Properties of Infinite Series
We now develop a toolkit of tests and properties that allow us to determine convergence of infinite series without computing the exact sum. Topics include algebraic properties, comparison tests, absolute vs. conditional convergence, and the alternating series test.
Theorem: Algebraic Operations on Series
If $\sum a_n = A$ and $\sum b_n = B$ both converge, then:
- $\sum (ca_n) = $ for any $c \in \mathbb{R}$;
- $\sum (a_n + b_n) = $ ;
- $\sum (a_n - b_n) = $ .
Caution: These operations require both series to converge individually. You cannot rearrange a divergent series.
Theorem: Divergence Test
If $\sum a_n$ converges, then $\lim\limits_{n \to \infty} a_n = $ .
Contrapositive (useful for testing): If $\lim\limits_{n \to \infty} a_n \neq 0$, then .
Proof
Suppose $\sum a_n$ converges to $S$. Then the partial sums $S_N = \sum\limits_{n=1}^{N} a_n$ satisfy $S_N \to S$.
Also, $S_{N-1} = \sum\limits_{n=1}^{N-1} a_n \to S$ (as $N-1 \to \infty$).
Now observe:
$$a_N = S_N - S_{N-1}$$Taking the limit as $N \to \infty$:
$$\lim_{N \to \infty} a_N = \lim_{N \to \infty} (S_N - S_{N-1}) = S - S = 0$$Therefore, $\lim\limits_{n \to \infty} a_n = 0$. $\square$
Example: Using the Divergence Test
Does $\sum\limits_{n=1}^{\infty} \dfrac{2n}{n+1}$ converge?
Solution:
Compute the limit of the terms:
$$\lim_{n \to \infty} \frac{2n}{n+1} = \lim_{n \to \infty} \frac{2}{1 + \frac{1}{n}} = $$Since the limit is not zero, by the Divergence Test: .
Theorem: Comparison Test
Suppose $0 \leq a_n \leq b_n$ for all $n \geq N_0$. Then:
- If $\sum b_n$ converges, then $\sum a_n$ ;
- If $\sum a_n$ diverges, then $\sum b_n$ .
Theorem: Limit Comparison Test
Suppose $a_n, b_n > 0$ for all $n$, and
$$\lim_{n \to \infty} \frac{a_n}{b_n} = L \quad (0 < L < \infty)$$Then $\sum a_n$ and $\sum b_n$ or together.
Example: Limit Comparison Test
Determine if $\sum\limits_{n=1}^{\infty} \dfrac{n}{n^2 + 3}$ converges or diverges.
Strategy:
Compare with $\sum\limits_{n=1}^{\infty} \dfrac{1}{n}$ (which diverges).
Compute the limit ratio:
$$\lim_{n \to \infty} \frac{\dfrac{n}{n^2+3}}{\dfrac{1}{n}} = \lim_{n \to \infty} \frac{n^2}{n^2+3} = $$By the Limit Comparison Test, since the limit is (nonzero) and $\sum \dfrac{1}{n}$ diverges, we have: $\sum \dfrac{n}{n^2+3}$ .
Definition
A series $\sum a_n$ is:
- Absolutely convergent if $\sum |a_n|$ ;
- Conditionally convergent if $\sum a_n$ converges but $\sum |a_n|$ .
Theorem: Absolute Convergence Implies Convergence
If $\sum |a_n|$ converges, then $\sum a_n$ .
Absolute convergence is much easier to test (all terms are non-negative after taking absolute values). If you can show absolute convergence, convergence follows automatically.
Theorem: Alternating Series Test
Suppose $(a_n)$ is a sequence of positive numbers satisfying:
- $a_n$ is (decreasing);
- $\lim\limits_{n \to \infty} a_n = $.
Then the alternating series $\sum\limits_{n=1}^{\infty} (-1)^n a_n$ .
Example 1: Alternating Harmonic Series
Show that $\sum\limits_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n}$ converges.
Check the conditions:
Let $a_n = \dfrac{1}{n}$. Then:
- $a_n$ is decreasing: $\dfrac{1}{n} > \dfrac{1}{n+1}$ ✓
- $\lim\limits_{n \to \infty} a_n = \lim\limits_{n \to \infty} \dfrac{1}{n} = $ ✓
By the Alternating Series Test: $\sum\limits_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n}$ .
Note: This series converges, but $\sum\limits_{n=1}^{\infty} \dfrac{1}{n}$ (harmonic series) diverges. So this is a conditionally convergent series.
Example 2: Absolute vs. Conditional Convergence
Classify: $\sum\limits_{n=1}^{\infty} \dfrac{(-1)^n}{n^2}$
Check absolute convergence:
$\sum\limits_{n=1}^{\infty} \left|\dfrac{(-1)^n}{n^2}\right| = \sum\limits_{n=1}^{\infty} \dfrac{1}{n^2}$. This is a -series with $p = 2 > 1$, so it .
Conclusion:
Since $\sum\limits_{n=1}^{\infty} \dfrac{(-1)^n}{n^2}$ is absolutely convergent, it is .
Theorem: Rearrangement Behavior
| Convergence Type | Rearrangement Property |
|---|---|
| Absolutely Convergent | Any rearrangement converges to the same sum |
| Conditionally Convergent | Rearrangements can converge to different values (or diverge) |
Warning: For conditionally convergent series (like the alternating harmonic series), rearranging the terms can change the sum or even make it diverge. This is a profound and counterintuitive fact!
| Test | Condition | Conclusion |
|---|---|---|
| Divergence | $\lim a_n \neq 0$ | $\sum a_n$ diverges |
| $p$-Series | $\sum \dfrac{1}{n^p}$ | Converges ⟺ $p > 1$ |
| Comparison | $a_n \leq b_n$, $\sum b_n$ converges | $\sum a_n$ converges |
| Limit Comparison | $\lim \dfrac{a_n}{b_n} = L \in (0, \infty)$ | $\sum a_n$ and $\sum b_n$ same behavior |
| Alternating | $a_n \searrow 0$ | $\sum (-1)^n a_n$ converges |
| Absolute | $\sum |a_n|$ converges | $\sum a_n$ converges |
Exercise 1
For each series, determine whether it converges absolutely, converges conditionally, or diverges.
- $\sum\limits_{n=1}^{\infty} \dfrac{(-1)^n \sqrt{n}}{n+1}$
- $\sum\limits_{n=1}^{\infty} \dfrac{n^2}{2^n}$
- $\sum\limits_{n=1}^{\infty} \dfrac{(-1)^n}{n \ln n}$
Exercise 2
Use the Limit Comparison Test to determine convergence or divergence:
$$\sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^2 - 1}$$You now have a toolkit: Divergence test, $p$-series, comparison tests, alternating series test, and absolute convergence. Use the most efficient test for each situation. Absolute convergence is the "strongest" condition (implies ordinary convergence); conditional convergence is delicate and context-dependent.
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