A `p`-series, or hyperharmonic series, is a positive-term series (a series Σ`a`_{n} such that `a`_{n}>0 for every `n`) in the form of:

∞

Σ 1/`n`^{ p} = 1 + 1/2^{ p} + . . . + 1/`n`^{ p} + . . .,

`n`=1

where `p` is a positive real number.

(If `p` = 1, then it is the harmonic series.)

Theorem:

The `p`-series

∞

Σ 1/`n`^{ p}

`n`=1

- converges if
`p` > 1
- diverges if
`p` ≤ 1 (but remember; p must also be greater than zero, or it would be 1/`n`^{ -p} = `n`^{ p})

Proof: Using the integral test (all of the conditions are satisfied; f is positive for `x` ≥ 1 and it is decreasing)

∞

∫ (1/`x`^{ p})`dx` =

1

`t`
lim ∫ (`x`^{(-p)})`dx` =
(`t`→∞) 1

(1/1-`p`)lim(`t`^{(1-p)} -1).
(`t`→∞)

If `p` > 1, then `p` - 1 > 0 and the last expression can be written as:

(1/1-`p`)lim(1/(`t`^{(p-1)}) - 1) =
(`t`→∞)
(1/1-`p`)(0-1) = 1/`p`-1.

Thus, the `p`-series converges if `p` > 1. If 0 < `p` < 1, then 1 - `p` > 0 and

(1/1-`p`)lim (`t`^{(1-p)} - 1) = ∞.
(`t`→∞)
Thus, the `p`-series diverges. If `p` ≤ 0, then
lim(1/`n`^{ p})!= 0
(`n`→∞)
and, by the `n`th-term test, the series diverges(see the above parenthetic on 1/`n`^{-p}).

Application for the `p`-series theorem is to determine if the infinite series (adding each term of the infinite sequence) converges

or diverges. If a series converges, it means that the series has a sum. If a series diverges, it means that the series will add

to infinity, thus, it does not have a sum. An example I like to explain this with is:

consider the series: .3 + .03 + .003 + .0003 ...(to infinity.) This infinite series will go to the sum of 1/3 (.3333333 repeating.)

Although this is not a `p`-series, it helps to understand adding numbers to infinity. But if this was a `p`-series, we could have

determined whether it would converge before we found the actual sum (though, the sum of some series is difficult to find).

Examples:

∞

Σ 1/`n`^{ 2} ........ Converges since `p` = 2 > 1

`n`=1

∞

Σ 1/`√(n)` ........ Diverges since `p` = 1/2 > 1

`n`=1