When the Terms Grow, the Series Cannot Settle
A series is built by adding its terms in order. If those terms grow without bound, the additions cannot settle toward a finite sum. The reason is direct: each term is the difference between two consecutive partial sums. If the partial sums approach a finite value, their differences must approach zero. Terms whose magnitudes grow without bound cannot meet that requirement.
This gives a decisive use of the Nth-term test. A series can be shown to diverge by examining its individual terms, without finding a formula for all its partial sums. It is useful, however, to distinguish terms that grow without bound from terms that have a finite nonzero limit. Both situations rule out convergence, but they are different cases.
Series Terms and Partial Sums
For a series \(\sum_{n=1}^{\infty} a_n\), its \(N\)th partial sum is the sum of its first \(N\) terms:
The infinite series converges to a finite sum \(S\) if the sequence of partial sums \(S_N\) approaches \(S\) as \(N\) grows. If the partial sums do not approach any finite number, the series diverges. A key identity follows by subtracting consecutive partial sums:
Thus each term measures the change from one partial sum to the next.
Why the Nth-Term Test Works
The theorem follows from the definition of convergence and the difference identity. If \(S_N\) converges to \(S\), then \(S_{N-1}\) also converges to \(S\): shifting the index by one does not change the limit. Therefore, the difference between these two sequences must approach \(S-S=0\). Since that difference is \(a_N\), the terms must approach zero.
Then its partial sums satisfy \(S_N\to S\) for some finite number \(S\).
As \(N\to\infty\), both \(S_N\) and \(S_{N-1}\) approach \(S\).
\(a_N=S_N-S_{N-1}\to S-S=0\), so the terms must approach zero.
This also proves the contrapositive: if the terms fail to approach zero, the series cannot converge. In particular, if their magnitudes grow without bound, they do not approach zero, so the series diverges.
There is a slightly broader observation. If the terms are unbounded, meaning that their magnitudes are not bounded above, then they cannot approach zero either. A sequence that approaches zero is eventually bounded, and its finitely many earlier terms cannot make the whole sequence unbounded. So unbounded terms, even without a limit of \(+\infty\) or \(-\infty\), are enough to rule out convergence of the series.
Worked Examples
Worked Example: Terms That Grow to Positive Infinity
Determine whether the series \(\sum_{n=1}^{\infty}\frac{n^2+1}{n}\) converges.
First simplify its \(n\)th term:
For every \(n\ge1\), \(1/n>0\), so \(a_n>n\). Given any \(M>0\), choose an integer \(N>M\). For all \(n\ge N\),
Therefore \(a_n\to+\infty\), and in particular \(a_n\) does not approach zero. The Nth-term test shows that the series diverges.
Check: If its partial sums converged, the difference of consecutive partial sums would approach zero. But that difference is \(a_n=n+1/n\), which grows without bound. This contradicts the necessary condition, confirming divergence.
Worked Example: Terms That Grow to Negative Infinity
Determine whether the series \(\sum_{n=1}^{\infty}(-2n)\) converges.
Its \(n\)th term is \(a_n=-2n\). Its magnitude is
Thus the terms do not approach zero, and the Nth-term test gives divergence. We can also check directly using the finite sum:
As \(N\to\infty\), \(S_N\to-\infty\), so the partial sums do not approach a finite sum.
Check: The partial-sum formula gives \(S_N-S_{N-1}=-N(N+1)+(N-1)N=-2N=a_N\), as required. Both the terms and the partial sums confirm divergence.
Worked Example: A Finite Nonzero Term Limit
Determine whether the series \(\sum_{n=1}^{\infty}\frac{3n+1}{n+2}\) converges, and compare its terms with the unbounded cases.
Rewrite the term by expressing its numerator using the denominator:
Since \(5/(n+2)\to0\), the terms approach \(3\), a finite nonzero number. The Nth-term test therefore shows that the series diverges. Here the terms do not grow without bound; they settle toward a nonzero value.
Check: If the series converged, its terms would have to approach zero. Their limit is \(3\), not \(0\), so convergence is impossible. This is distinct from the first two examples, where the magnitudes of the terms grow without bound.
How to State the Test Precisely
The term test is often stated in its general form: if \(\lim_{n\to\infty}a_n\) exists and is not zero, then \(\sum a_n\) diverges. The unbounded case is another direct way to use the necessary condition. If \(a_n\to+\infty\) or \(a_n\to-\infty\), then \(a_n\) certainly does not approach zero. Likewise, if \(|a_n|\to\infty\), the terms cannot approach zero, and the series diverges.
The finite nonzero case and the unbounded case share the same logic: neither has terms approaching zero. They differ in the behavior of the terms themselves. A finite nonzero limit means the terms approach a particular nonzero value; unbounded terms have magnitudes that exceed every fixed bound eventually. The earlier material on divergence from a nonzero term limit treats the first case; this tutorial emphasizes the second.
The test is one-way. It identifies divergence when the terms fail to approach zero. If the terms do approach zero, this necessary condition is satisfied, but that fact alone does not establish that the series converges. The terms and the series are related, but they are not the same object: convergence of the series concerns the limit of the partial sums.
A Common Mistake
A common error is to treat “the terms are getting large” as if it were merely an informal impression, without connecting it to the necessary condition for series convergence. Another is to confuse a finite nonzero limit with an unbounded term sequence. Both cases prove divergence, but one should name the actual behavior correctly.
Worked Example: Correcting a Misclassification
Suppose a student considers the terms \(a_n=5-\frac{1}{n}\) and writes: “These terms grow without bound, so the series diverges.” The conclusion is right, but the stated reason is wrong. Compute the limit:
The terms do not grow without bound; they approach the finite nonzero value \(5\). The correct reason for divergence is that the terms do not approach zero, which is necessary for convergence of the series.
Check: If the series converged, the difference of consecutive partial sums would approach zero. That difference is \(a_n=5-1/n\), which approaches \(5\). The contradiction confirms divergence, while the computed limit confirms the correct classification.
Using the Test Efficiently
When applying the Nth-term test, first simplify the expression for \(a_n\) and determine its limiting behavior. If its magnitude grows without bound, state that it does not approach zero and conclude that the series diverges. If it approaches a finite nonzero number, the same conclusion follows, but the reason is a nonzero finite limit rather than unbounded growth.
Do not try to infer convergence from a single term or from a small number of terms. The test concerns the behavior as \(n\) tends to infinity. And do not claim that a series converges merely because its terms approach zero: the theorem proved here gives a necessary condition, not a sufficient one. The correct conclusion from this test is decisive when the condition fails, and limited when it holds.
Check Your Understanding
For each question, focus on the behavior of the terms and the necessary condition for series convergence.
- Why does convergence of the partial sums imply that \(a_N=S_N-S_{N-1}\) approaches zero?
- If \(|a_n|\to\infty\), what does the Nth-term test conclude about \(\sum a_n\), and why?
- What is the difference between terms approaching a finite nonzero number and terms growing without bound?
- If the terms of a series approach zero, does the necessary condition alone prove convergence? Explain.
- For \(a_n=4n+1\), identify the behavior of the terms and state what it implies for \(\sum_{n=1}^{\infty}a_n\).