Why the Integral Test Needs Justification
A series of positive terms can sometimes be studied by comparing its terms with the area under a curve. The Integral Test makes this connection precise, but it is not enough to find a function whose formula resembles the terms. The function must satisfy specific conditions on an interval containing the relevant integers.
In particular, you must verify positivity, continuity, and decrease. These conditions allow the values \(f(n)\) to be related to areas over intervals of length one. Once the conditions are established, convergence of the improper integral determines convergence of the series, and divergence of the integral determines divergence of the series.
The Integral Test and Its Conditions
Suppose \(a_n=f(n)\) for every integer \(n\geq N\), where \(N\) is a positive integer. The function must be positive and continuous on \([N,\infty)\), and decreasing there. In practice, showing that \(f'(x)<0\) for every \(x\geq N\) is a common way to establish that \(f\) is decreasing.
The conditions are required on a tail of the domain, not necessarily from the very first term of a series. Omitting or adding finitely many terms does not change whether a series converges: the partial sums are changed by only a fixed finite amount. Similarly, changing the lower endpoint of a convergent or divergent improper integral by a finite amount does not change its behavior at infinity, provided the integral over the finite interval is well-defined.
Why the Test Works
If \(f\) is decreasing, then on each interval \([k,k+1]\), its values lie between its endpoint values:
Integrating these inequalities over an interval of width one gives:
Adding the inequalities across consecutive intervals connects sums of terms to areas. This is the whole reason the decrease condition matters: it ensures that the graph on each unit interval stays between the heights at its endpoints.
Verify that \(f\) is positive, continuous, and decreasing for every \(x\geq N\), and that \(a_n=f(n)\) for integers \(n\geq N\).
Use \(f(k+1)\leq\int_k^{k+1}f(x)\,dx\leq f(k)\), then add over consecutive intervals.
If the integral is finite, the positive series has bounded partial sums; if the series has bounded partial sums, the integral is bounded. In each case the relevant increasing quantities converge.
Here is the complete partial-sum comparison behind the conclusion. For an integer \(M>N\), adding the interval inequalities gives:
If \(\int_N^\infty f(x)\,dx\) converges, the series partial sums on the middle are bounded above by that finite integral, so the positive-term series converges. Conversely, if the series converges, the middle sums are bounded above, and the integral on the left is bounded as \(M\) increases. Because \(f\) is positive, the improper integral is the limit of an increasing collection of integrals; boundedness therefore gives a finite limit. The initial term \(f(N)\) is just one finite term and does not affect convergence.
Worked Examples
Worked Example: A Convergent Series from a Rational Function
Use the Integral Test to determine whether \(\sum_{n=1}^{\infty}\frac{1}{n^2+4}\) converges.
Choose \(f(x)=\frac{1}{x^2+4}\), so \(f(n)=\frac{1}{n^2+4}\). For \(x\geq1\), the denominator is positive, so \(f(x)>0\). The rational function is continuous wherever its denominator is nonzero; since \(x^2+4>0\) for all real \(x\), it is continuous on \([1,\infty)\).
Differentiate to check decrease:
For \(x\geq1\), the numerator is negative and the denominator is positive, so \(f'(x)<0\). Thus \(f\) is decreasing on the required interval.
Now evaluate the improper integral:
This is finite, so the Integral Test says the series converges. As a check on the antiderivative, differentiating \(\frac12\arctan(x/2)\) gives \(\frac12\cdot\frac{1}{1+(x/2)^2}\cdot\frac12=\frac{1}{x^2+4}\), as required.
Worked Example: A Divergent Series with a Logarithmic Function
Determine whether \(\sum_{n=2}^{\infty}\frac{1}{n\ln n}\) converges.
Let \(f(x)=\frac{1}{x\ln x}\). For \(x\geq2\), \(x>0\) and \(\ln x>0\), so \(f(x)>0\). The denominator is nonzero on \([2,\infty)\), so \(f\) is continuous there. Also \(f(n)=\frac{1}{n\ln n}\) for each integer \(n\geq2\).
The derivative is:
For \(x\geq2\), the numerator \(\ln x+1\) is positive and the denominator is positive, so \(f'(x)<0\). Therefore, \(f\) is decreasing on \([2,\infty)\).
Evaluate the integral using \(u=\ln x\), so \(du=\frac{1}{x}\,dx\):
The series diverges by the Integral Test. The antiderivative checks because \(\frac{d}{dx}\ln(\ln x)=\frac{1}{x\ln x}\) for \(x>1\).
Worked Example: A Decreasing Function with an Exponential Factor
Determine whether \(\sum_{n=1}^{\infty}(n+1)e^{-n}\) converges.
Set \(f(x)=(x+1)e^{-x}\), which satisfies \(f(n)=(n+1)e^{-n}\). For \(x\geq1\), both factors \(x+1\) and \(e^{-x}\) are positive, so \(f(x)>0\). Both are continuous, so their product is continuous.
Using the product rule:
For \(x\geq1\), \(x>0\) and \(e^{-x}>0\); hence \(f'(x)<0\), and \(f\) is decreasing.
An antiderivative is \(-(x+2)e^{-x}\), because:
Therefore:
The limit term is zero: exponential decay dominates the linear factor, so \((b+2)e^{-b}\to0\). The integral is finite, and the Integral Test shows that the series converges.
Common Pitfall: Ignoring the Starting Point
A function need not satisfy the hypotheses before the interval on which the test is applied. For example, in the logarithmic example, \(f(x)=1/(x\ln x)\) is not defined at \(x=1\), because \(\ln1=0\). It would be incorrect to claim that this function is continuous on \([1,\infty)\) or to write the Integral Test integral starting at \(1\).
The series in that example starts at \(n=2\), and \(f\) is positive, continuous, and decreasing on \([2,\infty)\). Thus the test applies with \(N=2\). The correction is to use the valid interval:
not an integral beginning at \(1\). More generally, check the function on the actual tail where its values represent the terms. A finite number of earlier terms cannot alter convergence, but a condition failure on the claimed interval must not be ignored.
Writing a Complete Integral-Test Justification
A sound justification makes the logical chain explicit. Identify the function and the index from which its values match the terms. Then establish positivity, continuity, and decrease on the same interval before evaluating the improper integral.
- State \(f(x)\) and verify \(a_n=f(n)\) for all integers \(n\geq N\).
- Explain why \(f(x)>0\) and why \(f\) is continuous on \([N,\infty)\).
- Show that \(f\) is decreasing there, often by proving \(f'(x)<0\).
- Evaluate or classify \(\int_N^\infty f(x)\,dx\), including its limiting behavior.
- State the corresponding conclusion about the series and name the Integral Test.
The conditions are not decorative. Positivity lets the partial sums and integrals increase as their upper endpoints grow; continuity supports the interval integrals; and decrease provides the rectangle bounds connecting terms to areas. If any condition has not been checked on the chosen tail, the Integral Test conclusion has not yet been justified.
Check Your Understanding
For each question, focus on the hypotheses as well as the final convergence conclusion.
- State the positivity, continuity, and monotonicity conditions required by the Integral Test.
- Why does decrease imply \(f(k+1)\leq\int_k^{k+1}f(x)\,dx\leq f(k)\) when \(f\) is positive?
- For \(f(x)=1/(x\ln x)\), why is \(N=2\) a valid starting point, while \(N=1\) is not?
- If \(\int_N^\infty f(x)\,dx\) diverges and the Integral Test hypotheses hold, what follows about \(\sum_{n=N}^\infty f(n)\)?
- Why does adding or removing a finite number of terms not change whether a series converges?
