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Convergence Tests · Tutorial 826 of 999

Why a Nonzero Term Limit Proves a Series Diverges

Use the behavior of a series’ terms to rule out convergence, while recognizing that terms approaching zero is only a necessary condition.

Advanced 9 min read

What You'll Learn

  • Define the partial sums that determine whether a series converges.
  • Derive why convergence of a series requires its terms to approach zero.
  • Apply the nth-term divergence test when the term limit is nonzero or does not exist.
  • Distinguish a necessary condition for convergence from a sufficient condition.
  • Show by grouping terms that the harmonic series diverges even though its terms approach zero.

Why Inspect the Terms of a Series?

A series is defined through the behavior of its partial sums, not by adding infinitely many terms all at once. Yet the individual terms can provide a quick way to rule out convergence. If the terms fail to approach zero, the partial sums cannot settle toward a finite value.

This is a one-way test. A nonzero term limit, or a term sequence with no limit, proves divergence. A zero term limit does not prove convergence; it only leaves that possibility open.

Series and Partial Sums

For a sequence of real numbers \(a_1,a_2,a_3,\dots\), the infinite series \(\sum_{n=1}^{\infty}a_n\) is defined by the sequence of partial sums

$$ S_N=\sum_{n=1}^{N}a_n=a_1+a_2+\cdots+a_N. $$

The series converges if the sequence of partial sums \(S_N\) approaches a finite real number as \(N\to\infty\). If the partial sums do not approach a finite limit, the series diverges. The terms \(a_n\) and the partial sums \(S_N\) are related, but they are not the same sequence.

Definition: A series \(\sum_{n=1}^{\infty}a_n\) converges to \(S\) when its partial sums satisfy \(\lim_{N\to\infty}S_N=S\). Its terms are the numbers \(a_n\) being added.

That distinction is central to the term test. The test does not calculate the sum. It asks whether the terms have a behavior that makes convergence impossible.

The Necessary Condition for Convergence

Each term after the first is the difference between two consecutive partial sums. In particular, for \(n\ge 2\),

$$ a_n=S_n-S_{n-1}. $$

If the series converges, then both sequences \(S_n\) and \(S_{n-1}\) approach the same limit. Their difference must therefore approach zero. The following derivation makes that implication explicit.

1
Assume the series converges.
Then there is a finite number \(S\) such that \(\lim_{n\to\infty}S_n=S\).
2
Use the same limit for shifted partial sums.
As \(n\to\infty\), the index \(n-1\) also tends to infinity, so \(\lim_{n\to\infty}S_{n-1}=S\).
3
Subtract consecutive partial sums.
Since \(a_n=S_n-S_{n-1}\), the limit laws give \(\lim_{n\to\infty}a_n=S-S=0\).
Theorem: If the series \(\sum_{n=1}^{\infty}a_n\) converges, then \(\lim_{n\to\infty}a_n=0\). Equivalently, if the terms do not approach zero, the series diverges.

This theorem is a necessary-condition test: every convergent series must pass the test \(\lim a_n=0\). Its contrapositive gives a useful way to prove divergence. If the terms have a nonzero limit, have no limit, or otherwise fail to approach zero, the series cannot converge.

Applying the nth-Term Divergence Test

The test is also called the nth-term test for divergence. It applies to series with positive, negative, or alternating terms; no assumption that the terms are positive is needed. First find the limit of \(a_n\), the general term. Then compare the result with the necessary condition.

Test for divergence: For \(\sum_{n=1}^{\infty}a_n\), if \(\lim_{n\to\infty}a_n=L\) for a finite \(L\ne 0\), the series diverges. If the terms do not approach zero because their limit does not exist or they grow without bound, the series also diverges.

Worked Example: A Nonzero Term Limit

Determine whether the series \(\sum_{n=1}^{\infty}\frac{4n-1}{3n+2}\) converges or diverges.

Its general term is \(a_n=\frac{4n-1}{3n+2}\). Divide the numerator and denominator by \(n\):

$$ a_n=\frac{4-\frac{1}{n}}{3+\frac{2}{n}}. $$

As \(n\to\infty\), \(\frac{1}{n}\to 0\) and \(\frac{2}{n}\to 0\). Therefore,

$$ \lim_{n\to\infty}a_n =\frac{4-0}{3+0} =\frac{4}{3}. $$

The term limit is \(\frac{4}{3}\), which is not zero. The necessary condition for series convergence fails, so the series diverges.

Check: The computed limit is nonzero, as required for this conclusion. In particular, the terms do not become arbitrarily close to zero.

Worked Example: Terms with No Limit

Determine whether \(\sum_{n=1}^{\infty}(-1)^{n+1}\) converges.

The terms are \(a_n=(-1)^{n+1}\). For odd indices \(n=2k-1\), \(a_{2k-1}=1\). For even indices \(n=2k\), \(a_{2k}=-1\). Thus one subsequence of terms is constantly \(1\), while another is constantly \(-1\). The terms do not approach a single limit, and in particular do not approach zero.

The partial sums confirm the failure directly. Starting with \(S_0=0\), each pair of terms adds to zero:

$$ S_{2k}=0,\qquad S_{2k-1}=1. $$

For example, \(S_1=1\), \(S_2=0\), \(S_3=1\), and \(S_4=0\). The even partial sums remain at \(0\), and the odd partial sums remain at \(1\), so the partial sums do not approach one common value. The series diverges.

Check: The partial-sum formulas follow from grouping the first \(2k\) terms into \(k\) cancelling pairs, or the first \(2k-1\) terms into \(k-1\) such pairs plus one \(1\). They agree with the first four partial sums shown.

Why a Zero Term Limit Is Not Enough

The theorem gives an implication in one direction: if a series converges, then its terms approach zero. It does not say that terms approaching zero force the series to converge. In logic, reversing an implication without proof is invalid. To see the issue concretely, consider the harmonic series.

Worked Example: Zero Terms but a Divergent Series

Consider \(\sum_{n=1}^{\infty}\frac{1}{n}\). Its terms satisfy

$$ \lim_{n\to\infty}\frac{1}{n}=0. $$

So the nth-term divergence test does not decide this series. To establish divergence, examine partial sums in blocks whose endpoints are powers of \(2\). The first term is \(1\). For each integer \(j\ge 1\), the block from \(n=2^{j-1}+1\) through \(n=2^j\) contains \(2^{j-1}\) terms. Every term in that block is at least \(\frac{1}{2^j}\), because \(n\le 2^j\). Hence the block sum is at least

$$ 2^{j-1}\cdot\frac{1}{2^j}=\frac{1}{2}. $$

Consequently, the partial sum through \(2^m\) satisfies

$$ S_{2^m} =1+\sum_{j=1}^{m}\sum_{n=2^{j-1}+1}^{2^j}\frac{1}{n} \ge 1+\sum_{j=1}^{m}\frac{1}{2} =1+\frac{m}{2}. $$

As \(m\to\infty\), the lower bound \(1+\frac{m}{2}\) grows without bound. Thus these partial sums cannot approach a finite limit, and the harmonic series diverges even though its terms approach zero.

Check: For \(m=2\), the blocks are \(n=2\) and \(n=3,4\). Their sums are \(\frac12\) and \(\frac13+\frac14=\frac7{12}\), respectively; each is at least \(\frac12\). The displayed bound gives \(S_4\ge 2\), and indeed \(S_4=1+\frac12+\frac13+\frac14=\frac{25}{12}\ge 2\).

A Common Mistake and Its Correction

A common incorrect argument is: “For \(\sum_{n=1}^{\infty}\frac{1}{n}\), the terms approach zero, so the series converges.” The calculation \(\lim_{n\to\infty}\frac1n=0\) is correct, but the conclusion does not follow. It treats a necessary condition as though it were sufficient.

The correction is to state exactly what the term limit tells you: a nonzero limit or failure to approach zero proves divergence; a zero limit gives no conclusion by itself. For the harmonic series, the block estimate above supplies the additional argument needed to prove divergence.

Using the Test Effectively

The nth-term divergence test is often most valuable before using a more involved method. It can quickly eliminate a series from consideration when its terms do not approach zero. But if the term limit is zero, stop short of claiming convergence: another argument is needed to decide the series.

  • If \(\lim a_n=L\ne 0\), conclude that \(\sum a_n\) diverges.
  • If the terms have no limit and do not approach zero, conclude that the series diverges.
  • If \(\lim a_n=0\), record that the test is inconclusive; the series may converge or diverge.

The reason for this asymmetry is built into the definition of a series: convergence concerns the partial sums. When those sums converge, consecutive sums become close, forcing their difference \(a_n\) to approach zero. But differences approaching zero alone do not guarantee that the sums settle to a finite value.

Check Your Understanding

For each question, distinguish what the term limit proves from what it leaves undecided.

  1. State the necessary condition on \(a_n\) when \(\sum_{n=1}^{\infty}a_n\) converges.
  2. If \(\lim_{n\to\infty}a_n=-2\), what can you conclude about \(\sum a_n\), and why?
  3. Why does a term sequence with no limit rule out convergence of its series?
  4. If \(\lim_{n\to\infty}a_n=0\), does the nth-term divergence test establish convergence? Explain.
  5. In the harmonic-series grouping argument, how many terms are in the block from \(2^{j-1}+1\) to \(2^j\), and why is its sum at least \(\frac12\)?

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