Series Convergence Calculator
Test whether an infinite series converges using the ratio test, alongside partial sums and the necessary check that terms approach zero.
Series Convergence Calculator
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Formula: Ratio Test: L = |a(n+1)/a(n)| → if L<1 converges, L>1 diverges, L=1 inconclusive
Worked example — Converges. Sum = 1 when the index starts at n = 1 (and 2 when it starts at n = 0)
Formula
Ratio Test: L = |a(n+1)/a(n)| → if L<1 converges, L>1 diverges, L=1 inconclusive
Tests for series convergence: Ratio test computes the limit of consecutive term ratios. Also checks: partial sums trend, term magnitude approaching zero (necessary condition), and root test.
Worked Examples
Example 1: Geometric series with ratio 1/2
Problem:Does the series with terms a(n) = 1/2^n, starting at n = 1, converge? If so, to what?
Solution:Ratio test: a(n+1)/a(n) = (1/2^(n+1)) / (1/2^n) = 1/2 for every n, so L = 0.5. Since L = 0.5 < 1 the series converges absolutely. For a geometric series the exact sum is a/(1 - r), where a is the FIRST term actually included. Starting at n = 1 the first term is 1/2, so the sum is 0.5/(1 - 0.5) = 1. Starting at n = 0 the first term is 1, so the sum would instead be 1/(1 - 0.5) = 2. Partial sums confirm it: 0.5, 0.75, 0.875, 0.9375, ... climbing toward 1.
Result:Converges. Sum = 1 when the index starts at n = 1 (and 2 when it starts at n = 0)
Example 2: Harmonic series - terms go to zero but the sum does not converge
Problem:Does the series with terms a(n) = 1/n converge?
Solution:The terms do shrink to zero, which is necessary for convergence but never sufficient. Ratio test: a(n+1)/a(n) = n/(n+1), whose limit is L = 1, so the ratio test is inconclusive. Integral test: the integral of 1/x from 1 to N equals ln(N), which grows without bound, so the series diverges. Numerically the growth is brutally slow: the partial sum passes 10 only around n = 12,367, which is why a truncated numerical check can look convergent.
Result:Diverges. Terms to 0 is necessary but not sufficient; the ratio test returns L = 1 (inconclusive)
Example 3: p-series and a factorial series
Problem:Compare a(n) = 1/n^2 with a(n) = 2^n/n!.
Solution:For 1/n^2 the ratio test again gives L = 1 and is inconclusive, but the p-series rule settles it: sum of 1/n^p converges for p > 1, and here p = 2. The exact value is pi^2/6 = 1.644934. For 2^n/n! the ratio test is decisive: a(n+1)/a(n) = 2^(n+1)/(n+1)! x n!/2^n = 2/(n+1), whose limit is 0. L = 0 < 1, so the series converges. Its sum from n = 0 is e^2 = 7.389056, and from n = 1 it is e^2 - 1 = 6.389056.
Result:Both converge: sum of 1/n^2 = pi^2/6 = 1.6449 | sum of 2^n/n! from n=1 = e^2 - 1 = 6.3891
Frequently Asked Questions
What are the main convergence tests?
Ratio test, root test, comparison test, integral test, alternating series test, p-series test (converges if p>1), and geometric series test (converges if |r|<1).
Can a series converge if terms do not approach zero?
No. If lim a(n) ≠ 0, the series MUST diverge (nth term test). However, terms approaching zero is necessary but NOT sufficient — the harmonic series 1/n diverges despite 1/n → 0.
When is the ratio test inconclusive and what should I use instead?
The ratio test is inconclusive when L = 1, which often occurs with p-series and polynomial-over-polynomial terms. In these cases, use the p-series test (Σ 1/nᵖ converges iff p > 1), the comparison test against a known series, or the integral test. For alternating series like Σ (-1)ⁿ/n, use the alternating series test instead.
What is the difference between absolute and conditional convergence?
A series converges absolutely if Σ |aₙ| converges. It converges conditionally if Σ aₙ converges but Σ |aₙ| diverges. The alternating harmonic series Σ (-1)ⁿ/n is a classic example: it converges conditionally but not absolutely. Absolutely convergent series can be rearranged without changing the sum; conditionally convergent series cannot.
How does the root test compare with the ratio test?
The root test evaluates L as the limit of the nth root of the absolute value of a(n), and reads the answer the same way: L < 1 converges, L > 1 diverges, L = 1 is inconclusive. It is the stronger of the two, since whenever the ratio test settles a series the root test reaches the same verdict, and it also handles terms built from nth powers where ratios are awkward.
When is the integral test the right tool to reach for?
Use it when a(n) = f(n) for a function that is positive, continuous and eventually decreasing on an interval running to infinity, and that you can actually integrate. The series and the improper integral then converge or diverge together. It is what proves the p-series rule, because the integral of 1/xᵖ converges exactly when p > 1, and it settles borderline cases such as the divergent sum of 1/(n ln n).
How do I bound the error when I stop after a finite number of terms?
For an alternating series that satisfies the alternating series test, the remainder after n terms is no larger than the first omitted term, which makes truncation error trivial to control. For a positive decreasing series covered by the integral test, the tail is bracketed between integrals of f taken from n and from n + 1 to infinity. A partial sum with no such bound is a numerical hint, not a proof.
Why can partial sums look convergent even when the series diverges?
Because divergence can be extraordinarily slow. The harmonic series grows like the natural logarithm of n, so after fifty terms the running total is only about 4.5 and each new term adds under 0.02, which looks like settling down. Every numerical experiment on finitely many terms shares this blind spot, which is why an analytic test is needed before declaring a verdict.
What is the difference between the direct and limit comparison tests?
Direct comparison requires a term-by-term inequality: if 0 ≤ a(n) ≤ b(n) and the sum of b(n) converges, so does the sum of a(n). Limit comparison only needs the ratio a(n)/b(n) to approach a finite positive number, after which both series behave alike. Limit comparison is usually easier for rational terms, where you compare against the ratio of the leading powers.
How does the ratio test give the radius of convergence of a power series?
Apply the ratio test to the terms of the sum of c(n)(x − a)ⁿ while treating x as fixed. The limit takes the form |x − a| × L, and demanding that it stay below 1 gives |x − a| < 1/L, so the radius of convergence is R = 1/L. The test says nothing at the two endpoints where |x − a| = R, so each must be substituted in and checked separately.
References
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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