where $a_n > 0$.

Examples:

• $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = \sum (-1)^{n+1} \frac{1}{n}$ (alternating harmonic)
• $-1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \cdots = \sum (-1)^n \frac{1}{n}$

💡 Key Idea: Signs flip back and forth: positive, negative, positive, negative, ...

Alternating Series Test (AST)

The series $\sum (-1)^n a_n$ (or $\sum (-1)^{n+1} a_n$) converges if:

1. $a_n > 0$ for all $n$ (terms are positive before adding the sign)
2. $a_n \geq a_{n+1}$ for all $n \geq N$ (eventually decreasing)
3. $\lim_{n \to \infty} a_n = 0$ (terms approach 0)

🎯 Three conditions: positive, decreasing, limit is 0

Why It Works

The partial sums oscillate but get closer together:

$S_1 = a_1$ $S_2 = a_1 - a_2$ $S_3 = a_1 - a_2 + a_3$

Since $a_n$ is decreasing:

• Adding positive term moves right (but less each time)
• Subtracting negative term moves left (but less each time)

The partial sums converge to a limit!

Example 1: Alternating Harmonic Series

Test $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots$

Check conditions: Let $a_n = \frac{1}{n}$

1. ✓ $a_n = \frac{1}{n} > 0$ for all $n$

2. ✓ $a_n = \frac{1}{n}$ is decreasing (since $\frac{1}{n+1} < \frac{1}{n}$)

3. ✓ $\lim_{n \to \infty} \frac{1}{n} = 0$

Conclusion: By the Alternating Series Test, the series converges!

(Actually converges to $\ln 2 \approx 0.693$)

Note: The regular harmonic series $\sum \frac{1}{n}$ diverges, but the alternating version converges!

Example 2: Use AST

Determine if $\sum_{n=1}^{\infty} \frac{(-1)^n n}{n^2 + 1}$ converges.

Check conditions: Let $a_n = \frac{n}{n^2 + 1}$

1. ✓ $a_n > 0$ for all $n > 0$

1. Check if decreasing: Is $a_{n+1} \leq a_n$?

$a_{n+1} = \frac{n+1}{(n+1)^2 + 1}$

This is hard to compare directly. Try the derivative test:

Let $f(x) = \frac{x}{x^2 + 1}$

$f'(x) = \frac{(x^2+1)(1) - x(2x)}{(x^2+1)^2} = \frac{1 - x^2}{(x^2+1)^2}$

For $x > 1$: $1 - x^2 < 0$, so $f'(x) < 0$

✓ $a_n$ is decreasing for $n \geq 1$

1. Check limit:

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

Conclusion: All three conditions satisfied!

By AST, the series converges.

Example 3: AST Fails

Does $\sum_{n=1}^{\infty} \frac{(-1)^n n}{n + 1}$ converge?

Check conditions: Let $a_n = \frac{n}{n+1}$

1. ✓ $a_n > 0$

2. Let's assume it's eventually decreasing (we can verify)

3. Check limit:

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

Condition 3 fails!

Since $\lim a_n \neq 0$, we have $\lim (-1)^n a_n \neq 0$.

By the nth Term Test: the series diverges.

(Don't even need AST here - nth term test is quicker!)

Absolute vs Conditional Convergence

A series $\sum a_n$ is:

Absolutely convergent if $\sum |a_n|$ converges

Conditionally convergent if $\sum a_n$ converges but $\sum |a_n|$ diverges

💡 Important: Absolute convergence is stronger than regular convergence!

Theorem: Absolute Convergence Implies Convergence

If $\sum |a_n|$ converges, then $\sum a_n$ converges.

Contrapositive: If $\sum a_n$ diverges, then $\sum |a_n|$ diverges.

Example 4: Absolute vs Conditional

The alternating harmonic series $\sum \frac{(-1)^{n+1}}{n}$:

Converges (by AST)

But: $\sum \left|\frac{(-1)^{n+1}}{n}\right| = \sum \frac{1}{n}$ (harmonic series) diverges!

This series is conditionally convergent (converges, but not absolutely).

Example 5: Absolutely Convergent

Does $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$ converge absolutely, conditionally, or diverge?

Step 1: Test absolute convergence

$\sum \left|\frac{(-1)^n}{n^2}\right| = \sum \frac{1}{n^2}$

This is a p-series with $p = 2 > 1$, so it converges!

Conclusion: Since $\sum |a_n|$ converges, the series is absolutely convergent.

(It's also convergent, since absolute convergence implies convergence)

Alternating Series Estimation Theorem

If $\sum (-1)^n a_n$ converges by AST, and $S$ is the sum, then:

The error after $n$ terms satisfies:

$|R_n| = |S - S_n| \leq a_{n+1}$

💡 Key Idea: Error is at most the first omitted term!

Also: The true sum is between $S_n$ and $S_{n+1}$ (partial sums bracket the answer).

Example 6: Estimate Error

For $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$, estimate the error if we use the first 10 terms.

The error is:

$|R_{10}| \leq a_{11} = \frac{1}{11} \approx 0.091$

So the sum is within 0.091 of $S_{10} = 1 - \frac{1}{2} + \frac{1}{3} - \cdots + \frac{1}{10}$.

Example 7: How Many Terms?

For $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$, how many terms are needed for error less than 0.01?

Need: $a_{n+1} < 0.01$

$\frac{1}{(n+1)^2} < 0.01 = \frac{1}{100}$

$(n+1)^2 > 100$

$n+1 > 10$

$n \geq 10$

Need at least 10 terms.

Strategy for Alternating Series

Step 1: Recognize it's alternating (look for $(-1)^n$ or $(-1)^{n+1}$)

Step 2: Apply AST (check three conditions)

Step 3: If AST works, test for absolute convergence:

• Check if $\sum |a_n|$ converges (using p-series, comparison, etc.)
• If yes: absolutely convergent
• If no: conditionally convergent

⚠️ Common Mistakes

Mistake 1: Forgetting to Check Decreasing

WRONG: "Since $\lim a_n = 0$, AST says it converges"

RIGHT: Must check ALL three conditions! Decreasing is crucial.

Counterexample: $a_n = \begin{cases} \frac{1}{n} & n \text{ odd} \\ \frac{1}{n^2} & n \text{ even} \end{cases}$

Limit is 0, but not decreasing!

Mistake 2: Testing Wrong Series for Absolute Convergence

WRONG: Test $\sum (-1)^n a_n$ for absolute convergence

RIGHT: Test $\sum |(-1)^n a_n| = \sum a_n$ (drop the $(-1)^n$!)

Mistake 3: Using AST on Non-Alternating Series

AST only works for series that alternate signs!

Can't use it on $\sum \frac{1}{n}$ (all positive).

Mistake 4: Confusing Conditional and Absolute

Conditionally convergent: Converges, but not absolutely (more fragile)

Absolutely convergent: Both $\sum a_n$ and $\sum |a_n|$ converge (stronger!)

If absolutely convergent, then also convergent (but not vice versa).

Summary Table

| Series | Converges? | $\sum |a_n|$ Converges? | Type | |--------|------------|------------------------|------| | $\sum \frac{(-1)^n}{n}$ | Yes (AST) | No (harmonic) | Conditional | | $\sum \frac{(-1)^n}{n^2}$ | Yes (AST) | Yes (p-series) | Absolute | | $\sum \frac{(-1)^n n}{n+1}$ | No (nth term) | No | Divergent |

📝 Practice Strategy

1. Look for $(-1)^n$ pattern - signals alternating series
2. Check nth term test first: If $\lim a_n \neq 0$, diverges immediately!
3. Apply AST: Check positive, decreasing (use derivative!), limit is 0
4. For decreasing: Can use $f'(x) < 0$ or direct comparison
5. Test absolute convergence: Drop the $(-1)^n$, test $\sum a_n$
6. For error estimates: Use $|R_n| \leq a_{n+1}$
7. Memorize: Absolute convergence → convergence (but not reverse!)

This is geometric with $r = \frac{1}{3} < 1$.

Converges! (Sum is $\frac{1/3}{1-1/3} = \frac{1}{2}$)

Step 2: Conclusion on convergence

Since $\sum |a_n|$ converges:

The series is absolutely convergent (and therefore convergent).

Step 3: Estimate error using 5 terms

By Alternating Series Estimation Theorem:

$|R_5| \leq a_6 = \frac{1}{3^6} = \frac{1}{729} \approx 0.00137$

The error using 5 terms is at most 0.00137.

Step 4: Calculate $S_5$ (optional)

$S_5 = \frac{1}{3} - \frac{1}{9} + \frac{1}{27} - \frac{1}{81} + \frac{1}{243}$

$= \frac{81 - 27 + 9 - 3 + 1}{243} = \frac{61}{243} \approx 0.251$

The true sum is approximately $0.251 \pm 0.00137$.

1. ✓ $a_n = \frac{1}{\ln n} > 0$ for $n \geq 2$ (since $\ln n > 0$ for $n > 1$)

1. Check if decreasing: As $n$ increases, $\ln n$ increases, so $\frac{1}{\ln n}$ decreases ✓

Or use derivative: $f(x) = \frac{1}{\ln x}$

$f'(x) = -\frac{1}{x(\ln x)^2} < 0$ for $x > 1$ ✓

1. Check limit:

$\lim_{n \to \infty} \frac{1}{\ln n} = 0$ ✓

(Since $\ln n \to \infty$)

By AST, the series converges.

Step 2: Test for absolute convergence

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

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

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

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

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

Since the limit is $\infty$ and $\sum \frac{1}{n}$ diverges, $\sum \frac{1}{\ln n}$ diverges.

Step 3: Conclusion

The series converges (by AST) but does not converge absolutely.

The series is conditionally convergent.