Integral Test for MATH 141
Exam Relevance for MATH 141
Integral test appears in MATH 141 for p-series type problems. Remember the three conditions to verify.
When to Use the Integral Test
The Integral Test connects infinite series to improper integrals. If you can integrate the function, you can determine if the series converges.
Use this test when:
- The terms $a_n = f(n)$ come from a function you can integrate
- Other tests (geometric, p-series, divergence) don't immediately apply
- The function involves $\ln n$, $\frac{1}{n^p}$, or similar "nice" forms
The Integral Test
Let $f(x)$ be a function that is:
- Continuous on $[1, \infty)$
- Positive on $[1, \infty)$
- Decreasing on $[1, \infty)$
If $a_n = f(n)$, then:
$$\sum_{n=1}^{\infty} a_n \text{ and } \int_1^{\infty} f(x)\,dx \text{ both converge or both diverge}$$
Key point: The series and integral have the same convergence behavior, but NOT the same value!
Checking the Conditions
Before using the Integral Test, verify:
Continuous and Positive
Usually obvious from the formula. If $f(x) = \frac{1}{x^2}$, it's continuous and positive for $x \geq 1$.
Decreasing
Method 1: Show $f'(x) < 0$ for $x \geq 1$
Method 2: Show $f(x+1) < f(x)$ for all $x \geq 1$
Method 3: Obvious from the form (e.g., $\frac{1}{x^2}$ clearly decreases)
The Classic Example: P-Series Proof
The Integral Test is how we prove the p-series rule!
Problem: For what values of $p$ does $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^p}$ converge?
Let $f(x) = \frac{1}{x^p}$. Check conditions:
- Continuous for $x \geq 1$ ✓
- Positive for $x \geq 1$ ✓
- Decreasing for $x \geq 1$ (when $p > 0$) ✓
Evaluate the integral:
$$\int_1^{\infty} \frac{1}{x^p}\,dx = \int_1^{\infty} x^{-p}\,dx$$
Case 1: $p \neq 1$
$$= \left[\frac{x^{-p+1}}{-p+1}\right]_1^{\infty} = \frac{1}{1-p}\left[x^{1-p}\right]_1^{\infty}$$
- If $p > 1$: $1-p < 0$, so $x^{1-p} \to 0$ as $x \to \infty$. Integral = $\frac{1}{p-1}$ ✓ Converges
- If $p < 1$: $1-p > 0$, so $x^{1-p} \to \infty$. Diverges
Case 2: $p = 1$
$$\int_1^{\infty} \frac{1}{x}\,dx = [\ln x]_1^{\infty} = \infty$$
Diverges (this is the harmonic series!)
$$\boxed{\sum \frac{1}{n^p} \text{ converges if } p > 1, \text{ diverges if } p \leq 1}$$
Problem: Determine if $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n \ln n}$ converges or diverges.
Let $f(x) = \frac{1}{x \ln x}$ for $x \geq 2$.
Check conditions:
- Continuous for $x \geq 2$ ✓
- Positive for $x \geq 2$ ✓
- Decreasing: $f'(x) = -\frac{\ln x + 1}{(x \ln x)^2} < 0$ for $x \geq 2$ ✓
Evaluate the integral:
$$\int_2^{\infty} \frac{1}{x \ln x}\,dx$$
Use substitution: $u = \ln x$, $du = \frac{1}{x}dx$
$$= \int_{\ln 2}^{\infty} \frac{1}{u}\,du = [\ln u]_{\ln 2}^{\infty} = \ln(\ln x)\Big|_2^{\infty} = \infty$$
The integral diverges, so by the Integral Test:
$$\boxed{\text{The series diverges}}$$
Problem: Determine if $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n (\ln n)^2}$ converges or diverges.
Let $f(x) = \frac{1}{x (\ln x)^2}$ for $x \geq 2$.
Conditions are satisfied (similar to Example 2).
Evaluate the integral:
$$\int_2^{\infty} \frac{1}{x (\ln x)^2}\,dx$$
Substitution: $u = \ln x$, $du = \frac{1}{x}dx$
$$= \int_{\ln 2}^{\infty} \frac{1}{u^2}\,du = \left[-\frac{1}{u}\right]_{\ln 2}^{\infty} = 0 - \left(-\frac{1}{\ln 2}\right) = \frac{1}{\ln 2}$$
The integral converges, so by the Integral Test:
$$\boxed{\text{The series converges}}$$
Problem: Determine if $\displaystyle\sum_{n=1}^{\infty} n e^{-n}$ converges or diverges.
Let $f(x) = x e^{-x}$.
Check conditions:
- Continuous ✓
- Positive for $x \geq 1$ ✓
- Decreasing: $f'(x) = e^{-x} - xe^{-x} = e^{-x}(1-x) < 0$ for $x > 1$ ✓
Evaluate the integral (integration by parts):
$$\int_1^{\infty} x e^{-x}\,dx$$
Let $u = x$, $dv = e^{-x}dx$, so $du = dx$, $v = -e^{-x}$
$$= \left[-xe^{-x}\right]_1^{\infty} + \int_1^{\infty} e^{-x}\,dx$$
$$= (0 - (-e^{-1})) + \left[-e^{-x}\right]_1^{\infty}$$
$$= \frac{1}{e} + (0 - (-e^{-1})) = \frac{1}{e} + \frac{1}{e} = \frac{2}{e}$$
The integral converges, so:
$$\boxed{\text{The series converges}}$$
Problem: Can we use the Integral Test on $\displaystyle\sum_{n=1}^{\infty} \frac{\sin n}{n^2}$?
Let $f(x) = \frac{\sin x}{x^2}$.
Check conditions:
- Continuous ✓
- Positive? NO! $\sin x$ is negative for some values.
The function is not always positive, so the Integral Test does not apply.
$$\boxed{\text{Integral Test cannot be used — try another test}}$$
(This series actually converges absolutely by comparison with $\sum \frac{1}{n^2}$.)
Remainder Estimates
A bonus of the Integral Test: we can estimate how far $S_n$ is from the true sum!
If $\sum a_n$ converges, the remainder $R_n = S - S_n$ satisfies:
$$\int_{n+1}^{\infty} f(x)\,dx \leq R_n \leq \int_n^{\infty} f(x)\,dx$$
Problem: Estimate the error if we approximate $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^3}$ using the first 10 terms.
$$R_{10} \leq \int_{10}^{\infty} \frac{1}{x^3}\,dx = \left[-\frac{1}{2x^2}\right]_{10}^{\infty} = 0 + \frac{1}{200} = 0.005$$
$$R_{10} \geq \int_{11}^{\infty} \frac{1}{x^3}\,dx = \frac{1}{2(11)^2} = \frac{1}{242} \approx 0.00413$$
$$\boxed{0.00413 \leq R_{10} \leq 0.005}$$
The first 10 terms give us accuracy to within 0.005 of the true sum.
Quick Reference: Common Integrals
| Series | Integral | Result |
|---|---|---|
| $\sum \frac{1}{n^p}$ | $\int \frac{1}{x^p}dx$ | Converges if $p > 1$ |
| $\sum \frac{1}{n \ln n}$ | $\int \frac{1}{x \ln x}dx = \ln(\ln x)$ | Diverges |
| $\sum \frac{1}{n (\ln n)^p}$ | $\int \frac{1}{x (\ln x)^p}dx$ | Converges if $p > 1$ |
| $\sum n^k e^{-n}$ | $\int x^k e^{-x}dx$ | Converges (all $k$) |
Common Mistakes and Misunderstandings
❌ Mistake: Forgetting to check conditions
Wrong: "I'll just integrate $\frac{\cos n}{n}$ to test $\sum \frac{\cos n}{n}$."
Why it's wrong: $\cos x$ is not always positive, so the Integral Test doesn't apply.
Correct: Always verify: continuous, positive, AND decreasing before using the test.
❌ Mistake: Thinking the sum equals the integral
Wrong: "$\int_1^{\infty} \frac{1}{x^2}dx = 1$, so $\sum \frac{1}{n^2} = 1$."
Why it's wrong: The test says they have the same convergence behavior, not the same value! Actually, $\sum \frac{1}{n^2} = \frac{\pi^2}{6} \approx 1.645$.
Correct: Use the Integral Test only to determine convergence/divergence, not to find the sum.
❌ Mistake: Starting at the wrong index
Wrong: For $\sum_{n=2}^{\infty} \frac{1}{n \ln n}$, integrating from 1.
Why it's wrong: $\ln 1 = 0$, so $\frac{1}{x \ln x}$ is undefined at $x = 1$. Start your integral where the series starts.
Correct: Integrate from 2 to match the series: $\int_2^{\infty} \frac{1}{x \ln x}\,dx$.
❌ Mistake: Using Integral Test when simpler tests work
Wrong: Using the Integral Test on $\sum \frac{1}{n^3}$.
Why it's inefficient: This is a p-series with $p = 3 > 1$. You can immediately say it converges!
Correct: Check for geometric series, p-series, and divergence test first. Use Integral Test when those don't apply.
The Integral Test
The series and integral have the same convergence behavior. IMPORTANT: They don't have the same value — just the same convergence/divergence status.
Variables:
- $f(x)$:
- must be continuous, positive, and decreasing
- $f(n)$:
- the nth term of the series
Integral Test Remainder Estimate
Bounds the error when approximating a convergent series with partial sums. R_n is the remainder (true sum minus partial sum S_n).
Variables:
- $R_n$:
- remainder = S - S_n
- $S_n$:
- partial sum of first n terms
- $S$:
- true sum of the series
Logarithm Series Convergence
Similar to p-series but with (ln n)^p instead of n^p. The boundary is still p = 1. Note: starts at n = 2 since ln(1) = 0.
Variables:
- $p$:
- the exponent on ln(n)
- $n ≥ 2$:
- series must start at n=2 or higher
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