Calculus II: Integral Test for Series

Determining whether an infinite series converges or diverges is a fundamental skill in advanced calculus, with direct applications in engineering fields from signal analysis to computational physics. The Integral Test provides a powerful bridge between the discrete world of sums and the continuous world of integrals, allowing you to leverage your integration skills to solve series problems. This test is particularly valuable for series whose terms come from a function you can integrate, offering not just a yes/no answer on convergence but also a method for estimating the sum's value—a crucial capability for approximations in real-world design and modeling.

The Integral Test: Statement and Logical Conditions

The Integral Test establishes a direct link between an infinite series and an improper integral. Formally, let ${\sum }_{n=1}^{\infty }{a}_n$ be a series with positive terms. Suppose $a_n=f(n)$, where $f$ is a function that is continuous, positive, and decreasing on the interval $[1,\infty )$. Under these conditions, the infinite series and the corresponding improper integral share the same convergence behavior:

$$\sum _{n=1}^{\infty }{a}_n\text{ converges if and only if }{\int }_1^{\infty }f(x)dx\text{ converges}.$$

The logic hinges on a visual comparison using areas. Imagine plotting the points of the sequence $a_n$ and the curve $f(x)$. You can trap the sum of the series between two integrals. By constructing left-endpoint and right-endpoint Riemann sums, you get the key inequality:

$${\int }_1^{N+1}f(x)dx\le \sum _{n=1}^N{a}_n\le a_1+{\int }_1^{N}f(x)dx.$$

If the improper integral ${\int }_1^{\infty }f(x)dx$ converges to a finite number $L$, the partial sums of the series are bounded above and must converge. Conversely, if the integral diverges to infinity, the partial sums are unbounded below, forcing the series to diverge as well. The three function conditions are not mere formalities; if $f$ is not eventually decreasing, the inequality fails, and the test is invalid.

Applying the Test to p-Series and Standard Examples

The classic and most important application of the Integral Test is to the p-series, which has the general form ${\sum }_{n=1}^{\infty }\frac{1}{n^p}$. Here, $f(x)=1/x^p$. This function meets the test's conditions for $x\ge 1$ when $p>0$ (it's positive and decreasing). For $p\le 0$, the terms do not approach zero, so the series diverges by the Test for Divergence.

To apply the Integral Test for $p>0$, you evaluate the improper integral:

$${\int }_1^{\infty }\frac{1}{x^p}dx={\int }_1^{\infty }{x}^{-p}dx.$$

The antiderivative is $\frac{x^{1-p}}{1-p}$ for $p\ne 1$. Evaluating the limit as $b\to \infty$:

• If $p>1$, then $1-p<0$, so ${\lim }_{b\to \infty }{b}^{1-p}=0$. The integral converges to $\frac{1}{p-1}$.
• If $0<p<1$, then $1-p>0$, so ${\lim }_{b\to \infty }{b}^{1-p}=\infty$. The integral diverges.

The special case $p=1$ (the harmonic series) requires separate handling: $${\int }_1^{\infty }\frac{1}{x}dx=\underset{b\to \infty }{\lim }\ln (b)=\infty ,$$ which diverges. Therefore, the p-series converges if $p>1$ and diverges if $p\le 1$, a result proven decisively by the Integral Test.

Consider a more engineering-focused example: ${\sum }_{n=1}^{\infty }\frac{n}{e^{n^2}}$. Here, $f(x)=x{e}^{-x^2}$. This function is positive and decreasing for $x\ge 1$ (you can verify the derivative is negative). The associated integral is solved using substitution, letting $u=x^2$: $${\int }_1^{\infty }x{e}^{-x^2}dx=\underset{b\to \infty }{\lim }{\left[-\frac{1}{2}{e}^{-x^2}\right]}_1^b=\frac{1}{2e}.$$ Since this improper integral converges to a finite value, the original series converges by the Integral Test.

Estimating Sums and the Remainder Estimation Theorem

One of the test's most practical strengths is its ability to provide numerical bounds for a series' sum, which is essential when a closed-form expression doesn't exist. The inequalities used in the test's proof lead directly to estimation techniques.

Let $S={\sum }_{n=1}^{\infty }{a}_n$ be a convergent series satisfying the Integral Test conditions, and let $S_N={\sum }_{n=1}^N{a}_n$ be its $N$th partial sum. The remainder, or error, when approximating $S$ by $S_N$ is $R_N=S-S_N={\sum }_{n=N+1}^{\infty }{a}_n$.

The Remainder Estimation Theorem for the Integral Test provides bounds for this error. Because $f$ is decreasing, you can once again trap the tail of the series between integrals:

$${\int }_{N+1}^{\infty }f(x)dx\le R_N\le {\int }_N^{\infty }f(x)dx.$$

This allows you to estimate $S$ with controlled precision: $S_N+{\int }_{N+1}^{\infty }f(x)dx\le S\le S_N+{\int }_N^{\infty }f(x)dx$.

Example: Estimate the sum of ${\sum }_{n=1}^{\infty }\frac{1}{n^3}$ with an error less than 0.005. You first find an $N$ such that $R_N<0.005$. Using the upper bound, you need ${\int }_N^{\infty }\frac{1}{x^3}dx<0.005$. Evaluating, ${\int }_N^{\infty }{x}^{-3}dx={\lim }_{b\to \infty }{\left[-\frac{1}{2x^2}\right]}_N^b=\frac{1}{2N^2}$. Solving $\frac{1}{2N^2}<0.005$ gives $N^2>100$, so $N>10$. Using $N=11$, you compute $S_{11}\approx 1.2010$. The bounds are $S_{11}+{\int }_{12}^{\infty }f(x)dx\le S\le S_{11}+{\int }_{11}^{\infty }f(x)dx$. Calculating these integrals gives $1.2010+0.00347\le S\le 1.2010+0.00413$, or $1.2045\le S\le 1.2051$. You can confidently report $S\approx 1.205$ within your error tolerance.

Strategic Test Selection: Integral vs. Comparison

You now have multiple tools for testing series convergence. Choosing efficiently saves time and avoids algebraic dead ends. The Integral Test is your primary candidate when the series term $a_n$ is clearly generated by a function $f(x)$ that is easy to integrate and you can verify the three conditions (positive, continuous, decreasing). It is especially powerful for series involving logarithms, exponentials, and rational functions that resemble $1/x^p$.

In contrast, the Comparison Tests (Direct and Limit) are often faster when you can easily identify a simpler, benchmark series (like a p-series or geometric series) with known behavior. Use this strategic flowchart:

1. First, always apply the Test for Divergence. If ${\lim }_{n\to \infty }{a}_n\ne 0$, you're done.
2. Is the series a recognizable type? Geometric or p-series? Apply their known rules immediately.
3. Does the term involve factorials or $n$th powers? The Ratio or Root Test is likely the best first attempt.
4. For other positive-term series, ask: "Is $a_n$ easily integrable?" If yes, and you can confirm $f(x)$ is decreasing (often by checking that $f^{\prime }(x)<0$), proceed with the Integral Test. This test is particularly advantageous if you also need to estimate the sum.
5. If integration is messy or the decreasing condition is hard to verify, try comparison. Look at the dominant term as $n\to \infty$. For example, for $\sum \frac{n^2+5}{2n^4-n}$, the behavior is like $\frac{1}{n^2}$ (convergent p-series). The Limit Comparison Test with $b_n=1/n^2$ is a swift, one-line calculation.

A hybrid strategic insight: The Integral Test proved the p-series rule, so you are often using the result of the Integral Test indirectly when you apply the p-series as a benchmark in a Comparison Test. The Integral Test is the theory; the Comparison Test is the efficient practice for similar series.

Common Pitfalls

1. Neglecting to Verify All Conditions: The most common error is applying the test to a function that is not eventually decreasing. For $f(x)=\frac{x}{x^2+1}$, the derivative $f^{\prime }(x)$ is positive for small $x$, so the function increases before it decreases. You must verify that $f^{\prime }(x)<0$ for all $x\ge N$ (often for $x\ge 1$). If in doubt, compute the derivative.
2. Misapplying the Test to Alternating Series: The Integral Test is only for series with positive terms. Applying it directly to something like $\sum (-1{)}^n/n$ is incorrect. For alternating series, you must use the Alternating Series Test.
3. Incorrect Integral Bounds in Estimation: When using the Remainder Estimation Theorem, the bounds are ${\int }_{N+1}^{\infty }f(x)dx\le R_N\le {\int }_N^{\infty }f(x)dx$. A frequent mistake is to use ${\int }_N^{\infty }$ for the lower bound, which is false for a decreasing function. Remember, the leftover sum is larger than the area under the curve from $N+1$ onward.
4. Choosing the Wrong Test Strategically: Attempting the Integral Test on $\sum \frac{n!}{n^n}$ is possible but leads to a very difficult integral involving the gamma function. The Ratio Test is clearly the simpler choice here. Always assess the complexity of the resulting integral before committing.

Summary

• The Integral Test states that for a positive, continuous, eventually decreasing function $f$ where $a_n=f(n)$, the series $\sum a_n$ converges if and only if the improper integral ${\int }_1^{\infty }f(x)dx$ converges.
• A primary application is proving the p-series rule: $\sum 1/n^p$ converges for $p>1$ and diverges for $p\le 1$.
• Beyond convergence checks, the test provides powerful estimation tools. The Remainder Estimation Theorem gives concrete bounds for the error $R_N$ when approximating the infinite sum by a partial sum, allowing for precise calculations to a desired accuracy.
• Strategically, the Integral Test is best suited for series with integrable terms where you need an error estimate. For simple convergence questions on series that resemble p-series, the Comparison Tests are often more efficient.
• Always meticulously verify the three conditions—especially that the function is decreasing—before applying the test, and never use it on series with negative or alternating terms.