Example 1: Convergent p-series
Σ 1/n²
Since p = 2 > 1, the series converges.
Determine whether common infinite series converge or diverge using established mathematical tests. Choose a convergence test, enter its required parameter or limit, and review the result with a step-by-step explanation.
Choose the test that matches your series. Some tests require you to enter a limit that you have already calculated. The calculator applies the test to that value; it does not automatically derive limits from an arbitrary expression.
Load one of these examples to see how the calculator applies a convergence test.
Σ 1/n²
Since p = 2 > 1, the series converges.
Σ 1/n
Since p = 1, the series diverges.
Σ (1/2)ⁿ, n = 0 to infinity
The ratio is 1/2 and the sum is 2.
Σ 1/n!
The ratio limit is 0, so the series converges absolutely.
Σ 1/n²
The ratio limit is 1, so use another test.
Σ (-1)ⁿ⁺¹/n
The magnitudes decrease to zero, so the series converges.
An infinite series is the sum of the terms of an infinite sequence. For a series Σ an, its partial sums are defined by SN = a1 + a2 + ... + aN.
The series converges if its partial sums approach a finite limit as N tends to infinity. Otherwise, it diverges. Convergence tests help establish this behavior without necessarily calculating the exact infinite sum.
| Test | Convergence condition | What happens otherwise? |
|---|---|---|
| p-Series | p > 1 | Diverges when p ≤ 1. |
| Geometric Series | |r| < 1 | Diverges for nonzero first term when |r| ≥ 1. |
| nth-Term Test | Term limit must be zero for convergence. | A nonzero term limit proves divergence. |
| Ratio Test | L < 1 proves absolute convergence. | L > 1 proves divergence; L = 1 is inconclusive. |
| Root Test | L < 1 proves absolute convergence. | L > 1 proves divergence; L = 1 is inconclusive. |
| Alternating Series Test | Non-negative magnitudes eventually decrease to zero. | Failure of these conditions does not by itself prove divergence. |
Start by calculating lim an. If the limit is nonzero or does not exist, the series diverges. If it is zero, continue with another test.
If the series is a geometric series, examine its common ratio. If its terms are 1/np, use the p-series test.
For series containing factorials or products of successive factors, the ratio an+1/an may simplify considerably. Take the limit of its absolute value.
The root test is often useful when terms contain expressions raised to the nth power. Calculate the limit of the nth root of |an|.
For alternating terms, check whether their non-negative magnitudes decrease eventually and approach zero. If so, the alternating series test establishes convergence.
No single convergence test works for every series. In particular, a ratio or root limit equal to 1 requires a different method. Comparison, limit comparison, and integral tests can be useful alternatives when their conditions hold.
Yes, but not always. Terms approaching zero are necessary for convergence, but the harmonic series Σ 1/n is a well-known example of a divergent series whose terms approach zero.
It means the selected test does not establish convergence or divergence. It does not mean the series fails to converge. Another test may settle the question.
A series Σ an converges absolutely when Σ |an| converges. Absolute convergence implies convergence of the original series.
A series converges conditionally when it converges but the series of absolute values diverges. The alternating harmonic series is a standard example.
No. This version applies selected convergence criteria to supplied parameters and known limits. It does not symbolically derive arbitrary limits or prove the hypotheses of a test automatically.
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