# 🎯 The Integral Test ## The Idea If you can integrate a function $f(x)$, you can use the integral to test whether $\sum f(n)$ converges! > **💡 Key Idea**: Compare the series (sum of rectangles) to the integral (area under curve). --- ## Integral Test Theorem Let $f(x)$ be a function that is: 1. **Continuous** for $x \geq 1$ 2. **Positive**: $f(x) > 0$ 3. **Decreasing**: $f(x+1) < f(x)$ Then the series $\sum_{n=1}^{\infty} f(n)$ and the integral $\int_1^{\infty} f(x)\,dx$: **Both converge or both diverge together!** --- ## Visual Understanding Imagine rectangles with: - Width: 1 - Heights: $f(1), f(2), f(3), \ldots$ The sum $\sum f(n)$ is the total area of rectangles. The integral $\int f(x)\,dx$ is the area under the curve. Since $f$ is decreasing: - Rectangles (left endpoints) > curve area - Rectangles (right endpoints) < curve area So: $\int_1^{\infty} f(x)\,dx < \sum_{n=1}^{\infty} f(n) < f(1) + \int_1^{\infty} f(x)\,dx$ --- ## How to Use the Integral Test **Step 1**: Verify $f(x)$ is continuous, positive, and decreasing for $x \geq N$ (some starting point) **Step 2**: Evaluate $\int_N^{\infty} f(x)\,dx = \lim_{b \to \infty} \int_N^b f(x)\,dx$ **Step 3**: - If integral converges (finite) → series converges - If integral diverges (infinite) → series diverges --- ## Example 1: Harmonic Series Test $\sum_{n=1}^{\infty} \frac{1}{n}$ using the Integral Test. **Step 1: Check conditions** Let $f(x) = \frac{1}{x}$ for $x \geq 1$. ✓ Continuous for $x \geq 1$ ✓ Positive: $\frac{1}{x} > 0$ for $x > 0$ ✓ Decreasing: $f'(x) = -\frac{1}{x^2} < 0$ --- **Step 2: Evaluate improper integral** $$\int_1^{\infty} \frac{1}{x}\,dx = \lim_{b \to \infty} \int_1^b \frac{1}{x}\,dx$$ $$= \lim_{b \to \infty} [\ln|x|]_1^b = \lim_{b \to \infty} (\ln b - \ln 1)$$ $$= \lim_{b \to \infty} \ln b = \infty$$ --- **Step 3: Conclusion** The integral diverges, so by the Integral Test: **The harmonic series $\sum \frac{1}{n}$ diverges!** ✓ --- ## The p-Series A **p-series** has the form: $$\sum_{n=1}^{\infty} \frac{1}{n^p}$$ where $p$ is a constant. --- ## p-Series Convergence Test $$\sum_{n=1}^{\infty} \frac{1}{n^p} \begin{cases} \text{converges} & \text{if } p > 1 \\ \text{diverges} & \text{if } p \leq 1 \end{cases}$$ > **🎯 MEMORIZE THIS!** Most commonly tested series! --- ## Proof Using Integral Test **Case 1**: $p > 1$ $$\int_1^{\infty} \frac{1}{x^p}\,dx = \lim_{b \to \infty} \left[\frac{x^{-p+1}}{-p+1}\right]_1^b$$ $$= \lim_{b \to \infty} \left[\frac{1}{(1-p)x^{p-1}}\right]_1^b$$ $$= \frac{1}{1-p}\lim_{b \to \infty} \left(\frac{1}{b^{p-1}} - 1\right)$$ Since $p > 1$, we have $p - 1 > 0$, so $\frac{1}{b^{p-1}} \to 0$. $$= \frac{1}{1-p}(0 - 1) = \frac{1}{p-1}$$ (finite!) **Converges when $p > 1$.** --- **Case 2**: $p = 1$ This is the harmonic series (we showed it diverges). --- **Case 3**: $p < 1$ $$\int_1^{\infty} \frac{1}{x^p}\,dx = \lim_{b \to \infty} \left[\frac{x^{1-p}}{1-p}\right]_1^b = \infty$$ **Diverges when $p < 1$.** --- ## Example 2: Test $\sum_{n=1}^{\infty} \frac{1}{n^2}$ This is a p-series with $p = 2 > 1$. **By p-series test: converges!** (Actually converges to $\frac{\pi^2}{6}$, discovered by Euler!) --- ## Example 3: Test $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ Rewrite: $\sum \frac{1}{n^{1/2}}$ This is a p-series with $p = \frac{1}{2} < 1$. **By p-series test: diverges!** --- ## Example 4: Use Integral Test Test $\sum_{n=1}^{\infty} \frac{n}{n^2 + 1}$ for convergence. **Step 1: Check conditions** $f(x) = \frac{x}{x^2 + 1}$ ✓ Continuous for $x \geq 1$ ✓ Positive for $x > 0$ Check decreasing: $f'(x) = \frac{(x^2+1)(1) - x(2x)}{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2}$ For $x > 1$: $1 - x^2 < 0$, so $f'(x) < 0$ ✓ --- **Step 2: Evaluate integral** $$\int_1^{\infty} \frac{x}{x^2+1}\,dx$$ **Use substitution**: $u = x^2 + 1$, $du = 2x\,dx$ $$= \frac{1}{2}\int_2^{\infty} \frac{1}{u}\,du = \frac{1}{2}\lim_{b \to \infty} [\ln u]_2^b$$ $$= \frac{1}{2}\lim_{b \to \infty} (\ln b - \ln 2) = \infty$$ --- **Step 3: Conclusion** The integral diverges, so **the series diverges**. --- ## Example 5: Test $\sum_{n=1}^{\infty} \frac{\ln n}{n}$ **Step 1: Check conditions** (for $n \geq 2$ to avoid $\ln 1 = 0$) $f(x) = \frac{\ln x}{x}$ for $x \geq 2$ ✓ Continuous ✓ Positive for $x > 1$ Check decreasing: Use quotient rule or note that for large $x$, denominator grows faster. $f'(x) = \frac{\frac{1}{x} \cdot x - \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2}$ For $x > e \approx 2.718$: $\ln x > 1$, so $f'(x) < 0$ ✓ --- **Step 2: Evaluate integral** $$\int_2^{\infty} \frac{\ln x}{x}\,dx$$ **Use substitution**: $u = \ln x$, $du = \frac{1}{x}\,dx$ When $x = 2$: $u = \ln 2$ When $x \to \infty$: $u \to \infty$ $$= \int_{\ln 2}^{\infty} u\,du = \lim_{b \to \infty} \left[\frac{u^2}{2}\right]_{\ln 2}^b = \infty$$ --- **Step 3: Conclusion** **The series diverges.** --- ## Important Notes About Integral Test ### Note 1: Value Not Equal The Integral Test tells you about **convergence**, not the **value**! $$\sum_{n=1}^{\infty} f(n) \neq \int_1^{\infty} f(x)\,dx$$ (They're close, but not equal) --- ### Note 2: Can Start Anywhere You can start the series and integral at any point $N$: $$\sum_{n=N}^{\infty} f(n) \text{ and } \int_N^{\infty} f(x)\,dx$$ Both converge or both diverge (though values differ). --- ### Note 3: Must Be Decreasing If $f(x)$ is not eventually decreasing, can't use Integral Test! **Example**: $f(x) = \sin x$ oscillates, not decreasing. --- ### Note 4: Remainder Estimate If the series converges, the **remainder** (error after $n$ terms) satisfies: $$\int_{n+1}^{\infty} f(x)\,dx < R_n < \int_n^{\infty} f(x)\,dx$$ where $R_n = \sum_{k=n+1}^{\infty} f(k)$ --- ## Example 6: Estimate Error For $\sum_{n=1}^{\infty} \frac{1}{n^2}$, estimate the error if we use the first 10 terms. **Remainder is**: $$R_{10} = \sum_{n=11}^{\infty} \frac{1}{n^2}$$ --- **Upper bound**: $$R_{10} < \int_{10}^{\infty} \frac{1}{x^2}\,dx = \left[-\frac{1}{x}\right]_{10}^{\infty} = 0 - \left(-\frac{1}{10}\right) = \frac{1}{10} = 0.1$$ **Lower bound**: $$R_{10} > \int_{11}^{\infty} \frac{1}{x^2}\,dx = \left[-\frac{1}{x}\right]_{11}^{\infty} = \frac{1}{11} \approx 0.091$$ --- **Conclusion**: $0.091 < R_{10} < 0.1$ The error is less than 0.1! --- ## ⚠️ Common Mistakes ### Mistake 1: Using When Not Decreasing **WRONG**: Apply Integral Test to $\sum \frac{\sin n}{n}$ $f(x) = \frac{\sin x}{x}$ is not always decreasing (oscillates)! --- ### Mistake 2: Thinking Value Equals Integral **WRONG**: $\sum_{n=1}^{\infty} \frac{1}{n^2} = \int_1^{\infty} \frac{1}{x^2}\,dx = 1$ **RIGHT**: The series converges (by Integral Test), but its value is $\frac{\pi^2}{6} \approx 1.645$, not 1. --- ### Mistake 3: Forgetting p-Series Test Don't do complicated integral when you can use p-series test directly! For $\sum \frac{1}{n^p}$: just check if $p > 1$ (converges) or $p \leq 1$ (diverges). --- ### Mistake 4: Wrong Limits Make sure integral limits match series starting point! If series starts at $n = 2$, use $\int_2^{\infty}$. --- ## 📝 Practice Strategy 1. **Check p-series first**: Is it $\sum \frac{1}{n^p}$? Use p-series test! 2. **Verify conditions**: continuous, positive, decreasing (check derivative) 3. **Integrate**: Use substitution, partial fractions, etc. 4. **Improper integral**: Always use limits, check if finite or infinite 5. **For estimates**: Use remainder bounds $\int_{n+1}^{\infty} < R_n < \int_n^{\infty}$ 6. **Memorize p-series**: $p > 1$ converges, $p \leq 1$ diverges 7. **When stuck**: Try to compare to p-series instead of integrating!