A p-series, or hyperharmonic series, is a positive-term series (a series Σan such that an>0 for every n) in the form of:
Σ 1/n p = 1 + 1/2 p + . . . + 1/n p + . . .,
where p is a positive real number.
(If p = 1, then it is the harmonic series.)
Σ 1/n p
- 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 =
lim ∫ (x(-p))dx =
If p > 1, then p - 1 > 0 and the last expression can be written as:
(1/1-p)lim(1/(t(p-1)) - 1) =
(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) = ∞.
Thus, the p-series diverges. If p ≤ 0, then
lim(1/n p)!= 0
and, by the nth-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).
Σ 1/n 2 ........ Converges since p = 2 > 1
Σ 1/√(n) ........ Diverges since p = 1/2 > 1