P Series Test Calculator
p-Series Test Calculator
Slug: `/p-series-test-calculator/`
Primary keyword: p series test calculator
Meta title: p-Series Test Calculator: Converges Only When p Exceeds 1
Meta description: Test any p-series instantly. Enter p, get the verdict, and see why the harmonic series at p equal to 1 diverges while p equal to 1.01 converges.
Word count: approximately 1,000
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The cleanest result in the subject
For the series of one over n to the power p, convergence happens if and only if p is greater than 1.
No limits to evaluate. No cases. Read off p, compare it to 1, done.
The boundary is razor thin
Set p equal to 1 and you have the harmonic series, which diverges. Set p equal to 1.01 and it converges. Nothing about the terms looks meaningfully different, and yet the behaviour flips completely.
At p equal to 1, partial sums grow like the natural logarithm of n. Logarithmic growth is slow enough to look like flattening. Add two hundred thousand terms of the harmonic series and you reach about 12.78. Add two hundred thousand terms of one over n to the 1.01 and you get a number that genuinely is approaching a limit, though it takes a long while to look like it.
That similarity is why numerical evidence cannot settle p-series. Only the integral test can.
Values worth memorising
| p | Verdict | Sum |
|---|---|---|
| --- | --- | --- |
| 0.5 | diverges | unbounded |
| 1 | diverges | unbounded, grows like log n |
| 1.5 | converges | about 2.612 |
| 2 | converges | pi squared over six, about 1.644934 |
| 3 | converges | about 1.202057, no elementary form |
| 4 | converges | pi to the fourth over ninety, about 1.082323 |
Two entries there deserve comment. The p equal to 2 case is the Basel problem, solved by Euler in 1735. The p equal to 3 case is Apery's constant, proved irrational only in 1978, and still has no known closed form.
So convergence is easy to establish and the actual sum can remain out of reach for centuries.
Why the integral test settles it
Suppose f is positive, continuous and decreasing. Then the series of f(n) and the integral of f from 1 to infinity share the same fate.
Integrate one over x to the p. For p greater than 1 the antiderivative approaches a finite value, giving convergence. For p equal to 1 the antiderivative is the natural logarithm, which grows without bound. For p under 1 the integral also diverges.
That single computation produces the entire p-series rule, and it is the reason the rule has a strict inequality at 1.
Terms that hide a p-series
Exam questions rarely present a clean p-series. They disguise one.
- One over the square root of n is a p-series with p equal to 0.5, so it diverges.
- n over n cubed simplifies to one over n squared, so it converges.
- One over n to the two thirds has p under 1, so it diverges.
- n over n squared plus seven is not a p-series, but limit comparison against one over n shows it diverges.
Simplify first. Many series that look unfamiliar reduce to a power of n in two lines.
A common trap
The series of one over n log n is not a p-series, and it diverges, which surprises people because it shrinks faster than the harmonic series.
Its integral involves the log of a log, which grows without bound, just extremely slowly. Meanwhile one over n times log n squared converges. Neither result follows from the p-series rule, and both need the integral test.
Quick reference
If the term is a power of n, the p-series test finishes it. If it merely resembles a power of n, use limit comparison against the nearest p-series. If a logarithm appears, go to the integral test.
Questions
Does p have to be an integer?
No. Any real p works, and the rule is unchanged.
What happens at p equal to 1 exactly?
Divergence. This is the harmonic series, the standard counterexample to the idea that shrinking terms guarantee convergence.
Why does p equal to 2 give pi squared over six?
Euler proved it in 1735 by an argument about the roots of the sine function. The appearance of pi in a sum of reciprocal squares is one of the more striking results in mathematics.
Can the p-series test handle alternating signs?
Not directly. Use the alternating series test, which gives convergence for any p above 0.
Is there a p-series with a nice sum for odd p?
None is known for p equal to 3 or any larger odd number.