The Integral Test

Using integrals to test series convergence

🎯 The Integral Test

The Idea

If you can integrate a function f(x)f(x), you can use the integral to test whether ∑f(n)\sum f(n) converges!

💡 Key Idea: Compare the series (sum of rectangles) to the integral (area under curve).

Integral Test Theorem

Let f(x)f(x) be a function that is:

  1. Continuous for x≥1x \geq 1
  2. Positive: f(x)>0f(x) > 0
  3. Decreasing: f(x+1)<f(x)f(x+1) < f(x)

Then the series ∑n=1∞f(n)\sum_{n=1}^{\infty} f(n) and the integral ∫1∞f(x) dx\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),…f(1), f(2), f(3), \ldots

The sum ∑f(n)\sum f(n) is the total area of rectangles.

The integral ∫f(x) dx\int f(x)\,dx is the area under the curve.

Since ff is decreasing:

  • Rectangles (left endpoints) > curve area
  • Rectangles (right endpoints) < curve area

So: ∫1∞f(x) dx<∑n=1∞f(n)<f(1)+∫1∞f(x) dx\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)f(x) is continuous, positive, and decreasing for x≥Nx \geq N (some starting point)

Step 2: Evaluate ∫N∞f(x) dx=lim⁡b→∞∫Nbf(x) dx\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 ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{n} using the Integral Test.

Step 1: Check conditions

Let f(x)=1xf(x) = \frac{1}{x} for x≥1x \geq 1.

✓ Continuous for x≥1x \geq 1 ✓ Positive: 1x>0\frac{1}{x} > 0 for x>0x > 0 ✓ Decreasing: f′(x)=−1x2<0f'(x) = -\frac{1}{x^2} < 0

Step 2: Evaluate improper integral

∫1∞1x dx=lim⁡b→∞∫1b1x dx\int_1^{\infty} \frac{1}{x}\,dx = \lim_{b \to \infty} \int_1^b \frac{1}{x}\,dx

=lim⁡b→∞[ln⁡∣x∣]1b=lim⁡b→∞(ln⁡b−ln⁡1)= \lim_{b \to \infty} [\ln|x|]_1^b = \lim_{b \to \infty} (\ln b - \ln 1)

=lim⁡b→∞ln⁡b=∞= \lim_{b \to \infty} \ln b = \infty

Step 3: Conclusion

The integral diverges, so by the Integral Test:

The harmonic series ∑1n\sum \frac{1}{n} diverges! ✓

The p-Series

A p-series has the form:

∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p}

where pp is a constant.

p-Series Convergence Test

\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!

📚 Practice Problems

1Problem 1medium

❓ Question:

Use the Integral Test to determine if ∑n=2∞1nln⁡n\sum_{n=2}^{\infty} \frac{1}{n \ln n} converges or diverges.

💡 Show Solution

Step 1: Check conditions

f(x)=1xln⁡xf(x) = \frac{1}{x \ln x} for x≥2x \geq 2

✓ Continuous for x≥2x \geq 2 (since ln⁡x>0\ln x > 0 for x>1x > 1) ✓ Positive for x>1x > 1

Check decreasing: As xx increases, both xx and ln⁡x\ln x increase, so 1xln⁡x\frac{1}{x \ln x} decreases ✓

Step 2: Evaluate integral

∫2∞1xln⁡x dx\int_2^{\infty} \frac{1}{x \ln x}\,dx

Use substitution: u=ln⁡xu = \ln x, du=1x dxdu = \frac{1}{x}\,dx

When x=2x = 2: u=ln⁡2u = \ln 2 When x→∞x \to \infty: u→∞u \to \infty

=∫ln⁡2∞1u du=lim⁡b→∞[ln⁡∣u∣]ln⁡2b= \int_{\ln 2}^{\infty} \frac{1}{u}\,du = \lim_{b \to \infty} [\ln|u|]_{\ln 2}^b

=lim⁡b→∞(ln⁡b−ln⁡(ln⁡2))=∞= \lim_{b \to \infty} (\ln b - \ln(\ln 2)) = \infty

Step 3: Conclusion

The integral diverges.

By the Integral Test, the series diverges.

Note: Even though terms go to 0 (slowly), they don't go to 0 fast enough!

2Problem 2hard

❓ Question:

Use the Integral Test to determine whether the series converges or diverges:

∑n=1∞1n2+1\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}

💡 Show Solution

Solution:

Step 1: Verify conditions for Integral Test.

Let f(x)=1x2+1f(x) = \frac{1}{x^2 + 1}

  • Continuous for x≥1x \geq 1 ✓
  • Positive for x≥1x \geq 1 ✓
  • Decreasing for x≥1x \geq 1 (since f′(x)=−2x(x2+1)2<0f'(x) = \frac{-2x}{(x^2+1)^2} < 0 for x>0x > 0) ✓

Step 2: Evaluate the improper integral.

∫1∞1x2+1 dx=lim⁡b→∞∫1b1x2+1 dx\int_1^{\infty} \frac{1}{x^2 + 1} \, dx = \lim_{b \to \infty} \int_1^b \frac{1}{x^2 + 1} \, dx

Recall: ∫1x2+1 dx=arctan⁡x+C\int \frac{1}{x^2 + 1} \, dx = \arctan x + C

=lim⁡b→∞[arctan⁡x]1b= \lim_{b \to \infty} [\arctan x]_1^b

=lim⁡b→∞(arctan⁡b−arctan⁡1)= \lim_{b \to \infty} (\arctan b - \arctan 1)

=π2−π4=π4= \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}

Step 3: Conclusion.

Since the integral converges to π4\frac{\pi}{4}, by the Integral Test, the series converges.

3Problem 3easy

❓ Question:

Determine if ∑n=1∞2n+1n3\sum_{n=1}^{\infty} \frac{2n + 1}{n^3} converges or diverges.

💡 Show Solution

Step 1: Split the series

∑n=1∞2n+1n3=∑n=1∞2nn3+∑n=1∞1n3\sum_{n=1}^{\infty} \frac{2n + 1}{n^3} = \sum_{n=1}^{\infty} \frac{2n}{n^3} + \sum_{n=1}^{\infty} \frac{1}{n^3}

=∑n=1∞2n2+∑n=1∞1n3= \sum_{n=1}^{\infty} \frac{2}{n^2} + \sum_{n=1}^{\infty} \frac{1}{n^3}

=2∑n=1∞1n2+∑n=1∞1n3= 2\sum_{n=1}^{\infty} \frac{1}{n^2} + \sum_{n=1}^{\infty} \frac{1}{n^3}

Step 2: Identify p-series

∑1n2\sum \frac{1}{n^2}: p-series with p=2>1p = 2 > 1 → converges

∑1n3\sum \frac{1}{n^3}: p-series with p=3>1p = 3 > 1 → converges

Step 3: Use series properties

Sum of two convergent series converges.

The series converges.

Alternative Method (Integral Test):

Could also note that for large nn: 2n+1n3≈2nn3=2n2\frac{2n+1}{n^3} \approx \frac{2n}{n^3} = \frac{2}{n^2}

This behaves like 1n2\frac{1}{n^2} (p-series with p=2>1p = 2 > 1), so converges.

4Problem 4hard

❓ Question:

Use the Integral Test to show that ∑n=1∞ne−n2\sum_{n=1}^{\infty} ne^{-n^2} converges.

💡 Show Solution

Step 1: Check conditions

f(x)=xe−x2f(x) = xe^{-x^2} for x≥1x \geq 1

✓ Continuous for all xx ✓ Positive for x>0x > 0

Check decreasing: f′(x)=e−x2+x⋅(−2x)e−x2f'(x) = e^{-x^2} + x \cdot (-2x)e^{-x^2}

=e−x2(1−2x2)= e^{-x^2}(1 - 2x^2)

For x>12≈0.707x > \frac{1}{\sqrt{2}} \approx 0.707: 1−2x2<01 - 2x^2 < 0, so f′(x)<0f'(x) < 0 ✓

Decreasing for x≥1x \geq 1 ✓

Step 2: Evaluate integral

∫1∞xe−x2 dx\int_1^{\infty} xe^{-x^2}\,dx

Use substitution: u=−x2u = -x^2, du=−2x dxdu = -2x\,dx

So x dx=−12dux\,dx = -\frac{1}{2}du

When x=1x = 1: u=−1u = -1 When x→∞x \to \infty: u→−∞u \to -\infty

=∫−1−∞−12eu du=12∫−∞−1eu du= \int_{-1}^{-\infty} -\frac{1}{2}e^u\,du = \frac{1}{2}\int_{-\infty}^{-1} e^u\,du

=12lim⁡a→−∞[eu]a−1= \frac{1}{2}\lim_{a \to -\infty} [e^u]_a^{-1}

=12(e−1−0)=12e= \frac{1}{2}(e^{-1} - 0) = \frac{1}{2e} (finite!)

Step 3: Conclusion

The integral converges to 12e\frac{1}{2e}.

By the Integral Test, the series converges.

Note: The series value is NOT 12e\frac{1}{2e}, but we know it converges!

🎴

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