Series Convergence Calculator

How Series Convergence Works

An infinite series is an expression of the form Σ a_n, where you add infinitely many terms from a sequence. Since you cannot literally finish infinitely many additions, the definition of convergence uses partial sums. If the partial sums approach a finite number as you keep adding more and more terms, the series is said to converge. If the partial sums do not settle to a finite limit, the series diverges.

A partial sum is the sum of the first N terms: S_N = a₁ + a₂ + … + a_N. A series converges when S_N approaches a finite value as N grows. This idea is simple, but the behavior of series can be surprisingly subtle. Some series with tiny terms still diverge, while other series with alternating signs converge even when the absolute values do not.

The First Check: The Term Test

Before applying advanced tests, the most important necessary condition is that the terms must approach zero: if lim a_n ≠ 0 or the limit does not exist, then the series must diverge. This is sometimes called the divergence test or the nth-term test.

However, the reverse is not true. Even if lim a_n = 0, the series might still diverge. The harmonic series Σ 1/n is the classic example: terms go to zero, but partial sums grow without bound. That is why convergence testing needs more structure than checking the term limit alone.

Geometric Series

A geometric series has the pattern a, ar, ar², ar³, and so on. It is one of the most important families because it has a clean convergence rule and often appears after algebraic manipulation of more complicated terms.

When it converges, the infinite sum has a closed form. If you start at n = 0, the sum is a/(1 − r). If you start at n = 1, the sum becomes ar/(1 − r) or equivalently a₁/(1 − r) when a₁ is the first term. The Geometric tab in this tool shows the convergence decision, an infinite-sum value when applicable, and a partial-sum table.

p-Series and Power-Law Decay

A p-series is a benchmark for many comparisons: Σ 1/n^p. It converges only when terms decay faster than 1/n, meaning the exponent p must be greater than 1.

Multiplying by a constant k does not change convergence; it only scales the partial sums. Many real problems reduce to “how fast” a term decays, and p-series provide the measuring stick. If your terms behave like 1/n², you should expect convergence. If your terms behave like 1/n or 1/√n, you should expect divergence.

Alternating Series and Conditional Convergence

Alternating series flip sign from term to term, typically using (−1)ⁿ or (−1)ⁿ⁺¹. Alternation can force partial sums to “zig-zag” toward a limit even when the absolute values decay too slowly to converge.

The alternating series test says that Σ (−1)ⁿb_n converges if b_n is decreasing and b_n → 0. A common choice is b_n = 1/n^p. In that case, any p > 0 gives decreasing terms approaching zero, so the alternating series converges.

The next question is whether the series converges absolutely. If Σ |a_n| converges, the original series is absolutely convergent. If Σ |a_n| diverges but the alternating series still converges, then it is conditionally convergent.

For alternating p-style terms, absolute convergence happens only when p > 1. If 0 < p ≤ 1, the alternating series converges conditionally: it converges because of sign cancellation, but the magnitude-only series diverges.

Ratio and Root Tests

When terms contain factorials, exponentials, powers, or products, the ratio and root tests are often the fastest route. Instead of comparing to a benchmark series directly, these tests estimate how quickly terms shrink by measuring the “effective ratio” between successive terms or the “effective base” of exponential decay.

The important word is “inconclusive.” Some series sit exactly on the boundary L = 1, where the tests cannot decide. Many classic examples do this, including Σ 1/n and Σ 1/n². For those cases, you use a different test that captures polynomial decay rather than exponential decay.

Log-Adjusted p-Series and Integral-Style Rules

Some series are “almost” p-series, with an extra logarithm factor such as Σ 1/(n(ln n)^q). These are central in analysis and appear in algorithm complexity, probability tail bounds, and number theory.

A widely used classification is:

• If p > 1, Σ 1/(n^p(ln n)^q) converges for any real q.
• If p < 1, it diverges for any real q.
• If p = 1, it converges only when q > 1 and diverges when q ≤ 1.

This behavior mirrors an integral comparison: the logarithm factor is not strong enough to rescue the p < 1 case, but it can be strong enough to turn the borderline p = 1 case into a convergent series when the log power is large enough.

Telescoping Series

A telescoping series is one where most terms cancel after rewriting each term as a difference of simpler fractions. The partial sums collapse to a small expression involving only a few remaining boundary terms.

A common telescoping pattern is: 1/(n(n+k)) = (1/k)(1/n − 1/(n+k)). When you sum this from n = 1 to N, the 1/(n+k) terms cancel with later 1/n terms, leaving only a fixed finite remainder.

The Telescoping tab in this calculator uses the family c/(n(n+k)) because it both telescopes cleanly and gives a stable infinite sum that depends on the harmonic number H_k. For k fixed, the infinite sum equals c·H_k/k.

Why Partial Sums Are Still Useful

Even when you know whether a series converges, you often still want a numeric approximation to the sum. Many convergent series do not have an elementary closed form, and even when a closed form exists it may not be convenient for estimation. Partial sums show how quickly convergence happens and how large the truncation error might be.

Partial sums also reveal divergence patterns. If a series diverges slowly, the first few partial sums might look stable, which can be misleading. A table makes it easier to see the trend as N grows. This tool builds a table for each tab and offers CSV export so you can analyze the sequence in a spreadsheet.

Choosing the Right Convergence Test

Different tests detect different kinds of decay. A practical way to choose is to look at the structure of a_n.

• If terms involve a constant ratio from one term to the next, start with the geometric rule.
• If terms look like 1/n^p, use the p-series benchmark.
• If signs alternate, test alternating behavior first, then check absolute convergence separately.
• If you see factorials, exponentials, or powers like cⁿ, the ratio or root test is often best.
• If you see n and ln n together near a boundary case, consider the log-adjusted rule.
• If terms can be decomposed into differences like 1/n − 1/(n+k), look for telescoping cancellation.

Interpreting “Absolute” vs “Conditional” Results

Absolute convergence is stronger than ordinary convergence. If a series converges absolutely, rearranging terms does not change its sum, and numerical approximations are generally more stable. Conditional convergence is more delicate: sign cancellation is essential, and rearrangements can change the value or even force divergence. In applications, the difference matters in error analysis, numerical computation, and series manipulation.

Limitations and Assumptions

This tool focuses on common families where convergence rules are well-known and partial sums can be computed directly from parameters. It does not parse arbitrary symbolic formulas for a_n. For a fully general series given by a complicated expression, you typically apply a theoretical test by hand (ratio, comparison, integral, condensation, and so on), then use numeric partial sums only as supportive intuition.

Also remember that partial sums are not proofs. A convergent series can converge very slowly, and a divergent series can grow very slowly too. Always use a convergence test result as the primary decision, and treat partial sums as a diagnostic.

Practical Examples You Can Model Here

• Geometric: Σ 3(0.2)ⁿ converges quickly and has a closed-form sum.
• p-series: Σ 5/n^1.2 converges slowly but does converge because p > 1.
• Alternating: Σ (−1)ⁿ⁺¹/n converges conditionally (the alternating harmonic series).
• Ratio input: if you compute L = 0.6 from your algebra, the series is absolutely convergent.
• Log-adjusted: Σ 1/(n(ln n)²) converges, but Σ 1/(n ln n) diverges.
• Telescoping: Σ 1/(n(n+1)) converges and sums to 1.