Why a Lower Bound Can Prove Divergence
A series can be difficult to sum exactly, even when its terms are easy to compare. For divergence, an exact sum is often unnecessary: it can be enough to show that every term in the series is at least as large as the corresponding term of a simpler divergent series. The partial sums then inherit that lower bound.
The key is to compare sums of terms, not just individual terms in isolation. If the lower-bound series accumulates arbitrarily large partial sums, the larger series cannot have bounded partial sums. This reasoning gives a precise divergence test.
Partial Sums and Divergence
For a comparison on a tail beginning at \(N\), it is useful to write the partial sums from that point separately. Any finite number of initial terms contributes only a fixed finite amount. It cannot prevent the partial sums from becoming arbitrarily large if the tail sums do so.
The theorem requires the inequality and nonnegativity to hold from some index onward. The terms before \(N\) may be different or even negative: there are only finitely many of them, so their sum is a fixed constant. The essential requirement is that the lower-bound tail has unbounded partial sums.
Why the Rule Works
Let \(A_m=\sum_{n=1}^{m}a_n\) and \(B_m=\sum_{n=1}^{m}b_n\). For every \(m\ge N\), add the termwise inequalities from \(n=N\) to \(n=m\). This gives a comparison of the corresponding tail sums.
For \(m\ge N\), \( \sum_{n=N}^{m}a_n\ge \sum_{n=N}^{m}b_n\), since adding inequalities preserves their order.
Because \(b_n\ge0\) and its tail series diverges, the sums \(\sum_{n=N}^{m}b_n\) grow without bound as \(m\) increases.
\(A_m=\sum_{n=1}^{N-1}a_n+\sum_{n=N}^{m}a_n\). The first sum is fixed, while the second is at least an unbounded tail sum. Therefore \(A_m\) tends to \(+\infty\).
This proves the theorem. In particular, the conclusion does not depend on whether the terms of \(a_n\) match the lower-bound terms closely. Even a coarse lower bound suffices if its partial sums are unbounded.
One familiar lower-bound series is the harmonic series. Its divergence can be seen by grouping terms into blocks. For each integer \(k\ge1\), the block from \(2^{k-1}+1\) through \(2^k\) contains \(2^{k-1}\) terms, and each term is at least \(1/2^k\). Thus each such block has sum at least \(1/2\). Infinitely many blocks contribute at least \(1/2\) each, so the partial sums are unbounded. We will use this fact in the examples below.
Worked Examples
Worked Example: A Shifted Harmonic Lower Bound
Show that \(\sum_{n=1}^{\infty}\frac{1}{n+1}\) diverges by comparing it with a harmonic lower bound.
For \(n\ge1\), both \(n+1\) and \(2n\) are positive. The proposed termwise inequality is equivalent to:
So the inequality holds for every index in the series. The lower-bound series is \(\sum_{n=1}^{\infty}\frac{1}{2n}=\frac12\sum_{n=1}^{\infty}\frac1n\), which diverges because multiplying the harmonic series by the positive constant \(1/2\) leaves its partial sums unbounded. The divergence comparison theorem therefore proves that the original series diverges.
For a finite check, the first four terms of the original series sum to:
The corresponding lower-bound sum is:
Indeed, \(\frac{77}{60}\ge\frac{25}{24}\). The finite check agrees with the termwise comparison; the theorem establishes the conclusion for all partial sums, not just these first four.
Worked Example: A Square-Root Denominator
Show that \(\sum_{n=1}^{\infty}\frac{1}{\sqrt n+1}\) diverges.
For \(n\ge1\), \(\sqrt n\ge1\), so \(1\le\sqrt n\). Adding \(\sqrt n\) to both sides gives \(\sqrt n+1\le2\sqrt n\). These denominators are positive, so taking reciprocals reverses the inequality:
The lower-bound series is \(\frac12\sum_{n=1}^{\infty}1/n^{1/2}\). The \(p\)-series \(\sum 1/n^p\) diverges when \(0<p\le1\); here \(p=1/2\), so the lower-bound series diverges. The comparison theorem gives the desired divergence.
Check the inequality at \(n=4\):
The algebraic bound also checks directly at this index: \(\sqrt4+1=3\le4=2\sqrt4\). Thus the reciprocal inequality has the stated direction, and the comparison is valid.
Worked Example: A Comparison That Starts After the First Term
Define \(a_1=0\) and \(a_n=\frac{1}{n-1}\) for \(n\ge2\). Show that \(\sum_{n=1}^{\infty}a_n\) diverges.
For every \(n\ge2\), the denominators are positive and \(n\le n+1\) is not the needed comparison; instead compare \(n-1\) with \(n\). Since \(n-1<n\), taking positive reciprocals gives:
The lower-bound tail \(\sum_{n=2}^{\infty}1/n\) diverges: removing the first term from the harmonic series does not change the fact that its partial sums are unbounded. The comparison theorem applies with \(N=2\), so \(\sum a_n\) diverges.
For an explicit finite check, through \(n=4\) the original partial sum is:
The lower-bound tail through \(n=4\) is:
Since \(\frac{11}{6}=\frac{22}{12}\ge\frac{13}{12}\), the finite sums agree with the comparison. The missing first-term inequality causes no problem: the theorem only needs the termwise bound from \(n=2\) onward.
Common Mistake: Reversing the Inequality
When \(a_n\ge b_n\), adding the inequalities gives \(\sum a_n\ge\sum b_n\) for matching finite partial sums. A common error is to reverse this order after summing. For instance, take \(a_n=1/n+1/n^2\) and \(b_n=1/n\). Since \(1/n^2>0\), \(a_n>b_n\) for every \(n\). But the incorrect calculation would claim:
Already at \(m=1\), this would say \(2\le1\), which is false. The correction is to preserve the inequality direction when adding terms:
More generally, the divergence conclusion also depends on comparing against a lower-bound series that is nonnegative on the tail and has unbounded partial sums. A finite numerical check can catch an arithmetic or direction error, but it cannot replace the argument for every partial sum.
When to Use the Divergence Comparison Rule
Use the rule when a given series has terms that are hard to sum, but a simpler divergent series lies below them term by term. The bound need not hold at every index: verify it from some index \(N\) onward, and make sure the lower-bound series still diverges after discarding its finite initial segment.
The logic is one-way. A divergent lower bound forces the larger nonnegative tail to diverge. A convergent lower bound, however, does not decide whether the larger series converges or diverges. The comparison only supplies a conclusion when the lower bound itself accumulates without bound.
Check Your Understanding
Answer each question by identifying the relevant tail inequality and the lower-bound series.
- State the hypotheses of the divergence comparison theorem, including what must hold eventually.
- Why does adding \(a_n\ge b_n\) from \(n=N\) to \(n=m\) preserve the inequality?
- Suppose \(a_n\ge 1/(3n)\) for every \(n\ge4\). What comparison conclusion follows?
- Why does changing or removing finitely many terms not affect divergence?
- Does a convergent lower-bound series prove that the larger series converges? Explain.
