M
Math Coach Amy
Amy Ferguson Moncure

Calculus BC ยท Unit 10

Interval of Convergence

The ratio test gives the open interval. Each endpoint is its own series and needs its own test.

Jump to section:

Key Ideas Easy Medium Hard

\(|x-a|

What to Remember

Radius first

The ratio test on \(\sum c_n(x-a)^n\) produces an inequality \(|x-a|

Endpoints are separate series

At each endpoint, substitute the number and test that number series. A p-series, an alternating series, or the nth-term test is often enough.

Infinite and zero radius

If the limit is 0 for every \(x\), the interval is all real numbers. If the series converges only at the center, the radius is 0.

Inside the radius

Convergence inside the open interval is absolute. Conditional convergence can happen only at an endpoint.

Level 1

Easy

Find the open interval, then test both ends.

Example
Find the interval of convergence of \(\sum_{n=1}^{\infty}\dfrac{(x-2)^n}{n}\).

\(\left|\dfrac{a_{n+1}}{a_n}\right|=|x-2|\cdot\dfrac{n}{n+1}\)

The limit is \(|x-2|\).

Convergence needs \(|x-2|<1\), so \(1

At \(x=3\), the series is \(\sum 1/n\), which diverges.

At \(x=1\), the series is \(\sum (-1)^n/n\), which converges.

The interval is \([1,3)\).

Answer: \([1,3)\)

1
Find the interval of convergence of \(\sum_{n=0}^{\infty}\dfrac{x^n}{n!}\).
Show Answer

\(\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{|x|}{n+1}\)

The limit is 0 for every real \(x\).

The series converges for all real numbers.

Answer: all real numbers

2
Find the interval of convergence of \(\sum_{n=0}^{\infty} n!\,x^n\).
Show Answer

\(\left|\dfrac{a_{n+1}}{a_n}\right|=|x|(n+1)\)

For \(x\neq 0\), the limit is infinity, so the series diverges.

At \(x=0\), every term after the first is 0, so the series converges.

The interval is \(\{0\}\).

Answer: \(\{0\}\)

3
Find the interval of convergence of \(\sum_{n=1}^{\infty}\dfrac{x^n}{n^2}\).
Show Answer

The ratio limit is \(|x|\), so \(|x|<1\).

At \(x=1\), \(\sum 1/n^2\) converges.

At \(x=-1\), \(\sum (-1)^n/n^2\) converges absolutely.

The interval is \([-1,1]\).

Answer: \([-1,1]\)

4
Find the interval of convergence of \(\sum_{n=0}^{\infty}\dfrac{(x-1)^n}{3^n}\).
Show Answer

The ratio limit is \(|x-1|/3\).

\(|x-1|<3\), so \(-2

At \(x=4\), the terms are 1 and do not go to 0.

At \(x=-2\), the terms are \((-1)^n\) and do not go to 0.

The interval is \((-2,4)\).

Answer: \((-2,4)\)

5
What is the radius of convergence of \(\sum_{n=0}^{\infty}\left(\dfrac{x}{2}\right)^n\)?
A. \(1\)
B. \(2\)
C. \(1/2\)
D. the series converges for every x
Show Answer

\(|x/2|<1\)

\(|x|<2\)

The radius is 2.

Answer: B. \(2\)

Level 2

Medium

A coefficient that changes the radius, or an endpoint whose terms do not go to 0.

Example
Find the interval of convergence of \(\sum_{n=1}^{\infty}\dfrac{x^n}{n}\).

The ratio limit is \(|x|\), so \(|x|<1\).

At \(x=1\), \(\sum 1/n\) diverges.

At \(x=-1\), \(\sum (-1)^n/n\) converges.

The interval is \((-1,1]\).

Answer: \((-1,1]\)

1
Find the interval of convergence of \(\sum_{n=1}^{\infty}\dfrac{(2x)^n}{n}\).
Show Answer

The ratio limit is \(|2x|\).

\(|x|<\dfrac{1}{2}\)

At \(x=1/2\), the series is \(\sum 1/n\), which diverges.

At \(x=-1/2\), the series is \(\sum (-1)^n/n\), which converges.

The interval is \(\left[-\dfrac{1}{2},\dfrac{1}{2}\right)\).

Answer: \(\left[-\dfrac12,\dfrac12\right)\)

2
Find the interval of convergence of \(\sum_{n=1}^{\infty} n(x-3)^n\).
Show Answer

The ratio limit is \(|x-3|\), so \(2

At \(x=4\), the terms are \(n\), which do not go to 0.

At \(x=2\), the terms are \(n(-1)^n\), which do not go to 0.

The interval is \((2,4)\).

Answer: \((2,4)\)

3
Find the interval of convergence of \(\sum_{n=1}^{\infty}\dfrac{(x+1)^n}{n^3}\).
Show Answer

The ratio limit is \(|x+1|\), so \(-2

At \(x=0\), \(\sum 1/n^3\) converges.

At \(x=-2\), \(\sum (-1)^n/n^3\) converges.

The interval is \([-2,0]\).

Answer: \([-2,0]\)

4
Find the interval of convergence of \(\sum_{n=0}^{\infty}\dfrac{(x-5)^n}{4^n}\).
Show Answer

The ratio limit is \(|x-5|/4\).

\(|x-5|<4\), so \(1

At each endpoint the terms have absolute value 1 and do not go to 0.

The interval is \((1,9)\).

Answer: \((1,9)\)

5
Find the interval of convergence of \(\sum_{n=1}^{\infty}\dfrac{(x+4)^n}{n\,5^n}\).
Show Answer

The ratio limit is \(|x+4|/5\).

\(|x+4|<5\), so \(-9

At \(x=1\), \(\sum 1/n\) diverges.

At \(x=-9\), \(\sum (-1)^n/n\) converges.

The interval is \([-9,1)\).

Answer: \([-9,1)\)

Level 3

Hard

A polynomial coefficient, only even powers, or a radius you can only trap between two numbers.

Example
A power series \(\sum a_n x^n\) converges at \(x=4\) and diverges at \(x=-6\). What can you say about the radius \(R\)?

Convergence at distance 4 from the center forces \(R\ge 4\).

Divergence at distance 6 forces \(R\le 6\).

\(4\le R\le 6\)

Answer: \(4\le R\le 6\)

1
Using that same series, must it converge at \(x=2\)?
Show Answer

\(|2|<4\le R\)

Every point inside the radius converges.

Yes, it must converge.

Answer: yes

2
Must it converge at \(x=5\)?
Show Answer

Distance 5 sits between 4 and 6.

If \(R=4\), then \(x=5\) diverges. If \(R=6\), then \(x=5\) converges.

Not necessarily.

Answer: not necessarily

3
Must it diverge at \(x=7\)?
Show Answer

\(|7|>6\ge R\)

Every point outside the radius diverges.

Yes, it must diverge.

Answer: yes

4
Find the interval of convergence of \(\sum_{n=1}^{\infty}\dfrac{n^2(x-1)^n}{2^n}\).
Show Answer

The ratio limit is \(|x-1|/2\).

\(|x-1|<2\), so \(-1

At \(x=3\), the terms are \(n^2\) and do not go to 0.

At \(x=-1\), the terms are \((-1)^n n^2\) and do not go to 0.

The interval is \((-1,3)\).

Answer: \((-1,3)\)

5
Find the interval of convergence of \(\sum_{n=1}^{\infty}\dfrac{(-1)^n(x-2)^n}{\sqrt{n}}\).
Show Answer

The ratio limit is \(|x-2|\), so \(1

At \(x=3\), the series is \(\sum (-1)^n/\sqrt{n}\), which converges by the alternating series test.

At \(x=1\), \((x-2)^n=(-1)^n\), so the signs \((-1)^n(-1)^n\) cancel and the series is \(\sum 1/\sqrt{n}\), which diverges.

The interval is \((1,3]\).

Answer: \((1,3]\)

Need help with an interval of convergence?

Book a Session with Amy