Series Convergence Calculator

The series convergence calculator adds up the terms of \( \sum a_n \) and tells you whether the series appears to converge. It shows two partial sums, a numerical ratio test and the \( n \)th-term test, then gives a verdict and the approximate sum.

Use it to check a homework answer or estimate the value of a convergent series. One point up front: this is a numerical tool. It computes sums and ratios with real numbers and does not produce a formal proof, so treat its verdict as strong evidence. It is free and needs no sign-up.

How to use the series convergence calculator

1. Type the general term \( a_n \) using n as the variable. Use ^ for powers, n! for factorials, sqrt(), ln(), e^(n) and (-1)^n for alternating signs. Implicit multiplication like 3n works.
2. Enter the starting index (often 0 or 1). The term must be defined there, so start 1/n at 1.
3. Enter the number of terms \( N \) to add. The default of 1000 is a good start.
4. Press Test series. You get \( S_N \), the larger partial sum \( S_{10N} \), the estimated ratio \( L \), and a verdict.

How it works

The partial sum \( S_N \) adds \( N \) terms starting at your first index. The calculator also adds \( 10N \) terms (up to two million) so you can see whether the total is still moving. The verdict is decided in this order:

1. \( n \)th-term test. If \( |a_n| \) is still above 0.001 at \( n \) around 100,000 and 1,000,000, the terms aren’t going to 0, so the series diverges.
2. Ratio test. The tool estimates $$L = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$$ by computing the ratio at a large \( n \). If \( L < 1 \) the series converges, and if \( L > 1 \) it diverges.
3. When \( L = 1 \), the ratio test is inconclusive. The calculator then compares \( S_N \) with \( S_{10N} \). If they agree closely, it says the series appears to converge; if the sum keeps growing, it appears to diverge.

When the ratio test shows convergence and the sum matches a familiar number, such as a fraction or a multiple of \( e \), the exact form is shown too.

Worked example

Example 1: the ratio test settles it. Test \( \sum_{n=1}^{\infty} \frac{n^2}{2^n} \).

The first four terms already add to \( \frac{29}{8} \), but the ratio of consecutive terms is

$$\frac{a_{n+1}}{a_n} = \frac{(n+1)^2}{2n^2} \to \frac12 < 1.$$

So the series converges. Because the calculator measures the ratio at a large but finite \( n \), it shows a value just above 0.5 (about 0.502). The sum it reports is 6, which matches the known formula \( \sum n^2x^n = \frac{x(1+x)}{(1-x)^3} \) at \( x = \frac12 \). This is the example preloaded above.

Example 2: when \( L = 1 \). Compare \( \sum \frac1{n^2} \) with \( \sum \frac1n \), both starting at \( n = 1 \). For both, \( \frac{a_{n+1}}{a_n} \to 1 \), so the ratio test says nothing.

• For \( \frac1{n^2} \): \( S_{1000} \approx 1.6439346 \) and \( S_{10000} \approx 1.6448341 \). The sum moved by less than 0.001, so the verdict is “appears to converge”. The true sum is \( \frac{\pi^2}{6} \approx 1.6449341 \).
• For \( \frac1n \) (the harmonic series): \( S_{1000} \approx 7.4855 \) and \( S_{10000} \approx 9.7876 \). The sum jumped by about 2.30, close to \( \ln 10 \approx 2.3026 \), because harmonic sums grow like \( \ln N \). Verdict: “appears to diverge”, which is correct.

Limits of a numerical test

• Slow series can fool it. \( \sum \frac1{n^{1.1}} \) converges (it’s a \( p \)-series with \( p > 1 \), and \( \int_1^\infty x^{-1.1}\,dx = 10 \)), but it converges so slowly that \( S_{10000} - S_{1000} \approx 1.03 \). The calculator reports “appears to diverge”. Use the \( p \)-series rule or the integral test for series like this.
• It doesn’t separate absolute and conditional convergence. \( \sum \frac{(-1)^n}{n} \) “appears to converge” (to \( -\ln 2 \approx -0.6931 \)), which is right, but only conditionally. Test \( \sum \frac1n \) separately to see that it isn’t absolutely convergent.
• Huge terms can overflow. For \( \frac{3^n}{n!} \), both \( 3^n \) and \( n! \) exceed the largest number a computer stores long before \( n = 1000 \), and the tool reports that \( a_n \) is undefined. Lower \( N \) to 50 and it works, giving \( e^3 \approx 20.0855 \).

For written proofs, see the guides to the ratio test and geometric series. The integral test relies on improper integrals, which you can evaluate with the definite integral calculator using inf as the upper limit.

Related guides

• Ratio Test for Series Convergence: How to Use It
• Geometric Series Formula: Sum, Convergence and Examples
• Improper Integrals: How to Tell If They Converge or Diverge
• Taylor and Maclaurin Series: Formula and Examples

Further reading

• Convergence/Divergence of Series (Paul’s Online Math Notes) — partial sums and the divergence test.
• Ratio Test (Paul’s Online Math Notes) — how to apply it, and what \( L = 1 \) can hide.

FAQ

Does the series convergence calculator prove convergence?

No. It adds many terms and estimates the ratio numerically, which is reliable for most textbook series but isn’t a proof. Use its verdict to check the answer you prove with a standard test.

Which convergence tests does it use?

The \( n \)th-term (divergence) test and the ratio test, plus a comparison of partial sums when the ratio test is inconclusive. It doesn’t run the comparison, integral, root or alternating series tests.

Can it find the sum of a series?

Yes, approximately. \( S_N \) and \( S_{10N} \) are the partial sums. When they agree, that value is the sum to the digits shown. Familiar exact values like 6 or \( e \) are recognized.

How do I enter a factorial or an alternating series?

Type n! for factorials and (-1)^n for alternating signs, for example (-1)^n/n! starting at 0.