Series Convergence Calculator

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A series converges if its partial sums approach a finite number. For example, \(\sum \frac{1}{n^{2}}\) converges (a p-series with p = 2 > 1), while the harmonic series \(\sum \frac{1}{n}\) diverges. Enter aₙ to see which test settles it.

FreeStep by stepPartial sums graph

\(a_{n} = \frac{1}{n^{2}}\)Use n. Factorials (n!), (-1)^n, ln(n), sqrt(n) and powers are fine.

Try:
Convergence
\(\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2}}\ \text{converges}\)
The series converges (limit comparison with the p-series (p = 2)).
NS_N10020030040050060070080090010000.81.01.21.41.61.8

Partial sums S_N.

Test used Limit comparison with the p-series (p = 2)
Approximate sum (first 1,000 terms) 1.6439346
Partial sum of the first 10 terms 1.549767731
Partial sum of the first 100 terms 1.6349839
Partial sum of the first 1,000 terms 1.643934567
Show steps
  1. aₙ is a rational function of n; the denominator’s degree is 2 and the numerator’s is 0, so \(a_{n}\) behaves like \(\frac{1}{n^{2}}\) for large n.
  2. By the limit comparison test with \(\sum \frac{1}{n^{2}}\) (a p-series with p = 2 > 1), the series converges absolutely.

How to use the series convergence calculator

  1. Enter the general term aₙ using n, and the starting index.
  2. Press Test. The calculator tries the standard tests in order and reports the first one that decides the question.
  3. The partial sums and their graph show the behavior numerically. If no test is decisive, the result says so instead of guessing.

The main convergence tests

Test When to use Conclusion
nth-term (divergence) always first If \(a_{n} \not\to 0\), diverges
Geometric \(ar^{n}\) Converges iff \(|r| < 1\); sum \(\frac{a}{1-r}\)
p-series \(\frac{1}{n^{p}}\) Converges iff p > 1
(Limit) comparison rational terms Behaves like \(\frac{1}{n^{p}}\) with p = degree difference
Alternating series \((-1)^{n}b_{n}\) Converges if \(b_{n}\) decreases to 0
Ratio factorials, exponentials \(\lim\left|\frac{a_{n+1}}{a_{n}}\right| < 1\): converges; > 1: diverges

Absolute and conditional convergence

A series converges absolutely if \(\sum |a_{n}|\) converges. The alternating harmonic series \(\sum \frac{(-1)^{n+1}}{n}\) converges (to ln 2) but not absolutely, so it converges conditionally.

Common mistakes

  • Thinking \(a_{n} \to 0\) proves convergence; the harmonic series is the classic counterexample.
  • Using the ratio test on p-series: the limit is 1, so the test says nothing.
  • Forgetting that the starting index changes the sum (but not whether it converges).

Series convergence practice quiz: 10 questions

Test yourself with 10 multiple-choice questions. Choose an answer and press Submit to see if you are right and why. Your score appears at the end.

Question 1 of 10
1. Does Σ 1/n² converge?

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2. Does the harmonic series Σ 1/n converge?

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3. Σ (1/3)ⁿ from n = 0 sums to…

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4. If lim aₙ = 5, the series…

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5. If lim aₙ = 0, the series…

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6. Σ 1/√n is a p-series with p = …

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7. For Σ 2ⁿ/n!, the ratio |aₙ₊₁/aₙ| = 2/(n+1) tends to…

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8. Σ (−1)ⁿ⁺¹/n converges…

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9. Σ (n + 1)/(n³ + 2) behaves like…

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10. The ratio test gives a limit of exactly 1. Then…

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Frequently asked questions

How do I know if a series converges?

Check that the terms go to 0, then use the test that fits the form: geometric, p-series or comparison for powers, the alternating series test for alternating signs, and the ratio test for factorials and exponentials.

What is the difference between absolute and conditional convergence?

Absolute convergence means Σ|aₙ| converges. Conditional convergence means the series converges but Σ|aₙ| does not.

Can the calculator find the sum?

It gives the exact sum of geometric series and an estimate from partial sums for other convergent series.

Why does it say inconclusive?

Some series, such as Σ 1/(n ln n) or Σ cos(n)/n, need the integral test or more advanced tests. The calculator reports inconclusive rather than risk a wrong answer.

Does the starting index matter?

Not for convergence, since dropping finitely many terms does not change it. It does change the value of the sum.

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