Which of the Following Series Is Divergent?
The short version is – you can tell by looking at the terms, the growth rate, and a handful of quick tests.
Ever stared at a list of infinite series and wondered which one “blows up” and which one stays nicely bounded? You’re not alone. In a first‑year calculus class the question “Which of the following series is divergent?” pops up more often than a pop quiz on the chain rule. The answer isn’t always obvious, but the good news is that a few core ideas let you separate the runaway sums from the tame ones in seconds.
Below is the kind of cheat‑sheet you wish you’d had on exam day: a walk‑through of the main concepts, the most common pitfalls, and a handful of practical tips you can actually use when the professor throws a curveball. By the end you’ll be able to glance at a series and say, “That one diverges,” without breaking a sweat.
Worth pausing on this one.
What Is a Divergent Series?
When we talk about a series we mean an infinite sum
[ \sum_{n=1}^{\infty} a_n . ]
If the partial sums
[ S_N = a_1 + a_2 + \dots + a_N ]
settle down to a single number as (N) grows, the series converges. If the partial sums keep wandering off to infinity, oscillate without settling, or simply fail to approach any limit, the series diverges.
In plain language: a divergent series is one that refuses to “add up” to a finite value. It’s the mathematical equivalent of a conversation that never ends.
A quick sanity check
The nth term test is the first line of defense. If
[ \lim_{n\to\infty} a_n \neq 0, ]
the series is automatically divergent. The converse isn’t true – a term that goes to zero can still produce a divergent sum (think harmonic series). So the test is a one‑way street: non‑zero limit → divergent; zero limit → inconclusive.
Why It Matters
Knowing whether a series diverges isn’t just a homework exercise. In physics, engineering, and finance you often replace a complicated function with an infinite series to make calculations tractable. If the series diverges, the approximation collapses and your model can give nonsense results.
Take signal processing: Fourier series let you rebuild a wave from sines and cosines. Practically speaking, if the series you pick diverges, the reconstructed signal will blow up, ruining any downstream analysis. In economics, a divergent series in a present‑value calculation means the projected cash flow is infinite – a red flag for any investor Not complicated — just consistent..
In short, spotting divergence early saves time, money, and a lot of headaches.
How to Tell If a Series Diverges
Below is the toolbox you’ll reach for, organized by how “quick” each method feels. Most of the time you’ll combine a couple of them.
1. Nth‑Term Test (the “obvious” one)
If (\displaystyle\lim_{n\to\infty}a_n\neq0) the series diverges It's one of those things that adds up..
Example: (\displaystyle\sum_{n=1}^{\infty}\frac{2n+1}{n}).
Here (a_n\to 2\neq0). Divergent, plain and simple.
2. Comparison Test
You compare your series to another series whose behavior you already know.
- Direct comparison: If (0\le a_n\le b_n) for all large (n) and (\sum b_n) converges, then (\sum a_n) converges. Flip the inequality for divergence.
- Limit comparison: Compute
[ L=\lim_{n\to\infty}\frac{a_n}{b_n}. ]
If (0<L<\infty) both series share the same fate No workaround needed..
Example: (\displaystyle\sum_{n=1}^{\infty}\frac{1}{n\sqrt{n}}). Compare to (\sum 1/n^{3/2}) (a p‑series with (p=3/2>1)). The limit of the ratio is 1, so both converge Less friction, more output..
If you flip the exponent to (\displaystyle\sum\frac{1}{n^{0.9}}) you get divergence because the comparable p‑series has (p<1).
3. p‑Series Test
A series of the form (\displaystyle\sum\frac{1}{n^p}) converges iff (p>1). Anything else diverges.
That’s why the classic harmonic series (\displaystyle\sum\frac{1}{n}) (p=1) diverges, while (\displaystyle\sum\frac{1}{n^2}) converges.
4. Ratio Test (and Root Test)
For series with factorials, exponentials, or powers, the ratio test is a lifesaver Simple, but easy to overlook..
[ L=\lim_{n\to\infty}\Bigl|\frac{a_{n+1}}{a_n}\Bigr|. ]
- If (L<1) → convergent.
- If (L>1) or (L=\infty) → divergent.
- If (L=1) → inconclusive (go to another test).
The root test works similarly but uses the nth root:
[ L=\lim_{n\to\infty}\sqrt[n]{|a_n|}. ]
Same decision rule That's the part that actually makes a difference..
Example: (\displaystyle\sum\frac{n!}{2^n}). Ratio test gives
[ \frac{(n+1)!/2^{n+1}}{n!/2^n}= \frac{n+1}{2}\to\infty, ]
so the series diverges.
5. Integral Test
If (a_n = f(n)) where (f) is positive, continuous, and decreasing for (n\ge N), then
[ \sum a_n \text{ converges } \iff \int_N^{\infty} f(x),dx \text{ converges}. ]
Great for series that look like (\frac{1}{n\log n}) or (\frac{1}{n(\log n)^2}).
6. Alternating Series Test (Leibniz)
For series that flip sign: (\displaystyle\sum (-1)^{n}b_n) with (b_n\ge0) That's the part that actually makes a difference..
If (b_n) decreases monotonically to 0, the series converges (conditionally). If the decrease condition fails, you can’t conclude divergence, but many alternating series still diverge because the underlying absolute series diverges.
7. Cauchy Condensation Test
When the terms are decreasing and positive,
[ \sum a_n \text{ converges } \iff \sum 2^k a_{2^k} \text{ converges}. ]
It’s a neat shortcut for series like (\displaystyle\sum\frac{1}{n\log n}) The details matter here..
Common Mistakes (What Most People Get Wrong)
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Assuming the nth‑term test is enough
Many students stop after checking (\lim a_n =0) and call the series “maybe convergent.” That’s a classic dead‑end. The harmonic series shows the trap. -
Mixing up “greater than” and “less than” in comparison tests
If you compare to a known divergent series, you need the larger terms to guarantee divergence. Flip the inequality and you’ll get the opposite conclusion. -
Using the ratio test on a p‑series
Ratio test gives (L=1) for (\sum 1/n^p), which is inconclusive. Jumping to “it must converge” is a rookie move. -
Ignoring the monotonic requirement in the alternating series test
The sequence (b_n) must actually decrease. If it wiggles, the test fails and you need a different approach No workaround needed.. -
Treating conditional convergence as “good enough”
In applications where absolute convergence matters (e.g., rearranging terms, Fourier series), a conditionally convergent series can misbehave.
Practical Tips (What Actually Works)
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Start with the nth‑term test. It’s the fastest filter. If the limit isn’t zero, you’re done.
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Identify the shape: Does the term involve factorials? Exponentials? Powers of (n)? That will point you toward ratio/root or p‑series.
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Write the term in simplest form. Cancel common factors, pull out constants. A messy expression can hide a simple comparison Still holds up..
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Use the limit comparison test with a “nearby” p‑series. If your term looks like (\frac{1}{n^{p}}) times a slowly varying factor (log, root, etc.), compare to (\frac{1}{n^{p}}).
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When logs appear, remember the integral test shines: (\int \frac{dx}{x(\log x)^q}) converges only if (q>1) Most people skip this — try not to..
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For alternating series, check monotonicity first. If it fails, consider absolute convergence with the same tests you’d use for the non‑alternating version.
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Keep a “cheat sheet” of limits:
[ \lim_{n\to\infty}\frac{n^{a}}{b^{n}} = 0 \quad (b>1),\qquad \lim_{n\to\infty}\frac{b^{n}}{n^{a}} = \infty. ]
This helps decide ratio/root test outcomes instantly.
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Don’t forget the condensation test for series with (\frac{1}{n(\log n)^{p}}). It reduces the problem to a geometric‑like series.
FAQ
Q1: Does a series diverge if its terms don’t go to zero fast enough?
A: “Fast enough” is vague. The precise condition is the nth‑term test: if the limit isn’t zero, it diverges. If the limit is zero, you need another test; sometimes even a zero limit isn’t sufficient (harmonic series).
Q2: Can a series be both conditionally convergent and divergent?
A: No. “Conditionally convergent” means the series converges, but not absolutely. If it diverges, it’s simply divergent, not “conditionally divergent.”
Q3: How do I handle series with both factorials and powers, like (\displaystyle\sum\frac{n^3}{(2n)!})?
A: The ratio test is perfect here. Compute
[ \frac{(n+1)^3/(2n+2)!}{n^3/(2n)!} = \frac{(n+1)^3}{n^3}\cdot\frac{1}{(2n+2)(2n+1)}\to0, ]
so the series converges absolutely.
Q4: Is the integral test only for continuous functions?
A: Technically yes, but you can often bound a discrete term by a continuous function that matches it on integers. The test works as long as the function is positive, decreasing, and integrable on ([N,\infty)).
Q5: When should I use the root test instead of the ratio test?
A: The root test shines for series like (\displaystyle\sum (n^{2}+3)^{n}, /, 5^{n^{2}}) where the nth root simplifies the exponent. If the ratio test gives a messy limit or 1, try the root test But it adds up..
That’s it. The next time you see a list of series and the question “which of the following is divergent?”, run through the quick checklist: nth‑term test → identify the dominant growth → pick the right comparison/ratio/root test → double‑check monotonicity for alternating signs.
You’ll find that the “divergent” ones usually have a term that refuses to shrink fast enough, or they hide a harmonic‑style tail that sneaks past the zero‑limit test. In practice, with these tools in hand, you’ll spot the runaway sums faster than you can write “∞”. Happy summing!
