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p-series — Definition, Formula & Convergence

p-series

A series of the form 11p+12p+13p+⋯+1np+⋯\displaystyle \frac{1}{1^p} + \frac{1}{2^p} + \frac{1}{3^p} + \cdots + \frac{1}{n^p} + \cdots or ∑n=1∞1np,\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^p}\text{,} where p > 0. Often employed when using the comparison test and the limit comparison test.

Note: The harmonic series is a p-series with p =1.

p-series: ∑n=1∞1np=11p+12p+13p+⋯+1np+⋯\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^p} = \frac{1}{1^p} + \frac{1}{2^p} + \frac{1}{3^p} + \cdots + \frac{1}{n^p} + \cdots

If 0<p≤1,0 \lt p \le 1\text{,} the series diverges.
If p>1,p \gt 1\text{,} the series converges.

See also

Sigma notation, divergent series, convergent series

Key Formula

∑n=1∞1np=11p+12p+13p+⋯\sum_{n=1}^{\infty} \frac{1}{n^p} = \frac{1}{1^p} + \frac{1}{2^p} + \frac{1}{3^p} + \cdots
Where:
  • nn = The index of summation, starting at 1 and increasing through all positive integers
  • pp = A positive constant that determines convergence: the series converges if p > 1 and diverges if 0 < p ≤ 1

Worked Example

Problem: Determine whether the p-series ∑n=1∞1n3\sum_{n=1}^{\infty} \frac{1}{n^3} converges or diverges.
Step 1: Identify the series as a p-series and find the value of p.
∑n=1∞1n3  ⟹  p=3\sum_{n=1}^{\infty} \frac{1}{n^3} \implies p = 3
Step 2: Apply the p-series convergence rule: a p-series converges if and only if p > 1.
p=3>1p = 3 > 1
Step 3: Since p = 3 is greater than 1, the condition for convergence is satisfied.
∑n=1∞1n3 converges\sum_{n=1}^{\infty} \frac{1}{n^3} \text{ converges}
Answer: The series ∑n=1∞1n3\sum_{n=1}^{\infty} \frac{1}{n^3} converges because p=3>1p = 3 > 1.

Another Example

This example shows how a p-series is commonly used as a benchmark in the comparison test rather than being tested directly.

Problem: Use the p-series test and the comparison test to determine whether ∑n=1∞1n2+5\sum_{n=1}^{\infty} \frac{1}{n^2 + 5} converges or diverges.
Step 1: Notice that this is NOT itself a p-series because the denominator is n2+5n^2 + 5, not npn^p. However, for large n it behaves like 1n2\frac{1}{n^2}, which is a p-series with p=2p = 2.
1n2+5vs.1n2\frac{1}{n^2 + 5} \quad \text{vs.} \quad \frac{1}{n^2}
Step 2: Since n2+5>n2n^2 + 5 > n^2 for all n≥1n \geq 1, the given terms are smaller than the corresponding p-series terms.
0<1n2+5<1n2for all n≥10 < \frac{1}{n^2 + 5} < \frac{1}{n^2} \quad \text{for all } n \geq 1
Step 3: Confirm that the comparison series ∑1n2\sum \frac{1}{n^2} converges by the p-series test (p=2>1p = 2 > 1).
∑n=1∞1n2 converges (p=2>1)\sum_{n=1}^{\infty} \frac{1}{n^2} \text{ converges } (p = 2 > 1)
Step 4: By the direct comparison test, since each term of our series is positive and less than the corresponding term of a convergent series, our series also converges.
∑n=1∞1n2+5 converges by comparison\sum_{n=1}^{\infty} \frac{1}{n^2 + 5} \text{ converges by comparison}
Answer: The series ∑n=1∞1n2+5\sum_{n=1}^{\infty} \frac{1}{n^2 + 5} converges by the comparison test with the convergent p-series ∑1n2\sum \frac{1}{n^2}.

Frequently Asked Questions

What is the difference between a p-series and a geometric series?
A p-series has the form 1np\frac{1}{n^p} where the variable nn is in the base of the denominator. A geometric series has the form arnar^n where the variable nn is in the exponent. A geometric series converges when ∣r∣<1|r| < 1, while a p-series converges when p>1p > 1. They are tested with entirely different convergence criteria.
Why does the p-series diverge when p = 1?
When p=1p = 1, the p-series becomes the harmonic series ∑1n=1+12+13+⋯\sum \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots, which diverges. Even though the individual terms approach zero, they decrease too slowly for the partial sums to remain bounded. This can be proven using the integral test: ∫1∞1x dx=ln⁡x∣1∞=∞\int_1^{\infty} \frac{1}{x}\,dx = \ln x \big|_1^{\infty} = \infty.
How do you prove the p-series convergence rule?
The standard proof uses the integral test. You compare ∑1np\sum \frac{1}{n^p} with the improper integral ∫1∞1xp dx\int_1^{\infty} \frac{1}{x^p}\,dx. When p>1p > 1, this integral equals 1p−1\frac{1}{p-1}, which is finite, so the series converges. When p≤1p \leq 1, the integral diverges, so the series diverges.

p-Series vs. Geometric Series

p-SeriesGeometric Series
General form∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p}∑n=0∞arn\sum_{n=0}^{\infty} ar^n
Variable positionn is in the base of the denominatorn is in the exponent
Convergence conditionConverges when p>1p > 1Converges when ∣r∣<1|r| < 1
Exact sum known?Only for specific values of p (e.g., p=2p=2 gives π2/6\pi^2/6)Yes: sum =a1−r= \frac{a}{1-r}
Common useBenchmark for comparison and limit comparison testsDirect evaluation of sums; ratio test benchmark

Why It Matters

The p-series is one of the most important benchmark series in calculus. When you need to determine whether an unfamiliar series converges, you will frequently compare it to a p-series using the comparison test or the limit comparison test. Mastering the simple rule — converges if p>1p > 1, diverges if p≤1p \leq 1 — gives you a powerful tool for quickly resolving many series convergence problems on AP Calculus and college calculus exams.

Common Mistakes

Mistake: Thinking the p-series converges when p = 1 because the terms 1n\frac{1}{n} approach zero.
Correction: Terms approaching zero is necessary but not sufficient for convergence. The harmonic series (p=1p = 1) is the classic example of a divergent series whose terms tend to zero. You must have p>1p > 1 strictly.
Mistake: Confusing a p-series with a geometric series by misidentifying where the variable n appears.
Correction: In a p-series, nn is the base raised to a constant power: 1np\frac{1}{n^p}. In a geometric series, a constant base is raised to the variable power: rnr^n. Check whether the variable is in the base or the exponent before choosing your test.

Related Terms