Why the Choice of Test Matters
A series is an infinite sum of terms, written \(\sum_{n=1}^{\infty} a_n\). To determine whether it converges, first consider what the terms do, then look for a pattern that matches a test. The nth-term test, geometric-series test, and ratio test are useful in different situations, and each has a specific conclusion it permits.
In a free-response solution, naming a test is not enough. Show the limit or ratio the test requires, state the test’s condition, and connect that condition to your conclusion. If the test is inconclusive, say so rather than claiming convergence or divergence.
The nth-Term Test for Divergence
A necessary condition for \(\sum a_n\) to converge is that its terms approach zero. Therefore, if \(\lim_{n\to\infty}a_n\) is nonzero, or if that limit does not exist, the series diverges. This is called the nth-term test for divergence.
The test gives only a necessary condition for convergence, not a sufficient one. In particular, a term limit of zero does not prove that a series converges.
Worked Example: Using the nth-Term Test
Determine whether \(\sum_{n=1}^{\infty}\frac{3n}{n+2}\) converges or diverges.
Let \(a_n=\frac{3n}{n+2}\). Divide numerator and denominator by \(n\):
The terms approach \(3\), not \(0\). By the nth-term test, the series diverges. The calculation checks the required condition directly: \(\lim a_n=3\ne0\).
Geometric Series
A geometric series has a constant ratio between consecutive terms. In the standard form \(\sum_{n=0}^{\infty} ar^n\), \(a\) is the first term and \(r\) is the common ratio. Its behavior depends on the magnitude of \(r\).
To justify the formula, let \(S_N=a+ar+\cdots+ar^N\). Multiplying by \(r\) shifts the terms, so subtraction cancels all but the first and last terms:
For \(r\ne1\), this gives \(S_N=\frac{a(1-r^{N+1})}{1-r}\). When \(|r|<1\), \(r^{N+1}\to0\), so \(S_N\to\frac{a}{1-r}\). If \(|r|\ge1\) and \(a\ne0\), the terms \(ar^n\) do not approach zero, so the series diverges by the nth-term test.
Worked Example: Finding a Geometric Series Sum
Find the sum of \(\sum_{n=1}^{\infty}6\left(\frac13\right)^{n-1}\).
The first term is \(a=6\), and the common ratio is \(r=\frac13\). Since \(\left|\frac13\right|<1\), the geometric-series formula applies:
Check using the partial-sum formula: the first \(N\) terms sum to \(9\left(1-(\frac13)^N\right)\). As \(N\to\infty\), \((\frac13)^N\to0\), and the partial sums approach \(9\), as claimed.
The Ratio Test
The ratio test is particularly useful when terms contain factorials, powers, or products, because consecutive terms may simplify. Apply it to the absolute values of the terms. Assume the terms are nonzero for all sufficiently large \(n\), so the ratio is defined.
Here is why the two decisive cases work. If \(L<1\), choose a number \(q\) with \(L<q<1\). Eventually, each consecutive absolute-term ratio is at most \(q\). Thus, from some index \(N\) onward, \(|a_{N+k}|\le |a_N|q^k\). The sum of these bounds is a convergent geometric series, so the absolute tail is finite and the original series converges absolutely.
If \(L>1\), choose \(q\) with \(1<q<L\). Eventually \(|a_{n+1}|>q|a_n|\), so the terms’ magnitudes grow at least by a factor \(q\) at each step. They cannot approach zero. The nth-term test then implies divergence. When \(L=1\), neither argument works: the ratios are not eventually bounded above by a fixed number below \(1\), nor below by a fixed number above \(1\).
Worked Example: Applying the Ratio Test to Factorials
Determine whether \(\sum_{n=1}^{\infty}\frac{2^n}{n!}\) converges.
Set \(a_n=\frac{2^n}{n!}\). Each term is positive, and the consecutive ratio is
Therefore, \(L=\lim_{n\to\infty}\frac{2}{n+1}=0\), which is less than \(1\). The ratio test proves that the series converges absolutely. As a check on the algebra, \(a_1=2\), \(a_2=2\), and their ratio is \(1\), matching \(\frac{2}{1+1}=1\); later ratios decrease toward zero.
What a Ratio Limit of One Means
A ratio limit of \(1\) is not a verdict. It is a signal that the ratio test has not decided the question. Two series can both have ratio limit \(1\) but behave differently.
Worked Example: Seeing the Ratio Test’s Limit
Consider first \(\sum_{n=1}^{\infty}\frac1n\). Its consecutive-term ratio is
The ratio test gives no conclusion. This series does diverge: grouping terms in blocks gives \(1+\frac12+(\frac13+\frac14)+(\frac15+\cdots+\frac18)+\cdots\), and each block after the first contributes at least \(\frac12\). Thus the partial sums are unbounded.
Now consider \(\sum_{n=1}^{\infty}\frac{1}{n(n+1)}\). Its ratio is
Again, the ratio test gives no conclusion. But \(\frac{1}{n(n+1)}=\frac1n-\frac1{n+1}\), so the first \(N\) terms sum to \(1-\frac1{N+1}\), which approaches \(1\). This series converges. The contrasting outcomes confirm that a ratio limit of \(1\) does not determine convergence or divergence.
A Reliable Free-Response Strategy
Choose a test whose hypotheses match the form of the series, and state the result precisely. A short decision sequence can prevent unsupported conclusions.
Find \(\lim a_n\). If it is nonzero or does not exist, conclude divergence by the nth-term test. If it is zero, continue; that result alone decides nothing.
Identify the first term and constant ratio. Use the geometric-series result only when the terms really have a constant ratio.
Compute \(\left|a_{n+1}/a_n\right|\), simplify before taking the limit, and apply the ratio test’s three cases.
For the ratio test, \(L<1\) gives absolute convergence, \(L>1\) gives divergence, and \(L=1\) is inconclusive.
Common Pitfall: Treating an Inconclusive Result as a Verdict
A frequent error is to calculate \(L=1\) and announce that the series diverges. For example, for \(a_n=\frac{1}{n(n+1)}\), the ratio limit is \(1\). The incorrect conclusion “therefore the series diverges” does not follow from the ratio test—and, as the telescoping calculation above shows, the series actually converges.
The correction is to state: “The ratio limit is \(1\), so the ratio test is inconclusive.” A test’s failure to decide is not evidence for either outcome. Likewise, if the nth-term limit is zero, report only that the nth-term test is inconclusive; do not claim convergence from that fact alone.
Check Your Understanding
For each question, give the test’s conclusion and justify it with the relevant limit or calculation.
- What does the nth-term test say about a series if \(\lim a_n=4\)?
- Does \(\lim a_n=0\) prove that \(\sum a_n\) converges? Explain.
- Find the sum of \(\sum_{n=0}^{\infty}5(-\frac12)^n\), or explain why it diverges.
- For \(a_n=\frac{3^n}{n!}\), calculate \(\lim\left|a_{n+1}/a_n\right|\) and state the ratio-test conclusion.
- What does the ratio test conclude when its limit is \(1\)?
