Why does the p-Series Test require p > 1 and not p ≥ 1?

The value p = 1 is the exact boundary between convergence and divergence. For p = 1, you get the harmonic series \sum 1/n, which diverges slowly but definitely. For any p even slightly larger than 1, the terms shrink fast enough that the sum converges. This sharp transition at p = 1 is a striking feature of the p-Series family.

Can I use the p-Series Test on a series like sum of 1/(n^2 + 1)?

No, that series doesn't fit the form \sum 1/n^p. However, for large n, the terms 1/(n^2 + 1) behave like 1/n^2, so you can use the Ratio Test or comparison tests to show it converges by analogy.

What happens to the p-Series when p is negative?

If p is negative, then 1/n^p = n^{|p|} grows without bound as n increases, so the series diverges immediately. The p-Series Test only applies when the exponent is positive, and convergence only occurs when p > 1.

Does the p-Series Test work for alternating series like sum of (-1)^n / n^p?

No, the p-Series Test applies only to positive terms. For alternating series of that form, use the Alternating Series Test instead, which has a different convergence criterion.

p-Series Test

The p-Series Test determines convergence of series 1/n^p: the series converges if and only if the exponent p is strictly greater than 1.

\sum_{n=1}^{\infty} \frac{1}{n^p} \text{ converges} \iff p > 1

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What each symbol means

What p-Series Test takes

Symbol	Meaning
$n$	The positive integer index that counts each term in the series, starting from 1. If you treat n as a fixed constant rather than a variable, you lose the summation and the series loses its meaning.
$p$	The exponent applied to n in the denominator; its value is the key to convergence. If you misread p as a constant coefficient instead of an exponent, you will misapply the convergence criterion.

When to use it

When you encounter a series of the form sum of 1 over n to the p and need to determine whether it converges or diverges.

Level

Usually taught in: Calculus II

Worked examples

1. Recognize and apply the p-Series Test to a clean example

Problem

Determine whether the series

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

converges or diverges.

$p = 2$
Identify the exponent: the series is $\sum \frac{1}{n^p}$ with $p = 2$ .
$p > 1 \text{ for convergence}$
The p-Series Test says $\sum \frac{1}{n^p}$ converges if and only if $p > 1$ .
$2 > 1$
Check: is $2 > 1$ ? Yes, so the series converges.

Answer: $\text{Converges}$

This is a p-Series where we can immediately read off the exponent. Since $p = 2 > 1$ , the test tells us the series converges.

2. Apply the test when p is a fraction and the series diverges

Problem

Determine whether the series

\sum_{n=1}^{\infty} \frac{3}{n^{1/2}}

converges or diverges.

$\sum_{n=1}^{\infty} \frac{3}{n^{1/2}} = 3 \sum_{n=1}^{\infty} \frac{1}{n^{1/2}}$
Factor out the constant 3: constant factors don't affect convergence behavior.
$p = \frac{1}{2}$
Identify the exponent: $p = 1/2$ .
$\frac{1}{2} < 1$
Since $p = 1/2 < 1$ , the series diverges by the p-Series Test.

Answer: $\text{Diverges}$

The constant coefficient doesn't change convergence. The critical observation is that $p = 1/2 < 1$ , which means the terms shrink too slowly for the series to converge.

3. Apply p-Series Test to a real-world tier cost scenario

Problem

A streaming service charges a fixed monthly base cost plus additional tier costs. The

n

th tier costs

\frac{1}{n^2}

dollars per month. If the service must support infinitely many tiers, does the total additional tier cost converge?

$\text{Additional tier costs: } \sum_{n=1}^{\infty} \frac{1}{n^2}$
Model the variable tier costs as a series: each tier contributes $1/n^2$ to the total.
$p = 2$
This is a p-Series with exponent $p = 2$ .
$2 > 1 \therefore \text{series converges}$
The p-Series Test: since $2 > 1$ , the infinite sum of tier costs reaches a finite limit.

Answer: $\text{Yes, the tier cost converges to a finite total}$

The base monthly fee is constant and separate; only the variable tier costs matter for convergence. The p-Series Test shows that tier costs shrink fast enough (as $1/n^2$ ) that their sum remains bounded as the number of tiers grows indefinitely.

Common mistakes

Where p-Series Test usually goes wrong

The mistake	Why it is wrong	The fix
The series $\sum_{n=1}^{\infty} \frac{1}{n}$ converges because $p = 1$ .	The p-Series Test requires $p > 1$ for convergence, not $p \geq 1$ . The value $p = 1$ is the boundary case where the series (the harmonic series) diverges.	When $p = 1$ , the series diverges. The critical threshold is $p > 1$ , not $p = 1$ .
The series $\sum_{n=1}^{\infty} \frac{5}{n^2}$ diverges because there is a constant 5 in the numerator.	Confusing the test with other rules or thinking that a nonzero coefficient changes convergence behavior.	Constant multiples don't affect convergence: $\sum \frac{5}{n^2}$ has the same convergence behavior as $\sum \frac{1}{n^2}$ . This is still a p-Series with $p = 2 > 1$ , so it converges.
The series $\sum_{n=1}^{\infty} \frac{1}{\sqrt[3]{n}}$ diverges because a cube root is slower than a square root.	Confusing the comparison of roots with the p-Series criterion, or misidentifying the exponent.	Rewrite $\sqrt[3]{n} = n^{1/3}$ to identify $p = 1/3 < 1$ . Since $p < 1$ , the series diverges—but the reason is the exponent, not the type of root.

Tips and when to use something else

The p-Series Test gives a clean yes-or-no answer, so use it first when you recognize the form $\sum \frac{1}{n^p}$ —it is faster than other tests.
Remember that $p = 1$ is the critical boundary: $p = 1$ means divergence (the harmonic series), while any $p > 1$ means convergence.
Constant multiples and finitely many extra terms don't affect convergence: $\sum \frac{c}{n^p}$ and $\sum \frac{1}{n^p}$ behave the same way for any $c \neq 0$ .
If the series has extra terms or isn't exactly $\sum \frac{1}{n^p}$ , switch to the Integral Test or Ratio Test for a broader range of series.

Related formulas

Harmonic SeriesThe Harmonic Series, sum of reciprocals of positive integers, diverges infinitely—a foundational benchmark for determining whether other series converge.Integral TestTest series convergence by checking improper integral ∫₁^∞ f(x)dx when f is positive, continuous, and decreasing; the series ∑f(n) shares convergence behavior.Ratio TestDetermine whether an infinite series converges by checking if the ratio of consecutive terms approaches a value less than one.Alternating Series TestThe Alternating Series Test determines whether a series with alternating signs converges when its terms decrease monotonically to zero.

Frequently asked questions

Why does the p-Series Test require p > 1 and not p ≥ 1?: The value $p = 1$ is the exact boundary between convergence and divergence. For $p = 1$ , you get the harmonic series $\sum 1/n$ , which diverges slowly but definitely. For any $p$ even slightly larger than 1, the terms shrink fast enough that the sum converges. This sharp transition at $p = 1$ is a striking feature of the p-Series family.
Can I use the p-Series Test on a series like sum of 1/(n^2 + 1)?: No, that series doesn't fit the form $\sum 1/n^p$ . However, for large $n$ , the terms $1/(n^2 + 1)$ behave like $1/n^2$ , so you can use the Ratio Test or comparison tests to show it converges by analogy.
What happens to the p-Series when p is negative?: If $p$ is negative, then $1/n^p = n^{|p|}$ grows without bound as $n$ increases, so the series diverges immediately. The p-Series Test only applies when the exponent is positive, and convergence only occurs when $p > 1$ .
Does the p-Series Test work for alternating series like sum of (-1)^n / n^p?: No, the p-Series Test applies only to positive terms. For alternating series of that form, use the Alternating Series Test instead, which has a different convergence criterion.

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Reviewed 2026-09-18