The nth Term Test is the simplest divergence test: if the terms of the series do not approach zero, the series cannot converge. The Integral Test connects series to improper integrals and is especially useful for series like $\sum 1/n^p$ ($p$-series).
🎯 In this section you will learn
- The nth Term Test for divergence
- The Integral Test and its requirements ($f$ positive, continuous, decreasing)
- The $p$-series test: $\sum 1/n^p$ converges if $p > 1$, diverges if $p \le 1$
- How to apply the Integral Test to determine convergence
📌 nth Term Test for Divergence
$$ \text{If } \lim_{n \to \infty} a_n \neq 0, \text{ then } \sum a_n \text{ diverges} $$
If the terms don't approach zero, the series cannot converge
💡 Important: The converse is false
If $\lim a_n = 0$, the series may converge or diverge. The nth term test cannot prove convergence.
Show that $\sum_{n=1}^\infty \frac{n}{n+1}$ diverges.
- $\lim_{n \to \infty} \frac{n}{n+1} = 1 \neq 0$
- Therefore, the series diverges.
For $\sum_{n=1}^\infty \frac{1}{n}$, $\lim a_n = 0$, but the series diverges (harmonic series). The nth term test is inconclusive.
📌 The Integral Test
$$ \text{If } f(x) \text{ is positive, continuous, and decreasing for } x \ge 1, \text{ then } \sum_{n=1}^\infty f(n) \text{ converges } \iff \int_1^\infty f(x) dx \text{ converges} $$
📌 Requirements for Integral Test
- $f$ is positive for $x \ge 1$
- $f$ is continuous for $x \ge 1$
- $f$ is decreasing for $x \ge 1$ (i.e., $f'(x) < 0$)
Test $\sum_{n=1}^\infty \frac{1}{n^2}$ for convergence.
①
Check conditions
$f(x) = \frac{1}{x^2}$ is positive, continuous, and decreasing for $x \ge 1$.
②
Evaluate the integral
$\int_1^\infty \frac{1}{x^2} dx = \lim_{b \to \infty} \left[-\frac{1}{x}\right]_1^b = \lim_{b \to \infty} \left(-\frac{1}{b} + 1\right) = 1$
③
Conclusion
Since the integral converges, the series $\sum \frac{1}{n^2}$ converges.
Test $\sum_{n=1}^\infty \frac{1}{n}$ for convergence.
- $f(x) = 1/x$ is positive, continuous, decreasing for $x \ge 1$.
- $\int_1^\infty \frac{1}{x} dx = \lim_{b \to \infty} \ln b = \infty$ (diverges).
- Therefore, $\sum 1/n$ diverges.
📌 The $p$-Series Test
$$ \sum_{n=1}^\infty \frac{1}{n^p} \text{ converges if } p > 1, \text{ diverges if } p \le 1 $$
This follows directly from the Integral Test
- $\sum 1/n^{1.5}$ converges ($p = 1.5 > 1$)
- $\sum 1/\sqrt{n} = \sum 1/n^{0.5}$ diverges ($p = 0.5 < 1$)
- $\sum 1/n$ (harmonic) diverges ($p = 1$)
- $\sum 1/n^2$ converges ($p = 2 > 1$)
⚠️ Common Mistakes
- Using nth term test to prove convergence: $\lim a_n = 0$ does NOT guarantee convergence.
- Forgetting to check decreasing condition: The Integral Test requires $f$ to be decreasing.
- Misapplying $p$-series: The series must be of the form $\sum 1/n^p$; constants like $\sum 2/n^2$ also converge by constant multiple rule.
🔍 Key Takeaways
- nth Term Test: If $\lim a_n \neq 0$, the series diverges. If the limit is $0$, the test is inconclusive.
- Integral Test: $\sum f(n)$ converges iff $\int_1^\infty f(x) dx$ converges (requires $f$ positive, continuous, decreasing).
- $p$-series: $\sum 1/n^p$ converges if $p > 1$, diverges if $p \le 1$.
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