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Series Convergence Test Calculator

Decide whether a series converges or diverges, with the test and every step shown
does sum from n=1 to infinity of 2^n/n! converge?
test sum of (3n + 1)/(n^3 + 2n) for convergence
does sum from n=2 to infinity of 1/(n ln n) converge?
test sum of (-1)^n / sqrt(n) for absolute convergence

What Convergence Means

A series ∑an\sum a_n converges when its sequence of partial sums has a finite limit:

SN=∑n=1Nan,∑n=1∞an=lim⁡N→∞SNS_N = \sum_{n=1}^{N} a_n, \qquad \sum_{n=1}^{\infty} a_n = \lim_{N \to \infty} S_N

If that limit fails to exist, the series diverges. A test tells you which of these happens; it does not usually tell you the sum.

Two benchmark families anchor everything else:

∑n=1∞arn−1 converges  ⟺  ∣r∣<1,∑n=1∞1np converges  ⟺  p>1\sum_{n=1}^{\infty} ar^{n-1} \text{ converges} \iff |r| < 1, \qquad \sum_{n=1}^{\infty} \frac{1}{n^p} \text{ converges} \iff p > 1

The harmonic series ∑1n\sum \frac{1}{n} is the boundary case p=1p = 1, and it diverges — slowly, but definitely.

Absolute convergence: if ∑∣an∣\sum |a_n| converges then ∑an\sum a_n does too. The converse fails, and a series like ∑(−1)nn\sum \frac{(-1)^n}{n} is conditionally convergent.

The Tests and Their Hypotheses

Every test has conditions. Applying one outside its hypotheses produces a confident but meaningless answer.

TestRequiresConclusion
nth-termnonelim⁡an≠0⇒\lim a_n \neq 0 \Rightarrow diverges; =0= 0 proves nothing
p-seriesan=1/npa_n = 1/n^pconverges iff p>1p > 1
Integralff positive, continuous, decreasing on [N,∞)[N,\infty)∑an\sum a_n and ∫N∞f\int_N^{\infty} f share a fate
Comparison0≤an≤bn0 \le a_n \le b_n∑bn\sum b_n converges ⇒∑an\Rightarrow \sum a_n does
Limit comparisonan,bn>0a_n, b_n > 0 and lim⁡anbn=L\lim \frac{a_n}{b_n} = L with 0<L<∞0 < L < \inftyboth behave alike
RatioL=lim⁡∣an+1an∣L = \lim \left\lvert \frac{a_{n+1}}{a_n} \right\rvertL<1L<1 converges, L>1L>1 diverges, L=1L=1 inconclusive
RootL=lim⁡∣an∣nL = \lim \sqrt[n]{\lvert a_n \rvert}same thresholds as the ratio test
Alternatingbnb_n decreasing and bn→0b_n \to 0∑(−1)nbn\sum (-1)^n b_n converges

Choosing quickly

  1. Check lim⁡an\lim a_n first — if it is not 00, you are done.
  2. Factorials or nn-th powers ⇒\Rightarrow ratio or root test.
  3. A quotient of polynomials ⇒\Rightarrow limit comparison with 1/np1/n^{p}, using the difference of the leading degrees as pp.
  4. An easily integrated f(n)f(n) ⇒\Rightarrow integral test.
  5. Alternating signs ⇒\Rightarrow test ∑∣an∣\sum |a_n| first, then the alternating series test.

Common Mistakes to Avoid

  • Concluding convergence from an→0a_n \to 0: the harmonic series is the standing counterexample. The nth-term test can only prove divergence.
  • Using the ratio test on a rational function: it always returns L=1L = 1 there, which is inconclusive. Use limit comparison instead.
  • Skipping the decreasing hypothesis: the integral test and the alternating series test both require it, and it can fail for small nn even when it holds eventually.
  • Comparing in the wrong direction: a series larger than a divergent one diverges; a series larger than a convergent one tells you nothing.
  • Confusing sequence and series: an→0a_n \to 0 says the sequence converges, not the series.

示例题目

Step 1: A factorial suggests the ratio test
Step 2: an+1an=2n+1(n+1)!⋅n!2n=2n+1\dfrac{a_{n+1}}{a_n} = \dfrac{2^{n+1}}{(n+1)!} \cdot \dfrac{n!}{2^n} = \dfrac{2}{n+1}
Step 3: L=lim⁡n→∞2n+1=0L = \lim_{n\to\infty} \dfrac{2}{n+1} = 0
Step 4: Since L=0<1L = 0 < 1, the series converges absolutely
Answer: Converges (ratio test, L=0L = 0); the sum is in fact e2−1e^2 - 1

Step 1: Leading degrees differ by 3−1=23 - 1 = 2, so compare with bn=1n2b_n = \dfrac{1}{n^2}
Step 2: Both ana_n and bnb_n are positive for n≥1n \ge 1, as the limit comparison test requires
Step 3: anbn=(3n+1)n2n3+2n=3n3+n2n3+2n\dfrac{a_n}{b_n} = \dfrac{(3n+1)n^2}{n^3+2n} = \dfrac{3n^3+n^2}{n^3+2n}
Step 4: Divide through by n3n^3: the limit is 31=3\dfrac{3}{1} = 3, finite and positive
Step 5: ∑1/n2\sum 1/n^2 is a pp-series with p=2>1p = 2 > 1, so it converges — and therefore so does the original
Answer: Converges (limit comparison with 1/n21/n^2, L=3L = 3)

Step 1: Let f(x)=1xln⁡xf(x) = \dfrac{1}{x\ln x}, which is positive, continuous and decreasing for x≥2x \ge 2 — the integral test applies
Step 2: Substitute u=ln⁡xu = \ln x, du=dxxdu = \dfrac{dx}{x}, so ∫dxxln⁡x=∫duu=ln⁡∣u∣\displaystyle\int \frac{dx}{x\ln x} = \int \frac{du}{u} = \ln|u|
Step 3: Evaluate the improper integral: ∫2∞dxxln⁡x=lim⁡T→∞[ln⁡(ln⁡x)]2T\displaystyle\int_2^{\infty} \frac{dx}{x\ln x} = \lim_{T\to\infty}\left[\ln(\ln x)\right]_2^{T}
Step 4: ln⁡(ln⁡T)→∞\ln(\ln T) \to \infty, so the integral diverges
Step 5: Note an→0a_n \to 0 here, which is why the nth-term test was useless
Answer: Diverges (integral test)

常见问题

Start with the nth-term test, since a limit that is not zero settles the question immediately. After that, let the form guide you: factorials and nth powers point to the ratio or root test, rational expressions to limit comparison with a p-series, and integrable terms to the integral test.

No. It is necessary but not sufficient — the harmonic series 1 + 1/2 + 1/3 + ... has terms tending to zero yet diverges. The nth-term test can only ever prove divergence, never convergence.

The test is inconclusive and you must switch. This happens for every rational function and every p-series, so use limit comparison against 1/n^p, or the integral test. The root test almost always gives 1 in the same situations.

A series converges absolutely when the series of absolute values converges, which is the stronger property and permits rearranging terms freely. It converges conditionally when it converges but the absolute version does not — as with the alternating harmonic series, whose terms can be rearranged to sum to anything.

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