Why the Terms of a Series Matter
A series adds the terms of a sequence in order. For example, the series \(\sum_{n=1}^{\infty} a_n\) is studied through its sequence of partial sums, not by looking at one term in isolation. A basic question is whether those partial sums approach a finite value.
The behavior of the individual terms gives an important first check. If the terms fail to approach zero, the partial sums cannot settle to a finite limit. This fact is the Nth-Term Test for Divergence. Its other side is just as important: terms approaching zero do not by themselves prove that a series converges.
The Nth-Term Test for Divergence
The test examines the limit of the terms \(a_n\), rather than directly calculating the partial sums. It is a divergence test: it can establish that a series diverges when its terms do not approach zero, but it cannot establish convergence when they do approach zero.
When the limit exists and equals a number \(L\ne 0\), the terms fail to approach zero. When the term limit does not exist, the terms also fail to approach zero. In either case, the series diverges. In an AP-style justification, explicitly state the term limit and connect it to the test’s conclusion.
Why the Test Works
The proof follows from the definition of a convergent series. If the partial sums approach a finite limit, then consecutive partial sums must approach that same limit. But the difference of consecutive partial sums is exactly the next term.
Then its partial sums satisfy \(S_N\to S\) for some finite number \(S\).
For \(N\ge 2\), \(a_N=S_N-S_{N-1}\), because \(S_N=a_1+\cdots+a_{N-1}+a_N\) and \(S_{N-1}=a_1+\cdots+a_{N-1}\).
Since both \(S_N\) and \(S_{N-1}\) approach \(S\), their difference approaches \(S-S=0\). Thus \(a_N\to0\).
This proves that a convergent series must have terms tending to zero. The Nth-Term Test is the contrapositive: if the terms do not tend to zero, the series cannot converge. This argument proves a necessary condition for convergence, not a sufficient one.
Worked Examples: A Nonzero Term Limit
Worked Example: A Series Whose Terms Approach 1
Determine whether \(\sum_{n=1}^{\infty}\frac{n}{n+1}\) converges.
Let \(a_n=\frac{n}{n+1}\). Divide the numerator and denominator by \(n\):
The term limit is \(1\), which is not zero. By the Nth-Term Test, \(\sum_{n=1}^{\infty}\frac{n}{n+1}\) diverges. The test does not require calculating the partial sums once this nonzero term limit is established.
Check: The terms approach \(1\), not \(0\), so they satisfy the test’s condition for divergence.
When the Terms Have No Limit
The test also applies when the term sequence has no limit. A sequence that alternates between two distinct values does not approach zero; its terms keep moving between those values. The series then diverges, even if some of its partial sums repeat.
Worked Example: Alternating Terms That Do Not Settle
Determine whether \(\sum_{n=1}^{\infty}(-1)^n\) converges.
Its terms are \(a_n=(-1)^n\). For even \(n\), \(a_n=1\), and for odd \(n\), \(a_n=-1\). Thus the even-indexed terms approach \(1\), while the odd-indexed terms approach \(-1\). Since these are different, the full term sequence has no limit.
By the Nth-Term Test, the series diverges. This conclusion can also be checked from the partial sums: \(S_1=-1\), \(S_2=0\), \(S_3=-1\), and \(S_4=0\). More generally, \(S_{2k}=0\) and \(S_{2k+1}=-1\) for every positive integer \(k\), so the partial sums do not approach one finite limit.
Check: The terms alternate between \(-1\) and \(1\), so they do not approach zero; the partial sums alternate between \(-1\) and \(0\), confirming divergence.
A Zero Term Limit Is Not Enough
If \(\lim_{n\to\infty}a_n=0\), the Nth-Term Test gives no conclusion about whether \(\sum a_n\) converges. Some series with terms tending to zero converge, and others diverge. The limit condition is necessary for convergence, but it is not sufficient.
Here are two contrasting examples. The first diverges despite having terms that tend to zero. The second converges and also has terms that tend to zero. The difference cannot be decided by the Nth-Term Test alone.
Worked Example: A Zero Term Limit but a Divergent Series
Consider the harmonic series \(\sum_{n=1}^{\infty}\frac{1}{n}\). Its terms satisfy
This limit alone does not determine convergence. To show that this particular series diverges, group terms in its partial sums. For every positive integer \(m\), the partial sum through \(2^m\) is
In the group indexed by \(k\), there are \(2^{k-1}\) terms. Each denominator is at most \(2^k\), so every term in that group is at least \(\frac{1}{2^k}\). Therefore each group has sum at least
It follows that \(S_{2^m}\ge 1+\frac{m}{2}\). As \(m\to\infty\), this lower bound grows without bound, so the partial sums cannot approach a finite limit. The harmonic series diverges, even though its terms tend to zero.
Check: The term limit is zero, but the grouped partial sums grow at least as fast as \(1+\frac{m}{2}\); therefore the series diverges.
Worked Example: A Zero Term Limit and a Convergent Series
Consider \(\sum_{n=1}^{\infty}\frac{1}{n(n+1)}\). First, rewrite each term using partial fractions:
because \(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1-n}{n(n+1)}=\frac{1}{n(n+1)}\). The \(N\)th partial sum then telescopes:
Taking the limit gives \(\lim_{N\to\infty}S_N=1\), so the series converges to \(1\). Its terms also approach zero: since \(n(n+1)\to\infty\), \(\lim_{n\to\infty}\frac{1}{n(n+1)}=0\).
Check: The partial-sum formula gives \(S_N=1-\frac{1}{N+1}\), which approaches \(1\); the series converges, and its terms tend to zero as required.
A Common Mistake
A frequent error is to use the necessary condition as if it were a complete convergence test. For example, one might write:
The implication is invalid. The Nth-Term Test says that convergence requires a zero term limit; it does not say that a zero term limit guarantees convergence. The harmonic-series grouping above shows the correction: its terms tend to zero, but its partial sums are unbounded, so it diverges.
Writing the Justification Clearly
For a divergence claim based on this test, identify the term \(a_n\), evaluate its limit, and state the conclusion using the Nth-Term Test. For example: “Here \(a_n=\frac{n}{n+1}\), and \(\lim_{n\to\infty}a_n=1\ne0\). Therefore, by the Nth-Term Test, the series diverges.”
If the term limit equals zero, do not claim that the series converges on that basis. State that the test is inconclusive and use a different appropriate argument. This distinction is useful in free-response work because it makes clear both what the test establishes and what it does not establish.
Check Your Understanding
For each question, distinguish a conclusion the Nth-Term Test supports from one it cannot support.
- If \(\lim_{n\to\infty}a_n=4\), what can you conclude about \(\sum a_n\), and which test justifies the conclusion?
- If \(\lim_{n\to\infty}a_n=0\), does the Nth-Term Test establish convergence, divergence, or neither?
- Explain why convergence of the partial sums implies \(a_n\to0\), using \(a_n=S_n-S_{n-1}\).
- For \(\sum_{n=1}^{\infty}\frac{1}{n(n+1)}\), write the \(N\)th partial sum and its limit.
- Give a complete Nth-Term Test justification for \(\sum_{n=1}^{\infty}\frac{2n+1}{n}\) diverging.
