
If \(x=4\) you get the series
\[\begin{align} \sum\limits_{n=0}^\infty \left(\frac{x+3}{7}\right)^n &= \sum\limits_{n=0}^\infty \left(\frac{4+3}{7}\right)^n \\ &= \sum\limits_{n=1}^\infty 1 , \end{align}\]
which is also divergent.
Thus, the interval of convergence is \((-10,4)\).
Radius of Convergence - Key takeaways
- Let \(\sum\limits_{n=0}^\infty c_n(x-x_0)^n\) be a power series about \(x_0\).
- The radius of convergence of the series is a real value \(R>0\), for which the series converges for all \(x\) such that \(|x-x_0| < R\) and diverges for all \(x\) such that \(|x-x_0| > R\).
- If the series only converges for \(x=x_0\), then \(R=0\).
- If the series converges for all the values of \(x\), then \(R=\infty\).
- The interval of convergence of a power series is the interval composed by all the points \(x\) where the series converges.
- To calculate the radius of convergence, you can use the Ratio Test or the Root Test.
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Frequently Asked Questions about Radius of Convergence
How to find radius of convergence?
To find the radius of convergence, you can use the ratio test or the root test.
How to find radius of convergence of power series?
To find the radius of convergence, you can use the ratio test. You can find examples in our article Radius and Interval of convergence.
What is the radius of convergence?
Given a power series Σa_n(x-c), the radius of convergence is a real value R>0, such that if x satisfies |x-c|<R then the series converges, while if |x-c|>R then the series diverges.
Can the radius of convergence be infinity?
Yes, examples of series that have radius of convergence equal to infinity are sine, cosine, exponential, among others.
What does the radius of convergence tell us?
The radius of convergence of a series tells you for which values of x the series converges.
Is the radius of convergence always 1?
No, the radius of convergence depends on the series. In fact, for any real number R>0, one can construct a series with radius of convergence exactly equal to R.
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