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Integral and Direct Comparison · Tutorial 618 of 998

Common Errors in Integral and Direct Comparison

Learn to diagnose common comparison-test errors by checking the hypotheses, the direction of each inequality, and whether a claim applies to the tail of the series.

Intermediate 11 min read

What You'll Learn

  • State the conditions required for the integral and direct comparison tests.
  • Explain why monotonicity matters when comparing a series with an improper integral.
  • Identify when missing positivity invalidates a direct comparison argument.
  • Check the direction of an inequality before drawing a convergence conclusion.
  • Use an eventual bound correctly, even when it does not hold for the first terms.
  • Explain why finitely many terms cannot determine convergence or divergence.

Why Comparison Arguments Need Care

The integral and direct comparison tests are powerful because they let us determine the behavior of a series without summing it exactly. Their conclusions, however, depend on specific hypotheses. An inequality that points the wrong way, a function that is not decreasing, or a comparison applied to signed terms can make an apparently plausible argument invalid.

A reliable comparison proof has three parts: verify the conditions of the test, establish the needed inequality in the correct direction, and apply it to all terms or to a tail. A few early terms can be handled separately; they cannot replace any of these steps.

Definition: A series \(\sum_{n=1}^{\infty} a_n\) converges if its sequence of partial sums \(S_N=\sum_{n=1}^{N}a_n\) has a finite limit as \(N\to\infty\). Otherwise it diverges. A finite number of terms does not affect whether a series converges.

What the Tests Require

For direct comparison, the standard convergence rule concerns nonnegative terms: if \(0\le a_n\le b_n\) for all sufficiently large \(n\), and \(\sum b_n\) converges, then \(\sum a_n\) converges. The standard divergence rule also requires nonnegative terms: if \(0\le b_n\le a_n\) for all sufficiently large \(n\), and \(\sum b_n\) diverges, then \(\sum a_n\) diverges.

The inequalities in these two rules have different roles. To prove convergence, the series being studied must be bounded above by a convergent series. To prove divergence, it must be bounded below by a divergent series. Reversing either inequality removes the justification for the conclusion.

The integral test has different conditions. Suppose \(f\) is continuous, positive, and decreasing on \([N,\infty)\), where \(N\) is a positive integer, and \(a_n=f(n)\) for \(n\ge N\). Then \(\sum_{n=N}^{\infty}a_n\) and \(\int_N^\infty f(x)\,dx\) either both converge or both diverge. Decreasing here means nonincreasing: \(f(x_2)\le f(x_1)\) whenever \(x_1\le x_2\).

Important distinction: Monotonicity is a hypothesis of the integral test, not of direct comparison. Direct comparison instead requires the stated termwise inequality and nonnegative terms.

Why Monotonicity Matters for the Integral Test

If \(f\) is decreasing, then on each interval \([n,n+1]\) its values lie between \(f(n+1)\) and \(f(n)\). Consequently,

$$ f(n+1)\le \int_n^{n+1}f(x)\,dx\le f(n). $$

Adding these inequalities over consecutive intervals relates the series tail to the improper integral. In particular, for \(M\ge N\),

$$ \sum_{n=N+1}^{M+1} f(n) \le \int_N^{M+1}f(x)\,dx \le \sum_{n=N}^{M} f(n). $$

The inequalities depend on the function staying below \(f(n)\) and above \(f(n+1)\) throughout each interval. Without monotonicity, those rectangle bounds need not hold. This is why having a positive continuous function and matching its values at the integers is not, by itself, enough to invoke the integral test.

Worked Example: A Positive Function That Is Not Decreasing

Worked Example: A Positive Function That Is Not Decreasing

We construct a positive continuous function whose values at the integers give a divergent series, although its improper integral is finite. This shows why continuity and positivity alone do not justify the integral test.

For \(x\ge1\), define

$$ f(x)=e^{-x}+ \sum_{k=1}^{\infty}\frac{1}{k} \max\!\left(1-2k^3|x-k|,\,0\right). $$

Each term in the sum is a triangular bump centered at the integer \(k\), with height \(1/k\) and support from \(k-\frac{1}{2k^3}\) to \(k+\frac{1}{2k^3}\). The supports do not overlap, so on any bounded interval only finitely many bumps occur. The function is continuous and is positive because \(e^{-x}>0\).

At every positive integer \(n\), the bump centered at \(n\) has value \(1/n\), while all other bumps have value zero. Therefore

$$ f(n)=e^{-n}+\frac{1}{n}\ge\frac{1}{n}. $$

Thus the partial sums satisfy

$$ \sum_{n=1}^{M}f(n) \ge \sum_{n=1}^{M}\frac{1}{n}. $$

The harmonic series on the right diverges, so \(\sum f(n)\) diverges.

Now consider the integral. The exponential part has finite integral:

$$ \int_1^\infty e^{-x}\,dx=e^{-1}. $$

The full area of the triangular bump centered at \(k\) is one half times its base width \(1/k^3\) times its height \(1/k\), hence

$$ \frac{1}{2}\left(\frac{1}{k^3}\right)\left(\frac{1}{k}\right) =\frac{1}{2k^4}. $$

The first bump is only partly included on \([1,\infty)\), so its included area is no larger than this full area. Since \(\sum 1/k^4\) converges, the total bump area is finite. Hence \(\int_1^\infty f(x)\,dx\) is finite, while \(\sum_{n=1}^{\infty}f(n)\) diverges.

There is no contradiction with the integral test: this function is not decreasing. Its narrow peaks repeatedly rise as \(x\) passes through the integers, so the rectangle bounds used in the test do not apply.

Worked Example: Missing Positivity

Worked Example: Missing Positivity

A common mistake is to use the convergence comparison rule without checking that the terms of the series being studied are nonnegative. Consider

$$ a_n=-\frac{1}{n}, \qquad b_n=2^{-n}. $$

For every \(n\ge1\), \(-1/n<0<2^{-n}\), so \(a_n\le b_n\). Also,

$$ \sum_{n=1}^{M}2^{-n}=1-2^{-M} \longrightarrow 1, $$

and therefore \(\sum b_n\) converges. It would be invalid to conclude from \(a_n\le b_n\) that \(\sum a_n\) converges: the required lower bound \(0\le a_n\) is false.

Indeed, the partial sums of the series being studied are

$$ \sum_{n=1}^{M}a_n =-\sum_{n=1}^{M}\frac{1}{n}. $$

The harmonic partial sums grow without bound, so these partial sums decrease without bound. Thus \(\sum a_n\) diverges. The example verifies that the nonnegativity condition cannot simply be omitted from the direct comparison convergence rule.

Follow the Inequality in the Correct Direction

For nonnegative terms, the convergence rule uses an upper bound by a convergent series; the divergence rule uses a lower bound by a divergent series. A divergent upper bound does not establish divergence, and a convergent lower bound does not establish convergence.

1
To prove convergence.
Find a convergent comparison series \( \sum b_n \) with \(0\le a_n\le b_n\) eventually.
2
To prove divergence.
Find a divergent comparison series \( \sum b_n \) with \(0\le b_n\le a_n\) eventually.
3
Check the actual relation.
Do not infer that a series behaves like its comparison merely because both appear in the same inequality.

Worked Example: A Reversed Inequality Gives No Conclusion

Worked Example: A Reversed Inequality Gives No Conclusion

Consider \(a_n=1/[n(n+1)]\) and \(b_n=1/n\). For \(n\ge1\),

$$ 0<\frac{1}{n(n+1)}\le\frac{1}{n}, $$

because \(n(n+1)\ge n\). The series \(\sum b_n\) diverges, but the inequality is an upper bound on \(a_n\), not a lower bound. The divergence comparison rule therefore does not apply.

In fact, this series converges. Its terms can be rewritten as

$$ \frac{1}{n(n+1)} =\frac{1}{n}-\frac{1}{n+1}. $$

The \(M\)th partial sum is consequently

$$ \begin{aligned} \sum_{n=1}^{M}\frac{1}{n(n+1)} &=\left(1-\frac12\right) +\left(\frac12-\frac13\right) +\cdots +\left(\frac1M-\frac{1}{M+1}\right)\\ &=1-\frac{1}{M+1}. \end{aligned} $$

As \(M\to\infty\), \(1/(M+1)\to0\), so the partial sums tend to \(1\). This calculation confirms convergence and shows concretely why an upper bound by a divergent series proves nothing about divergence.

Eventual Bounds Are Enough

Comparison tests are often stated for every \(n\), but they remain valid when the required conditions hold only for \(n\ge N\). The first \(N-1\) terms contribute a fixed finite amount to each partial sum. Adding or removing a fixed finite amount cannot change whether the partial sums have a finite limit.

Takeaway: A comparison need not work at the beginning of a series. It must work for every term in some tail, and the conditions of the chosen test must hold on that tail.

Worked Example: A Lower Bound That Starts at the Second Term

Worked Example: A Lower Bound That Starts at the Second Term

Let \(a_n=1/(n+2)\). We will prove that \(\sum a_n\) diverges using a harmonic lower bound that is valid from \(n=2\), but not from \(n=1\).

For \(n\ge2\), all denominators are positive, and

$$ \frac{1}{n+2}\ge\frac{1}{2n} \quad\Longleftrightarrow\quad 2n\ge n+2 \quad\Longleftrightarrow\quad n\ge2. $$

At \(n=1\), the proposed inequality fails: \(1/3<1/2\). This does not matter because the bound is valid for every \(n\ge2\). For \(M\ge2\),

$$ \sum_{n=1}^{M}\frac{1}{n+2} \ge \sum_{n=2}^{M}\frac{1}{2n} = \frac12\sum_{n=2}^{M}\frac{1}{n}. $$

The right-hand side grows without bound as \(M\to\infty\), since removing the first term from the harmonic series does not affect its divergence. Therefore the partial sums of \(\sum 1/(n+2)\) grow without bound, and the series diverges. The argument checks the inequality at its actual starting index rather than silently applying it to the first term.

Why Finite Evidence Cannot Decide a Series

A series is determined by the behavior of all its terms, especially its tail. No matter how many terms have been calculated, a finite list alone cannot establish convergence or divergence. For instance, define two sequences by

$$ a_n=\frac1n\quad(n\ge1), \qquad b_n= \begin{cases} \frac1n,&1\le n\le100,\\[4pt] \frac1{n^2},&n\ge101. \end{cases} $$

Their first 100 terms agree exactly. Nevertheless, \(\sum a_n\) diverges because it is the harmonic series. The series \(\sum b_n\) converges: its first 100 terms have a finite sum, and its remaining tail is part of the convergent \(p\)-series \(\sum 1/n^2\). The same finite calculation is compatible with either outcome; the tail conditions decide the result.

A Practical Error-Checking Routine

Before accepting a comparison argument, check each of the following points. These checks are especially useful when a proof is short enough that a missing hypothesis can be easy to overlook.

  1. For the integral test, identify \(f\) and verify that it is continuous, positive, and decreasing on the interval being used.
  2. For direct comparison, verify that the terms meet the nonnegativity conditions of the rule being applied.
  3. Write the inequality explicitly and check its direction: upper bound for convergence, lower bound for divergence.
  4. Check the inequality algebraically on the claimed range of indices, including the stated starting index.
  5. If the bound holds only eventually, separate the finite initial terms from the tail.
  6. Do not infer a result from a finite number of terms or partial sums; establish a statement about the tail.

The central discipline is to match the conclusion to the hypotheses actually verified. A comparison is not valid because the expressions look similar or because a few numerical terms support it. It is valid when the required conditions and inequality hold on the relevant tail.

Check Your Understanding

For each question, identify the condition or comparison direction that controls the conclusion.

  1. What three properties of \(f\) are required for the integral test on a tail?
  2. If \(0\le a_n\le b_n\) eventually and \(\sum b_n\) diverges, what can you conclude about \(\sum a_n\)? Explain why.
  3. Why does \(a_n\le b_n\), together with convergence of \(\sum b_n\), not by itself guarantee convergence of \(\sum a_n\)?
  4. Suppose a valid lower bound by a divergent series holds only for \(n\ge7\). Why can it still prove divergence?
  5. Can two series with the same first 100 terms have different convergence behavior? Give the principle that explains your answer.

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