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Calculus BC · Unit 10 · 10.3
AP Calculus BC · Infinite Sequences and Series
The nth-Term Test for Divergence
One limit, one direction: it can prove divergence and never convergence.
key idea$\lim_{n\to\infty}a_n\neq 0\ \Longrightarrow\ \sum a_n\ \text{diverges}$
The test
This is the only test that can be applied to any series in a few seconds, and the only one whose conclusion runs in a single direction. It can prove divergence and it can never prove convergence — a limitation worth stating out loud every time you use it.
Applies to
any series — no positivity, monotonicity or other hypothesis
Conclusion
if $\displaystyle\lim_{n\to\infty}a_n\neq0$ (or fails to exist), the series diverges
If the limit is zero
the test is inconclusive — try something else
Cost
one limit; always worth trying first
Never says
“$a_n\to0$, therefore the series converges.” That is the converse error
| Series | $\displaystyle\lim a_n$ | Test says | Actual verdict |
|---|
| $\displaystyle\sum\frac{n}{2n+1}$ | $\tfrac12$ | diverges | diverges |
| $\displaystyle\sum(-1)^{n}$ | does not exist | diverges | diverges |
| $\displaystyle\sum\frac{1}{n}$ | $0$ | inconclusive | diverges |
| $\displaystyle\sum\frac{1}{n^{2}}$ | $0$ | inconclusive | converges |
| $\displaystyle\sum n e^{-n}$ | $0$ | inconclusive | converges |
Rows 3 and 4 are the whole argument for why this test cannot detect convergence: both have terms tending to zero and they end up on opposite sides. A limit of zero carries no information at all.
When to reach for it
| Symptom in $a_n$ | Likely limit | Do this |
|---|
| a ratio of polynomials of equal degree | the ratio of leading coefficients | apply the test — it diverges |
| a factor like $(-1)^{n}$ with no shrinking part | does not exist | apply the test — it diverges |
| $a_n\to$ a non-zero constant | that constant | apply the test — it diverges |
| $a_n\to0$ | $0$ | stop — choose a different test |
| $a_n$ contains $n!$ or $r^{n}$ | usually $0$ | expect the ratio test to do the real work |
Run this test first on every series, because it is cheap. Just be ready for it to return nothing: on most exam series the limit is zero and the real work has not started.
Worked examples
EX 1
Equal degrees
the limit is not zeroDetermine whether $\displaystyle\sum_{n=1}^{\infty}\dfrac{3n^{2}+1}{5n^{2}-n}$ converges.
1
Take the limit of the terms
Divide numerator and denominator by $n^{2}$: $\dfrac{3+1/n^{2}}{5-1/n}$.
2
Evaluate
$\displaystyle\lim_{n\to\infty}a_n=\dfrac{3}{5}$.
3
Compare with zero
$\dfrac35\neq0$.
4
Apply the test
By the $n$th-term test, the series diverges.
5
Interpret
Adding infinitely many terms each close to $0.6$ cannot possibly settle down.
EX 2
No limit at all
oscillationDetermine whether $\displaystyle\sum_{n=1}^{\infty}\cos(n\pi)$ converges.
1
Simplify the terms
$\cos(n\pi)=(-1)^{n}$.
2
Examine the limit
The terms alternate between $1$ and $-1$ forever, so $\displaystyle\lim a_n$ does not exist.
3
Note that this qualifies
The test requires only that the limit fail to be zero; failing to exist counts.
4
Conclude
The series diverges.
5
Head off a confusion
The alternating series test does not apply here, because $\lvert a_n\rvert=1$ never decreases toward zero.
EX 3
Inconclusive, then resolved
the limit is zeroA student computes $\displaystyle\lim_{n\to\infty}\dfrac{1}{\sqrt{n}}=0$ and concludes that $\displaystyle\sum\dfrac{1}{\sqrt{n}}$ converges. Correct them.
1
Name the error
The test has no converse. A limit of zero means inconclusive, never convergent.
2
Recognise the family
$\dfrac{1}{\sqrt{n}}=\dfrac{1}{n^{1/2}}$ is a $p$-series with $p=\dfrac12$.
3
Apply that criterion
Such a series converges only when $p\gt 1$.
4
Conclude
$p=\dfrac12\le1$, so the series diverges — the opposite of the student's claim.
5
State the habit
When the $n$th-term test returns zero, say ‘inconclusive’ explicitly and move on to another test.
EX 4
A limit worth computing carefully
an exponential formDetermine whether $\displaystyle\sum_{n=1}^{\infty}\left(1+\dfrac{2}{n}\right)^{n}$ converges.
1
Recognise the limit form
$\left(1+\dfrac{k}{n}\right)^{n}\to e^{k}$ as $n\to\infty$.
2
Apply it
$\displaystyle\lim_{n\to\infty}\left(1+\dfrac{2}{n}\right)^{n}=e^{2}$.
3
Compare with zero
$e^{2}\approx7.39\neq0$.
4
Conclude
The series diverges.
5
Note the risk
Each factor looks like it is heading to $1$, so the terms look harmless. Computing the limit properly is what exposes the divergence.
What it cannot tell you
It can never prove convergence
“$a_n\to0$” is compatible with both outcomes. Writing ‘the terms go to zero so the series converges’ earns no credit and is simply false.
A limit that fails to exist still forces divergence
Oscillating terms cannot tend to zero, so the test applies and the verdict is divergence.
Say ‘inconclusive’ out loud
When the limit is zero, name that explicitly before moving on. Free-response scoring rewards the correct reasoning, not just the final verdict.
Cheap, so run it first
One limit rules out a whole class of series in seconds. It just will not be the end of the story for most of them.