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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.

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

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

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.

Dr Chris Tisdell - What is a Taylor series? [min-secs]

video by Dr Chris Tisdell

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.

PatrickJMT - Radius of Convergence for a Power Series [min-secs]

video by PatrickJMT

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, determine the radius and interval of convergence for each series.

Basic Problems

\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{x^n}{n!} } }\)

Problem Statement

\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{x^n}{n!} } }\)

Solution

In this video, he writes the geometric series in a little different form than we use on this page. He writes it as \( \sum{c_n(x-a)^n} \).

293 video

video by Dr Chris Tisdell

close solution

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^n}{n} (x-5)^n } }\)

Problem Statement

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^n}{n} (x-5)^n } }\)

Solution

In this video, he writes the geometric series in a little different form than we use on this page. He writes it as \( \sum{ c_n(x-a)^n } \).

294 video

video by Dr Chris Tisdell

close solution

\(\displaystyle{ \sum_{n=1}^{\infty}{ nx^n } }\)

Problem Statement

\(\displaystyle{ \sum_{n=1}^{\infty}{ nx^n } }\)

Solution

In this video, he writes the geometric series in a little different form than we use on this page. He writes it as \( \sum{ c_n (x-a)^n } \).

295 video

video by Dr Chris Tisdell

close solution

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{x^n}{\sqrt{n}} } }\)

Problem Statement

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{x^n}{\sqrt{n}} } }\)

Solution

316 video

video by PatrickJMT

close solution

\(\displaystyle{ e^{x^2} = \sum_{n=0}^{\infty}{ \frac{x^{2n}}{n!} } }\)

Problem Statement

\(\displaystyle{ e^{x^2} = \sum_{n=0}^{\infty}{ \frac{x^{2n}}{n!} } }\)

Solution

317 video

video by PatrickJMT

close solution

\(\displaystyle{ \sum_{n=0}^{\infty}{ n^3 (x-5)^n } }\)

Problem Statement

\(\displaystyle{ \sum_{n=0}^{\infty}{ n^3 (x-5)^n } }\)

Solution

344 video

video by PatrickJMT

close solution

\(\displaystyle{ \sum{ \left(\frac{x}{2}\right)^n } }\)

Problem Statement

\(\displaystyle{ \sum{ \left(\frac{x}{2}\right)^n } }\)

Solution

345 video

video by MIT OCW

close solution

\(\displaystyle{ \sum{ \frac{x^n}{n~2^n} } }\)

Problem Statement

\(\displaystyle{ \sum{ \frac{x^n}{n~2^n} } }\)

Solution

346 video

video by MIT OCW

close solution

\(\displaystyle{ \sum{ \frac{x^{2n}}{(2n)!} } }\)

Problem Statement

\(\displaystyle{ \sum{ \frac{x^{2n}}{(2n)!} } }\)

Solution

347 video

video by MIT OCW

close solution

\(\displaystyle{ \sum_{n=1}^{\infty}{ (-1)^n n x^n } }\)

Problem Statement

\(\displaystyle{ \sum_{n=1}^{\infty}{ (-1)^n n x^n } }\)

Solution

348 video

video by MIP4U

close solution

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^n x^n}{n} } }\)

Problem Statement

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^n x^n}{n} } }\)

Solution

349 video

video by MIP4U

close solution

\( \displaystyle{ \sum_{n=1}^{\infty}{ \frac{(x+1)^{2n}}{n!} } }\)

Problem Statement

\( \displaystyle{ \sum_{n=1}^{\infty}{ \frac{(x+1)^{2n}}{n!} } }\)

Solution

351 video

video by MIP4U

close solution

\( \displaystyle{ \sum_{n=1}^{\infty}{ \frac{5^n x^n}{n!} } }\)

Problem Statement

\( \displaystyle{ \sum_{n=1}^{\infty}{ \frac{5^n x^n}{n!} } }\)

Solution

374 video

video by Krista King Math

close solution

\( \displaystyle{ \sum_{n=1}^{\infty}{ \frac{x^n 2^n}{n!} } }\)

Problem Statement

\( \displaystyle{ \sum_{n=1}^{\infty}{ \frac{x^n 2^n}{n!} } }\)

Solution

376 video

video by PatrickJMT

close solution

Maclaurin series for \( \cos(x^2) \)

Problem Statement

Maclaurin series for \( \cos(x^2) \)

Solution

1364 video

video by Krista King Math

close solution

\(\displaystyle{ \sum_{k=1}^{\infty}{ \frac{x^k}{3^kk^2} } }\)

Problem Statement

\(\displaystyle{ \sum_{k=1}^{\infty}{ \frac{x^k}{3^kk^2} } }\)

Final Answer

\( [-3,3] \)

Problem Statement

\(\displaystyle{ \sum_{k=1}^{\infty}{ \frac{x^k}{3^kk^2} } }\)

Solution

2000 video

video by Dr Chris Tisdell

Final Answer

\( [-3,3] \)

close solution

Intermediate Problems

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(x-2)^n}{(n+1)3^n} } }\)

Problem Statement

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(x-2)^n}{(n+1)3^n} } }\)

Solution

350 video

video by MIP4U

close solution

\( \displaystyle{ \sum_{n=1}^{\infty}{ \frac{(x-4)^n}{\sqrt[3]{n}} } }\)

Problem Statement

\( \displaystyle{ \sum_{n=1}^{\infty}{ \frac{(x-4)^n}{\sqrt[3]{n}} } }\)

Solution

373 video

video by Krista King Math

close solution

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^3 (x+5)^n}{6^n} } }\)

Problem Statement

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^3 (x+5)^n}{6^n} } }\)

Solution

375 video

video by PatrickJMT

close solution

\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(x+2)^n}{n+3} } }\)

Problem Statement

\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(x+2)^n}{n+3} } }\)

Solution

377 video

video by PatrickJMT

close solution

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{x^n 5^n}{n^n} } }\)

Problem Statement

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{x^n 5^n}{n^n} } }\)

Solution

378 video

video by PatrickJMT

close solution

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(x-2)^n}{\ln(n+4)} } }\)

Problem Statement

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(x-2)^n}{\ln(n+4)} } }\)

Solution

379 video

video by PatrickJMT

close solution

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(x+3)^n}{n^2+2n} } }\)

Problem Statement

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(x+3)^n}{n^2+2n} } }\)

Solution

380 video

video by PatrickJMT

close solution

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{ (-1)^n (x+1)^n }{ 5^n \sqrt{n} } } }\)

Problem Statement

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{ (-1)^n (x+1)^n }{ 5^n \sqrt{n} } } }\)

Solution

381 video

video by PatrickJMT

close solution

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{ (-1)^n (x+2)^n }{ (n+1)\ln(n+1) } } }\)

Problem Statement

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{ (-1)^n (x+2)^n }{ (n+1)\ln(n+1) } } }\)

Solution

382 video

video by PatrickJMT

close solution
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