Using the Nth-Term Test Carefully
A series can have terms that look complicated: fractions of polynomials, alternating signs, exponentials, or factorials. Before trying a more specialized method, it is often useful to ask a basic question: what happens to the individual terms as the index grows?
The answer can sometimes settle the classification immediately. If the terms do not approach zero, the series diverges. If they do approach zero, this test has reached its limit: the series might converge, or it might diverge. The distinction between those two conclusions is the central point of this workshop.
Definition and Test
For a sequence of terms \(a_n\), the infinite series \(\sum_{n=1}^{\infty}a_n\) is defined through its partial sums \(S_N=\sum_{n=1}^{N}a_n\). The series converges if the sequence of partial sums has a finite limit as \(N\to\infty\); otherwise, it diverges.
The condition \(a_n\to 0\) is necessary for convergence, but it is not sufficient. In practice, “inconclusive” means that the term-limit test alone cannot decide the series; it does not mean the series itself has an uncertain status.
Why the Test Works
The proof follows directly from the definition using partial sums. If the series converges to \(S\), then its partial sums satisfy \(S_N\to S\). For \(N\ge 2\), the \(N\)th term is the difference between two consecutive partial sums:
As \(N\to\infty\), both \(S_N\) and \(S_{N-1}\) approach \(S\). Their difference must therefore approach zero:
Thus, if the terms fail to approach zero, the partial sums cannot converge. This proves the divergence conclusions. The argument does not run in reverse: knowing that \(a_n\to0\) does not establish that the partial sums approach a finite limit.
Worked Examples with Rational Terms
Worked Example: A Rational Term with a Nonzero Limit
Classify the series \(\sum_{n=1}^{\infty}\frac{3n^2+1}{2n^2-5n}\) using the nth-term test. The denominator is zero at \(n=2\), so the displayed expression does not define a term there. To make a valid series, take the sum from \(n=3\) onward; omitting or changing finitely many initial terms does not affect convergence or divergence.
For \(n\ge3\), divide the numerator and denominator by \(n^2\), the highest power present:
As \(n\to\infty\), \(1/n^2\to0\) and \(5/n\to0\). The denominator tends to \(2\), which is not zero, so the quotient limit is
Since \(3/2\ne0\), the series diverges by the nth-term test. As a check on the calculation, the leading terms in the original fraction are \(3n^2\) and \(2n^2\); their ratio is \(3/2\), matching the limit found by division by \(n^2\).
For rational expressions, dividing by the highest power of \(n\) makes the limiting behavior visible. If the numerator and denominator have the same degree, their leading-coefficient ratio is the term limit. If the denominator has higher degree, the limit is zero. If the numerator has higher degree, the terms generally do not approach zero; check the resulting limit before classifying.
Oscillation and Exponential Terms
Worked Example: Oscillating Terms with No Limit
Consider \(\sum_{n=1}^{\infty}(-1)^n(1+1/n)\). The sign alternates, but the magnitude approaches \(1\), rather than zero. To verify the behavior, examine even and odd indices separately. For \(n=2k\), the sign is positive; for \(n=2k+1\), it is negative:
The even-indexed terms approach \(1\), while the odd-indexed terms approach \(-1\). Since these subsequences have different limits, the full sequence \(a_n\) has no limit. In particular, its terms do not approach zero. The series therefore diverges by the nth-term test.
Check the conclusion against the term formula: for every \(n\), \(|a_n|=1+1/n>1\). Thus the terms remain at least one unit from zero, consistent with the two subsequential limits.
Worked Example: An Exponential Term That Approaches Zero
Consider \(\sum_{n=1}^{\infty}\frac{2^n+1}{3^n+2}\). To determine the term limit, note that \(2^n+1\le2\cdot2^n\) for \(n\ge1\), and \(3^n+2\ge3^n\). Therefore
Because \(0<2/3<1\), the right-hand side approaches zero. The term is positive and squeezed between zero and a quantity tending to zero, so
The nth-term test is inconclusive: it does not prove convergence or divergence. The check is the explicit bound above, which confirms that the terms become arbitrarily small. Further information about the series is needed to decide its behavior.
Factorials: Simplify Before Taking the Limit
Factorials can make a term look complicated, but the identity \(n!=n(n-1)!\) often allows exact cancellation. Cancellation should be performed before estimating the limit, and only common factors may be cancelled.
Worked Example: A Factorial Ratio with Unbounded Terms
For \(n\ge2\), consider the series with terms \(a_n=\frac{n!}{(n-1)!}\). Using \(n!=n(n-1)!\), cancel the common factorial factor:
Thus \(\lim_{n\to\infty}a_n=+\infty\), so in particular the terms do not approach zero. The series diverges by the nth-term test. Checking the simplification at \(n=3\), for instance, gives \(3!/2!=6/2=3\), which agrees with \(a_n=n\).
Worked Example: A Factorial Ratio with Zero Term Limit
Consider \(\sum_{n=1}^{\infty}\frac{n!}{(2n)!}\). Expand only enough of the factorial in the denominator to compare factors:
There are \(n\) factors in this product, and each is at least \(n+1\). Consequently, for every positive integer \(n\),
The upper bound tends to zero, so the term limit is zero. The nth-term test is inconclusive. As a check, when \(n=2\), the term is \(2!/4!=2/24=1/12\); the product expression gives \(1/(3\cdot4)=1/12\), as required.
A Common Mistake: Treating a Necessary Condition as Sufficient
A frequent error is to see that \(a_n\to0\) and conclude that \(\sum a_n\) converges. That reasoning reverses the theorem. The theorem says that convergence requires terms to approach zero; it does not say that terms approaching zero guarantee convergence.
Worked Example: Correcting a Zero-Limit Conclusion
For the harmonic series \(\sum_{n=1}^{\infty}1/n\), the terms satisfy \(\lim_{n\to\infty}1/n=0\). A mistaken argument would be: “The terms approach zero, so the series converges.” The conclusion does not follow from the premise.
In fact, the harmonic series diverges, as shown by grouping terms into blocks:
In the block from \(2^{k-1}+1\) through \(2^k\), there are \(2^{k-1}\) terms, and each term is at least \(1/2^k\). The block therefore has sum at least \(2^{k-1}/2^k=1/2\). Adding arbitrarily many such blocks makes the partial sums unbounded, so the series diverges. The correction is that a zero term limit makes the nth-term test inconclusive, not that it proves convergence.
A Reliable Classification Routine
For a series with a mixed or unfamiliar term formula, use the following routine. It keeps the conclusion tied to what the limit actually establishes.
Identify \(a_n\) from the series, including its signs, index range, and any restrictions that make the formula defined.
Simplify rational expressions, compare subsequences when signs oscillate, and cancel or bound factorial expressions.
A nonzero limit, no limit, or unbounded terms proves divergence. A zero limit leaves the series undecided by this test.
This test matters because it is a quick necessary-condition check. It can rule out convergence before any more involved analysis, while also preventing a common overclaim when the terms do vanish. Its strength is decisive when the terms fail to approach zero; its limitation is equally important when they do.
Check Your Understanding
For each series, first find the limit of its terms, then state whether the nth-term test proves divergence or is inconclusive.
- \(\sum_{n=1}^{\infty}\frac{4n^2-1}{n^2+3n}\)
- \(\sum_{n=1}^{\infty}(-1)^n\frac{n+1}{n}\)
- \(\sum_{n=1}^{\infty}\frac{3^n+1}{5^n+2}\)
- \(\sum_{n=2}^{\infty}\frac{n!}{(n-1)!}\)
- In one sentence, explain why a zero term limit is not enough to conclude that a series converges.
