• 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

⚠️ 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

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

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

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

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

Step 2: Conclusion

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

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

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

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 $[-1, 1)$