Series Convergence Calculator

About the Series Convergence Calculator

Deciding whether an infinite series converges is a skill that relies on choosing the right test. Calculus courses teach at least seven of them: the nth-term test, geometric series, p-series, comparison and limit comparison, the ratio test, the root test, the integral test and the alternating series test. Each one works brilliantly on some series and says nothing at all about others, and the hard part is knowing which to reach for.

This calculator applies the tests in a fixed, sensible order and reports the first one that gives a definite answer. It recognises geometric series and sums them exactly, checks whether the terms tend to zero, handles alternating signs, and for positive terms tries the ratio test, the root test, comparison with a p-series, Cauchy condensation and the integral test. Every limit it needs is computed by the limit engine, which verifies itself numerically.

When a test relies on a property that cannot be proved symbolically, such as the terms eventually decreasing, the calculator checks it on many sample indices and says so openly. If no test is decisive, it tells you which tests were tried rather than guessing. For alternating series it also distinguishes absolute from conditional convergence.

What does it mean for a series to converge?

An infinite series $\sum_{n=1}^{\infty} a_n$ is the sum of infinitely many terms. To give it meaning, look at the partial sums $S_N = a_1 + a_2 + \cdots + a_N$ . If these approach a finite limit $S$ as $N$ grows, the series converges and its sum is $S$ . If they grow without bound or keep oscillating, the series diverges.

Convergence is subtle because it depends on how fast the terms shrink, not merely on whether they shrink. The terms of $\sum \frac1n$ go to zero, yet the partial sums grow past every number, so the harmonic series diverges. The terms of $\sum \frac{1}{n^2}$ shrink just a little faster, and that series converges to $\frac{\pi^2}{6}$ .

For most series you can decide convergence without finding the sum, and the sum often has no simple closed form anyway. Geometric series are the main exception: whenever the common ratio satisfies $|r| < 1$ , the sum is $\frac{a}{1-r}$ , where $a$ is the first term.

How to use the Series Convergence Calculator

Type the general term using n as the index, for example 1/n^2, (-1)^(n+1)/n or 2^n/n!.
Factorials (n!), powers, roots, logarithms written as log for ln, and (−1)^n are all supported.
Enter the start index, the first value of n in the sum. It defaults to 1.
Press Test convergence to see which tests were applied, the limits computed and the verdict.
For geometric series, read the exact sum from the final line.

Formulas

Geometric series

\sum_{n=0}^{\infty} a r^n = \frac{a}{1-r},\quad |r| < 1

nth-term test

\lim_{n\to\infty} a_n \ne 0 \;\Rightarrow\; \sum a_n \text{ diverges}

p-series

\sum \frac{1}{n^p} \text{ converges} \iff p > 1

Ratio test

L = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|;\ L < 1 \text{ converges},\ L > 1 \text{ diverges}

Root test

L = \lim_{n\to\infty}\sqrt[n]{|a_n|};\ L < 1 \text{ converges},\ L > 1 \text{ diverges}

Alternating series test

b_n \downarrow 0 \;\Rightarrow\; \sum (-1)^n b_n \text{ converges}

Worked example: p-series Σ 1/n²

The worked example below is computed live for the p-series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ . The terms tend to 0, so the nth-term test is inconclusive. The calculator then estimates how quickly the terms shrink and compares them with $\frac{1}{n^2}$ ; the limit of the ratio is 1, a positive finite number.

By the limit comparison test the series behaves exactly like the p-series with $p = 2$ , which converges because $p > 1$ . The verdict is that the series converges. Its sum, $\frac{\pi^2}{6} \approx 1.645$ , was found by Euler in 1734, but convergence can be settled without knowing it.

The solution below is generated by the same engine that powers the calculator above.

Final answer

\sum_{n=1}^{\infty} \frac{1}{{n}^{2}} \ \text{converges} \approx 1.644934067

How we read your problem

Decide whether Σ 1 / n² (n from 1 to ∞) converges or diverges.

\sum_{n=1}^{\infty} \frac{1}{{n}^{2}}

Step-by-step solution

Write the series
Σ a(n) for n = 1, 2, 3, … with a(n) = 1 / n^2. First terms: 1, 0.25, 0.111111, 0.0625, …
$\sum_{n=1}^{\infty} \frac{1}{{n}^{2}}$
nth-term test
lim a(n) = 0, so the divergence test is inconclusive; the terms do shrink, so a finer test is needed.
$\lim_{n\to\infty} \frac{1}{{n}^{2}} = 0$
Limit comparison with a p-series
Positive terms: the terms shrink roughly like 1/n^2 (estimated from a(2N)/a(N)), so compare with Σ 1/n^2: lim a(n)·n^2 = 1. The limit is a positive finite number, so the series behaves exactly like the p-series Σ 1/n^2, which converges because p > 1.
$\lim_{n\to\infty} \frac{a_n}{1/n^{2}} = 1$
Estimate the sum numerically
Partial sums S_N were computed for several N and extrapolated to N → ∞ (Richardson extrapolation; alternating terms are averaged first). Two independent extrapolations agree to better than 10⁻¹⁰, giving S ≈ 1.644934067. Any closed form shown is a numerical identification, not a proof.

Formulas used

nth-term (divergence) test
$\lim_{n\to\infty} a_n \ne 0 \Rightarrow \sum a_n \text{ diverges}$
p-series
$\sum \frac{1}{n^p} \text{ converges} \iff p > 1$

A series converges when its partial sums approach a finite number. The terms must tend to 0 for that to happen, but that alone is not enough (the harmonic series diverges), so a comparison, ratio, root, alternating or integral test is used to decide.

Verification

Partial sums (supporting evidence, not a proof): S₁₀ = 1.5497677, S₁₀₀ = 1.6349839, S₁₀₀₀ = 1.6439346, S₁₀₀₀₀ = 1.6448341

Detailed explanation

Start with the quick tests

Two checks should always come first because they are fast. If the series is geometric, the ratio $r$ decides everything: it converges when $|r| < 1$ and you get the sum immediately. For $\sum_{n=1}^{\infty} 3\left(-\frac23\right)^n$ the first term is $a = -2$ and $r = -\frac23$ , so the sum is $\frac{-2}{1 + 2/3} = -\frac65$ .

Next, the nth-term test. If the terms do not approach zero, the series must diverge, because you would be adding quantities that stay a fixed size forever. $\sum \frac{n}{n+1}$ diverges this way, since its terms approach 1. Be careful with the converse: terms that approach zero do not guarantee convergence, as the harmonic series shows.

Ratio and root tests

The ratio test compares each term with the one before it. If the ratio $\left|\frac{a_{n+1}}{a_n}\right|$ approaches a limit $L < 1$ , the terms eventually shrink at least as fast as a convergent geometric series, so the series converges. If $L > 1$ the terms grow and the series diverges. If $L = 1$ the test is inconclusive.

It is the natural choice for factorials and exponentials. For $\sum \frac{2^n}{n!}$ , the ratio simplifies to $\frac{2}{n+1}$ , which tends to 0, so the series converges. The root test, based on $\sqrt[n]{|a_n|}$ , is similar and works best when the whole term is raised to the nth power. Both tests always give $L = 1$ for p-series and rational functions of $n$ , so other tools are needed there.

Comparison with p-series

For terms that behave like a power of $n$ , compare with $\sum \frac{1}{n^p}$ , which converges exactly when $p > 1$ . The limit comparison test says that if $\frac{a_n}{b_n}$ approaches a positive finite number, the two series share the same fate. The calculator estimates the right $p$ from how the terms change when $n$ doubles, then computes that limit exactly.

So $\sum \frac{1}{\sqrt{n}}$ diverges by comparison with $\sum \frac1n$ , while $\sum \frac{\sin n}{n^2}$ converges absolutely, because $|\sin n| \le 1$ means its terms are bounded by $\frac{1}{n^2}$ .

Logarithms: condensation and the integral test

Series with logarithms, such as $\sum_{n=2}^{\infty} \frac{1}{n \ln n}$ , sit right on the boundary where ratio, root and simple comparison tests fail. Cauchy condensation handles them: for positive decreasing terms, $\sum a_n$ converges exactly when $\sum 2^n a_{2^n}$ does. Here the condensed series has terms of about $\frac{1.4427}{n}$ , a multiple of the harmonic series, so the original series diverges.

The integral test reaches the same kind of conclusion by comparing the sum with $\int f(x)\,dx$ , where $f(n) = a_n$ is positive and decreasing. Both techniques show how slowly a series can diverge: the partial sums of $\sum \frac{1}{n \ln n}$ grow like $\ln(\ln N)$ , which is below 4 even after a billion terms.

Alternating series and conditional convergence

When signs alternate, cancellation helps. The alternating series test says that if the magnitudes $b_n$ decrease to zero, the series converges. The alternating harmonic series $1 - \frac12 + \frac13 - \cdots$ converges by this test, to $\ln 2$ , even though $\sum \frac1n$ diverges.

That combination is called conditional convergence. If the series of absolute values also converges, as for $\sum \frac{(-1)^n}{n^2}$ , the convergence is absolute, which is stronger: absolutely convergent series can be rearranged without changing the sum, while conditionally convergent ones cannot. The calculator reports which kind you have.

Common mistakes

Concluding that a series converges because its terms approach zero.
Treating $L = 1$ in the ratio or root test as a verdict; it means the test is inconclusive.
Using the wrong first term in a geometric sum when the index starts at 1 instead of 0.
Applying the alternating series test without checking that the magnitudes decrease.
Confusing convergence of the sequence $a_n$ with convergence of the series $\sum a_n$ .
Comparing in the wrong direction, such as bounding a series above by a divergent one.

Tips

Run the nth-term test first; it settles many divergent series instantly.
Reach for the ratio test whenever you see factorials or $c^n$ .
For rational functions of $n$ , compare the degrees: the series behaves like $\sum \frac{1}{n^{q-p}}$ .
Use condensation or the integral test when logarithms appear.
For alternating series, test absolute convergence too, so you can say which kind it is.
Write out the first few partial sums numerically to get a feel for whether they are settling down.

Frequently asked questions

Which convergence test does the calculator use?

It tries them in order: geometric series, nth-term test, alternating series test, then for positive terms the ratio test, root test, limit comparison with a p-series, Cauchy condensation and the integral test. The first decisive test is reported.

Can it find the sum of the series?

It gives the exact sum for geometric series, such as −6/5 for Σ 3(−2/3)ⁿ from n = 1. For other series it decides convergence only, because most convergent series have no simple closed-form sum.

Why does the harmonic series diverge if its terms go to zero?

The terms shrink too slowly. Grouping them as 1/2, then 1/3 + 1/4 > 1/2, then 1/5 + … + 1/8 > 1/2 and so on shows the partial sums keep gaining at least 1/2, so they grow without bound.

What is the difference between absolute and conditional convergence?

A series converges absolutely if the sum of |aₙ| converges. It converges conditionally if it converges only because of sign cancellation, like the alternating harmonic series.

How do I type factorials and alternating signs?

Write n! for a factorial and (-1)^n or (-1)^(n+1) for alternating signs, for example (-1)^(n+1)/n or 2^n/n!. Use log(n) for the natural logarithm.

Does the start index affect convergence?

No. Changing where the sum starts adds or removes finitely many terms, which changes the sum but never whether it converges. It does matter for a geometric sum, and for terms such as 1/(n ln n) that are undefined at small n.

What happens if no test works?

The calculator lists the tests it tried and states that none was decisive. It never guesses. Oscillating terms like sin(n), whose limit cannot be determined, are also reported plainly.

What is a p-series?

A series of the form Σ 1/nᵖ. It converges when p > 1 and diverges when p ≤ 1. It is the standard benchmark for comparison tests, and Σ 1/n with p = 1 is the harmonic series.

Why does the ratio test fail for 1/n²?

The ratio n²/(n + 1)² approaches 1, which is the inconclusive case. The same happens for every p-series, which is why the calculator moves on to comparison tests for such terms.

Test your series for convergence

Enter the general term above to see which test decides it, every limit involved and the exact sum for geometric series.

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