Integral Test: Conditions and Conclusion

Why Compare a Series with an Integral?

A series adds the values of a function at whole-number inputs, while an improper integral accumulates the function’s values across a continuous interval. When the function is positive and decreases steadily, these two ways of accumulating values have the same basic convergence behavior. The integral test makes that relationship useful: an integral that is finite corresponds to a convergent series, and an integral that grows without bound corresponds to a divergent series.

The conditions matter. The function must match the series terms at integer inputs from some point onward, and it must be positive, continuous, and decreasing on the interval being considered. Earlier terms need not meet those conditions; a finite number of them cannot decide whether an infinite series converges.

Key idea: The integral test classifies convergence, not the exact value of a series. Its conclusion applies when the function satisfies the stated conditions on the tail of the series.

The Integral Test

Suppose a series has terms $a_n=f(n)$ for all integers $n$ from some index $N$ onward. Here $f$ is defined on the whole interval $[N,\infty)$. The relevant integral is improper: it is evaluated by taking a limit as its upper endpoint tends to infinity.

Theorem: Integral Test. Let $N$ be a positive integer, and suppose $f:[N,\infty)\to\mathbb{R}$ is continuous, positive, and nonincreasing. If $a_n=f(n)$ for every integer $n\ge N$, then $$ \sum_{n=N}^{\infty} a_n \text{ converges} \quad\Longleftrightarrow\quad \int_N^{\infty} f(x)\,dx \text{ converges}. $$ If the improper integral diverges, then the series diverges. Since $f(x)>0$, the integral can only diverge to $+\infty$. The same conclusion applies if the series has any finite number of additional initial terms.

“Nonincreasing” means that $f(y)\le f(x)$ whenever $N\le x\le y$. Some statements of the integral test use “decreasing” to mean strictly decreasing; the test also holds under the weaker nonincreasing condition. Positivity here means $f(x)>0$ throughout the interval.

For an improper integral of a positive function, convergence means that the limit below exists and is finite:

\int_N^{\infty} f(x)\,dx = \lim_{R\to\infty}\int_N^R f(x)\,dx.

Thus, computing an antiderivative is not by itself enough. One must evaluate the limit as $R\to\infty$ and check whether it is finite.

Why the Conditions Make Sense

The function values $f(n)$ are sampled at integer points, whereas the integral accumulates values between those points as well. Positivity prevents cancellation, and monotonicity keeps the function’s values on each unit interval controlled by its endpoint values. Consequently, sampling the function at successive integers cannot produce a series with a different convergence behavior from the total accumulated area. Continuity ensures the function is integrable on each bounded interval. The detailed inequalities behind this area comparison are the basis of the test.

Each condition has a role. If the function is not positive, positive and negative contributions could cancel in the integral or the series. If it is not eventually nonincreasing, its integer samples may not represent the area in the controlled way the test requires. If it is not continuous, the theorem’s hypotheses have not been checked. A conclusion from the integral test is justified only after the conditions hold on the interval from $N$ onward.

Worked Examples

Worked Example: A Convergent Series

Determine whether $\sum_{n=1}^{\infty}\frac{1}{n^2+1}$ converges by the integral test.

Set $f(x)=\frac{1}{x^2+1}$ for $x\ge 1$. Then $f(n)=\frac{1}{n^2+1}$. The denominator is positive, so $f(x)>0$; the function is continuous for $x\ge1$. Its derivative is

f'(x)=\frac{-2x}{(x^2+1)^2}<0 \qquad (x\ge1).

Therefore $f$ is decreasing on the required interval. Evaluate the improper integral using the antiderivative $\arctan x$:

\int_1^\infty \frac{1}{x^2+1}\,dx = \lim_{R\to\infty}\left[\arctan x\right]_1^R = \lim_{R\to\infty}\left(\arctan R-\frac{\pi}{4}\right) = \frac{\pi}{2}-\frac{\pi}{4} = \frac{\pi}{4}.

The integral is finite, so the series converges. Check the antiderivative: $\frac{d}{dx}\arctan x=\frac{1}{1+x^2}$, which is the integrand. The derivative check confirms the integration step; the finite limit supplies the convergence conclusion.

Worked Example: A Divergent Series

Use the integral test to classify $\sum_{n=1}^{\infty}\frac{1}{\sqrt{n+1}}$.

Take $f(x)=\frac{1}{\sqrt{x+1}}$ for $x\ge1$. It matches the terms because $f(n)=\frac{1}{\sqrt{n+1}}$. The function is positive and continuous on $[1,\infty)$, and

f'(x)=-\frac{1}{2(x+1)^{3/2}}<0 \qquad (x\ge1),

so it is decreasing. An antiderivative is $2\sqrt{x+1}$, since

\frac{d}{dx}\left(2\sqrt{x+1}\right)=\frac{1}{\sqrt{x+1}}.

Therefore, for $R\ge1$,

\int_1^R \frac{1}{\sqrt{x+1}}\,dx = 2\sqrt{R+1}-2\sqrt{2}.

As $R\to\infty$, this expression tends to $+\infty$. The improper integral diverges, so the series diverges. The antiderivative differentiates back to the integrand, and its limit confirms that the integral is unbounded.

Worked Example: A Tail That Starts After the Initial Terms

Let $a_1=7$, $a_2=-4$, and $a_n=e^{-n}$ for every integer $n\ge3$. Determine whether $\sum_{n=1}^{\infty}a_n$ converges.

The first two terms do not come from the positive function used for the tail. For $n\ge3$, set $f(x)=e^{-x}$ on $[3,\infty)$. Then $a_n=f(n)$. The function is continuous and positive, and

f'(x)=-e^{-x}<0 \qquad (x\ge3),

so it is decreasing. Its improper integral is

\int_3^\infty e^{-x}\,dx = \lim_{R\to\infty}\left[-e^{-x}\right]_3^R = \lim_{R\to\infty}(e^{-3}-e^{-R}) = e^{-3}.

This is finite, so the tail series $\sum_{n=3}^{\infty}e^{-n}$ converges by the integral test. Adding the finite initial sum $a_1+a_2=7+(-4)=3$ preserves convergence. The antiderivative $-e^{-x}$ differentiates to $e^{-x}$, and the limit verifies the integral’s value.

Why Finite Initial Terms Do Not Matter

A series converges when its partial sums approach a finite limit. Changing, removing, or adding finitely many terms changes every sufficiently late partial sum by the same fixed amount. That can shift the limit, but it cannot create or remove the existence of a finite limit.

Proof: Finite changes preserve convergence. Suppose the terms from index $N$ onward form a tail series, and let $C=\sum_{n=1}^{N-1}a_n$, a finite sum. For every $m\ge N$, the partial sum of the full series satisfies $$ \sum_{n=1}^{m}a_n = C+\sum_{n=N}^{m}a_n. $$ If the tail partial sums approach a finite limit $L$, the full partial sums approach $C+L$. Conversely, if the full partial sums approach a finite limit, subtracting the constant $C$ shows that the tail partial sums approach a finite limit. Thus the full series converges exactly when its tail does.

This is why the integral test needs its hypotheses only eventually. The values $a_1,\ldots,a_{N-1}$ may be different from $f(1),\ldots,f(N-1)$, or may fail to be positive. They have no effect on whether the series converges. The test classifies the tail, and the finite-change argument transfers that conclusion to the full series.

A Common Mistake: Treating the Integral as the Sum

The integral test does not say that the series sum equals the improper integral. It says that the two have the same convergence status under the theorem’s conditions. For example, the calculation

\int_1^\infty \frac{1}{x^2+1}\,dx=\frac{\pi}{4}

does not show that $\sum_{n=1}^{\infty}\frac{1}{n^2+1}=\frac{\pi}{4}$. That would be an incorrect conclusion: the integral accumulates values over a continuous interval, while the series adds values at integer inputs. The valid conclusion is only that the integral is finite, so the series converges.

A second common error is to apply the test before checking that the function is decreasing on the whole tail under consideration. A derivative that is negative only for sufficiently large $x$ may still be enough: choose an integer $N$ beyond that point, verify all the conditions on $[N,\infty)$, and treat the earlier terms as a finite prefix. But without establishing eventual monotonicity and the other hypotheses, the test has not yet been justified.

Practical Checklist

Before drawing a conclusion from the integral test, check each point:

Does a function $f$ satisfy $a_n=f(n)$ for every integer $n$ from some index onward?
Is $f$ continuous and positive on the interval beginning at that index?
Is $f$ nonincreasing on that entire interval?
Does the improper integral have a finite limit, or does it grow without bound?
Have you concluded only convergence or divergence, rather than claiming the integral equals the series?

If the hypotheses hold and the integral converges, the series converges. If the hypotheses hold and the integral diverges, the series diverges. If a condition fails, this theorem gives no conclusion; failure of a test’s conditions is not itself evidence that the series diverges or converges.

Check Your Understanding

Answer these questions using the theorem and its conditions.

State the three main conditions on $f$ in the integral test.
If the relevant improper integral is finite, what conclusion follows for the series tail?
Why can finitely many initial terms be negative without changing the convergence conclusion?
A function matches the terms at integers and is positive and continuous, but is not known to be eventually nonincreasing. Can the integral test conclusion be applied as stated?
Does the integral test assert that the series sum equals the improper integral? Explain.