Radius and Interval of Convergence

Finding where power series converge

🎯 Radius and Interval of Convergence

Review: Convergence Behavior

For power series $\sum_{n=0}^{\infty} c_n(x-a)^n$ :

Converges at center $x = a$ (always!)
May converge on an interval around $a$
Diverges outside that interval

The radius of convergence $R$ determines this interval.

Three Cases

Case 1: $R = 0$

Series converges only at $x = a$

Example: $\sum_{n=0}^{\infty} n! x^n$

Use Ratio Test: limit is $\infty$ unless $x = 0$ .

Case 2: $0 < R < \infty$

Series converges for $|x - a| < R$

Diverges for $|x - a| > R$

Endpoints $x = a \pm R$ need separate testing!

Case 3: $R = \infty$

Series converges for all $x$

Example: $\sum_{n=0}^{\infty} \frac{x^n}{n!}$ (this is $e^x$ )

Finding Radius Using Ratio Test

Step 1: Apply Ratio Test to $\sum c_n(x-a)^n$

$L = \lim_{n \to \infty} \left|\frac{c_{n+1}(x-a)^{n+1}}{c_n(x-a)^n}\right|$

$= |x-a| \cdot \lim_{n \to \infty} \left|\frac{c_{n+1}}{c_n}\right|$

Step 2: For convergence, need $L < 1$ :

$|x-a| \cdot \lim_{n \to \infty} \left|\frac{c_{n+1}}{c_n}\right| < 1$

$|x-a| < \frac{1}{\lim_{n \to \infty} \left|\frac{c_{n+1}}{c_n}\right|}$

Step 3: Radius is:

$R = \frac{1}{\lim_{n \to \infty} \left|\frac{c_{n+1}}{c_n}\right|}$

(If limit is 0, then $R = \infty$ ; if limit is $\infty$ , then $R = 0$ )

Example 1: Find Radius

Find the radius of convergence for $\sum_{n=1}^{\infty} \frac{x^n}{n}$ .

Use Ratio Test:

$c_n = \frac{1}{n}, \quad c_{n+1} = \frac{1}{n+1}$

$\lim_{n \to \infty} \left|\frac{c_{n+1}}{c_n}\right| = \lim_{n \to \infty} \frac{n}{n+1} = 1$

$R = \frac{1}{1} = 1$

Radius of convergence: $R = 1$

Series converges for $|x| < 1$ (since center is $a = 0$ ).

Finding the Interval of Convergence

Step 1: Find radius $R$ using Ratio Test

Step 2: The interval (before endpoints) is $(a - R, a + R)$

Step 3: Test endpoints $x = a - R$ and $x = a + R$ separately using:

Alternating Series Test
p-Series Test
Comparison Tests
etc.

Step 4: Write final interval using $($ or $[$ depending on endpoint convergence

Example 2: Find Interval

Find the interval of convergence for $\sum_{n=1}^{\infty} \frac{x^n}{n}$ .

From Example 1: $R = 1$ , center $a = 0$

Interval before endpoints: $(-1, 1)$

Test $x = 1$ :

$\sum_{n=1}^{\infty} \frac{1^n}{n} = \sum_{n=1}^{\infty} \frac{1}{n}$ (harmonic series)

Diverges!

Test $x = -1$ :

$\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$ (alternating harmonic)

By Alternating Series Test: Converges!

Interval of convergence: $[-1, 1)$

(Includes $-1$ , excludes $1$ )

Example 3: Both Endpoints Converge

Find the interval of convergence for $\sum_{n=1}^{\infty} \frac{x^n}{n^2}$ .

Step 1: Find radius

$\lim_{n \to \infty} \left|\frac{c_{n+1}}{c_n}\right| = \lim_{n \to \infty} \frac{n^2}{(n+1)^2} = 1$

$R = 1$

Step 2: Test $x = 1$

$\sum_{n=1}^{\infty} \frac{1}{n^2}$ (p-series with $p = 2$ )

Converges!

Step 3: Test $x = -1$

$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$

Converges absolutely (since $\sum \frac{1}{n^2}$ converges).

Converges!

Interval of convergence: $[-1, 1]$ (closed interval)

Example 4: Neither Endpoint Converges

Find the interval of convergence for $\sum_{n=0}^{\infty} n x^n$ .

Step 1: Find radius

$c_n = n, \quad c_{n+1} = n+1$

$\lim_{n \to \infty} \left|\frac{n+1}{n}\right| = 1$

$R = 1$

Step 2: Test $x = 1$

$\sum_{n=0}^{\infty} n$

This diverges (terms don't approach 0).

Diverges!

Step 3: Test $x = -1$

$\sum_{n=0}^{\infty} n(-1)^n$

Terms don't approach 0.

Diverges!

Interval of convergence: $(-1, 1)$ (open interval)

Example 5: Series Centered at $a \neq 0$

Find the interval of convergence for $\sum_{n=1}^{\infty} \frac{(x-3)^n}{n \cdot 2^n}$ .

Step 1: Find radius

$c_n = \frac{1}{n \cdot 2^n}, \quad c_{n+1} = \frac{1}{(n+1) \cdot 2^{n+1}}$

$\lim_{n \to \infty} \left|\frac{c_{n+1}}{c_n}\right| = \lim_{n \to \infty} \frac{n \cdot 2^n}{(n+1) \cdot 2^{n+1}}$

$= \lim_{n \to \infty} \frac{n}{2(n+1)} = \frac{1}{2}$

$R = \frac{1}{1/2} = 2$

Center: $a = 3$

Interval before endpoints: $(3-2, 3+2) = (1, 5)$

Test $x = 1$ :

$\sum_{n=1}^{\infty} \frac{(1-3)^n}{n \cdot 2^n} = \sum_{n=1}^{\infty} \frac{(-2)^n}{n \cdot 2^n}$

$= \sum_{n=1}^{\infty} \frac{(-1)^n \cdot 2^n}{n \cdot 2^n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n}$

Alternating harmonic series: Converges!

Test $x = 5$ :

$\sum_{n=1}^{\infty} \frac{(5-3)^n}{n \cdot 2^n} = \sum_{n=1}^{\infty} \frac{2^n}{n \cdot 2^n} = \sum_{n=1}^{\infty} \frac{1}{n}$

Harmonic series: Diverges!

Interval of convergence: $[1, 5)$

Example 6: Infinite Radius

Find the interval of convergence for $\sum_{n=0}^{\infty} \frac{x^n}{n!}$ .

Use Ratio Test:

$c_n = \frac{1}{n!}, \quad c_{n+1} = \frac{1}{(n+1)!}$

$\lim_{n \to \infty} \left|\frac{c_{n+1}}{c_n}\right| = \lim_{n \to \infty} \frac{n!}{(n+1)!} = \lim_{n \to \infty} \frac{1}{n+1} = 0$

$R = \frac{1}{0} = \infty$

Interval of convergence: $(-\infty, \infty)$

Series converges for all real $x$ !

(This is $e^x$ )

Summary Table

| Endpoint $x = a-R$ | Endpoint $x = a+R$ | Interval | |--------------------|--------------------|----------| | Diverges | Diverges | $(a-R, a+R)$ | | Converges | Diverges | $[a-R, a+R)$ | | Diverges | Converges | $(a-R, a+R]$ | | Converges | Converges | $[a-R, a+R]$ |

⚠️ Common Mistakes

Mistake 1: Forgetting to Test Endpoints

WRONG: "Radius is 1, so interval is $(-1, 1)$ "

RIGHT: Must test both endpoints separately! Could be $(-1,1)$ , $[-1,1)$ , $(-1,1]$ , or $[-1,1]$ .

Mistake 2: Wrong Endpoint Values

For $\sum c_n(x-3)^n$ with $R = 2$ :

WRONG: Endpoints are $\pm 2$

RIGHT: Endpoints are $3 \pm 2$ , i.e., $x = 1$ and $x = 5$

Center is 3, not 0!

Mistake 3: Testing Wrong Series at Endpoints

At $x = -1$ for $\sum \frac{x^n}{n}$ :

WRONG: Test $\sum \frac{x^n}{n}$ (still has $x$ in it!)

RIGHT: Substitute $x = -1$ : test $\sum \frac{(-1)^n}{n}$

Mistake 4: Assuming Symmetry

WRONG: "If $x = 1$ diverges, then $x = -1$ also diverges"

RIGHT: Must test each endpoint independently!

One might converge while the other diverges.

📝 Practice Strategy

Find radius $R$ : Use Ratio Test on coefficients
Find center $a$ : From $(x - a)^n$ term
Interval before endpoints: $(a - R, a + R)$
Test left endpoint $x = a - R$ : Substitute, use appropriate test
Test right endpoint $x = a + R$ : Substitute, use appropriate test
Write final interval: Use correct bracket notation
Special cases: $R = 0$ (only at center), $R = \infty$ (all $x$ )
Check your work: Make sure endpoints make sense with center!

📚 Practice Problems

1Problem 1medium

❓ Question:

Find the interval of convergence for $\sum_{n=1}^{\infty} \frac{(-1)^n (x-2)^n}{n \cdot 5^n}$ .

💡 Show Solution

Step 1: Find radius of convergence

$c_n = \frac{(-1)^n}{n \cdot 5^n}$

For ratio test, we use absolute values:

$\lim_{n \to \infty} \left|\frac{c_{n+1}}{c_n}\right| = \lim_{n \to \infty} \frac{\frac{1}{(n+1) \cdot 5^{n+1}}}{\frac{1}{n \cdot 5^n}}$

$= \lim_{n \to \infty} \frac{n \cdot 5^n}{(n+1) \cdot 5^{n+1}} = \lim_{n \to \infty} \frac{n}{5(n+1)} = \frac{1}{5}$

$R = \frac{1}{1/5} = 5$

Step 2: Find interval before endpoints

Center: $a = 2$

Interval: $(2-5, 2+5) = (-3, 7)$

Step 3: Test left endpoint $x = -3$

$\sum_{n=1}^{\infty} \frac{(-1)^n (-3-2)^n}{n \cdot 5^n} = \sum_{n=1}^{\infty} \frac{(-1)^n (-5)^n}{n \cdot 5^n}$

$= \sum_{n=1}^{\infty} \frac{(-1)^n \cdot (-1)^n \cdot 5^n}{n \cdot 5^n} = \sum_{n=1}^{\infty} \frac{(-1)^{2n}}{n} = \sum_{n=1}^{\infty} \frac{1}{n}$

This is the harmonic series: Diverges!

Step 4: Test right endpoint $x = 7$

$\sum_{n=1}^{\infty} \frac{(-1)^n (7-2)^n}{n \cdot 5^n} = \sum_{n=1}^{\infty} \frac{(-1)^n \cdot 5^n}{n \cdot 5^n}$

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

This is the alternating harmonic series.

By Alternating Series Test: Converges!

Answer: Interval of convergence is $(-3, 7]$

2Problem 2hard

❓ Question:

Find the radius and interval of convergence for:

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

💡 Show Solution

Solution:

This is a power series centered at $x = 3$ .

Use Ratio Test with $a_n = \frac{(x-3)^n}{n \cdot 2^n}$ :

$\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{(x-3)^{n+1}}{(n+1) \cdot 2^{n+1}} \cdot \frac{n \cdot 2^n}{(x-3)^n}\right|$

$= |x-3| \cdot \frac{n}{n+1} \cdot \frac{1}{2}$

$L = \lim_{n \to \infty} \frac{|x-3|}{2} \cdot \frac{n}{n+1} = \frac{|x-3|}{2}$

Converges when $L < 1$ : $\frac{|x-3|}{2} < 1$ , so $|x-3| < 2$

Radius: $R = 2$

This gives $1 < x < 5$ . Check endpoints:

At $x = 1$ : $\sum \frac{(-2)^n}{n \cdot 2^n} = \sum \frac{(-1)^n}{n}$ (alternating harmonic, converges)

At $x = 5$ : $\sum \frac{2^n}{n \cdot 2^n} = \sum \frac{1}{n}$ (harmonic series, diverges)

Interval of convergence: $[1, 5)$

3Problem 3easy

❓ Question:

Find the interval of convergence for $\sum_{n=0}^{\infty} \frac{(x+1)^n}{(n+1)!}$ .

💡 Show Solution

Step 1: Find radius

$c_n = \frac{1}{(n+1)!}$

$\lim_{n \to \infty} \left|\frac{c_{n+1}}{c_n}\right| = \lim_{n \to \infty} \frac{(n+1)!}{(n+2)!}$

$= \lim_{n \to \infty} \frac{1}{n+2} = 0$

$R = \frac{1}{0} = \infty$

Step 2: Conclusion

Since $R = \infty$ , the series converges for all $x$ .

Interval of convergence: $(-\infty, \infty)$

Note: No need to test endpoints when $R = \infty$ !

4Problem 4hard

❓ Question:

Find the interval of convergence for $\sum_{n=2}^{\infty} \frac{x^n}{(\ln n)^2}$ .

💡 Show Solution

Step 1: Find radius

$c_n = \frac{1}{(\ln n)^2}$

For the ratio:

$\lim_{n \to \infty} \left|\frac{c_{n+1}}{c_n}\right| = \lim_{n \to \infty} \frac{(\ln n)^2}{(\ln(n+1))^2}$

As $n \to \infty$ : $\ln(n+1) \approx \ln n$

$= \lim_{n \to \infty} \left(\frac{\ln n}{\ln(n+1)}\right)^2 = 1$

$R = 1$

Step 2: Interval before endpoints

Center: $a = 0$

Interval: $(-1, 1)$

Step 3: Test $x = 1$

$\sum_{n=2}^{\infty} \frac{1}{(\ln n)^2}$

Compare to $\sum \frac{1}{n}$ using Limit Comparison Test:

$\lim_{n \to \infty} \frac{\frac{1}{(\ln n)^2}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n}{(\ln n)^2}$

This is $\frac{\infty}{\infty}$ form. Use L'Hôpital's (twice):

$= \lim_{n \to \infty} \frac{1}{\frac{2\ln n}{n}} = \lim_{n \to \infty} \frac{n}{2\ln n}$

(Still $\frac{\infty}{\infty}$ , apply L'Hôpital's again)

$= \lim_{n \to \infty} \frac{1}{2/n} = \lim_{n \to \infty} \frac{n}{2} = \infty$

Since limit is $\infty$ and $\sum \frac{1}{n}$ diverges:

$\sum \frac{1}{(\ln n)^2}$ diverges!

Step 4: Test $x = -1$

$\sum_{n=2}^{\infty} \frac{(-1)^n}{(\ln n)^2}$

Check Alternating Series Test:

$a_n = \frac{1}{(\ln n)^2} > 0$ ✓
Decreasing: as $n$ increases, $\ln n$ increases, so $\frac{1}{(\ln n)^2}$ decreases ✓
$\lim_{n \to \infty} \frac{1}{(\ln n)^2} = 0$ ✓

Converges!

Answer: Interval of convergence is $[-1, 1)$

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Radius and Interval of Convergence

🎯 Radius and Interval of Convergence

Review: Convergence Behavior

Three Cases

Case 1: R=0R = 0R=0

Case 2: 0<R<∞0 < R < \infty0<R<∞

Case 3: R=∞R = \inftyR=∞

Finding Radius Using Ratio Test

Example 1: Find Radius

Finding the Interval of Convergence

Example 2: Find Interval

Example 3: Both Endpoints Converge

Example 4: Neither Endpoint Converges

Example 5: Series Centered at a≠0a \neq 0a=0

Example 6: Infinite Radius

Summary Table

⚠️ Common Mistakes

Mistake 1: Forgetting to Test Endpoints

Mistake 2: Wrong Endpoint Values

Mistake 3: Testing Wrong Series at Endpoints

Mistake 4: Assuming Symmetry

📝 Practice Strategy

📚 Practice Problems

1Problem 1medium

❓ Question:

2Problem 2hard

❓ Question:

3Problem 3easy

❓ Question:

4Problem 4hard

❓ Question:

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Case 1: $R = 0$

Case 2: $0 < R < \infty$

Case 3: $R = \infty$

Example 5: Series Centered at $a \neq 0$