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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition) > Unit 7

Lesson 8: Basic convergence tests# Worked example: p-series

p-series are infinite sums Σ(1/xᵖ) for some positive p. In this video you will see examples of identifying whether a p-series converges or diverges.

## Want to join the conversation?

- Why can't the variable p be a negative value? If it is negative, wouldn't that mean the overall number would be diverging because a negative exponent in the denominator will create a whole number?(9 votes)
- Well if the number is negative at the bottom it isn't considered a p-series problem at that point.(6 votes)

- Is there a way to find the sum of a p-series? For example, I believe this is a p-series and since it converges, I need to find a sum.

Σ 1/n^2 + 5n + 6(5 votes)- Very good question. At https://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/SeriesTests/p-series.html

it is written that "Unfortunately, there is no simple theorem to give us the sum of a p-series. For instance, the sum of the

1 + 1/4 + 1/9 + ... series is (pi^2)/6".

Also, you might find interesting the Euler - Mascheroni constant, one of the numbers that we are not sure whether they are rational or irrational:

https://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant(2 votes)

- what about a p-series that is negative on all terms?(2 votes)
- If all the terms of the p-series are negative, then the series is simply equal to -1 times the same series of positive terms.(1 vote)

- For P<0. it is logical that series diverges, but why not include those values in this explanation? Is there a deeper reason for that?(2 votes)
- Is that the Riemann Zeta function? I mean of course zeta(s)=sum of 1/n^s from s=1 all the way to infinity. So zeta is p-series?

What is the Todd function? I heard that RH had been proved using something called the Todd function.(1 vote)- If the input into the Riemann Zeta function is a complex number with real part >1, then yes, the Zeta function at that point is given as a p-series. If the real part is ≤1, the sum wouldn't converge, so we define the Zeta function another way for all other inputs.

On Monday, (9/24/2018), prominent mathematician Michael Atiyah presented what he claimed was a proof of the Riemann Hypothesis. What he presented was not a proof, and was not anything that might help guide us to a proof.

In the presentation, he used a function called the Todd function, which he defined elsewhere, in another paper he hasn't published. The 'proof' he presented was only four lines, wherein he didn't even use any properties of the Zeta function.(2 votes)

- I actually have several conceptual questions: what is harmonic series and what is p-series? Is p-series just 1/x^p?(1 vote)
- p-series: Σ(1/xᵖ), where p > 0.

General harmonic series: Σ(1/(ax + b)), where a > 0 and b >= 0. Note that it's not a p-series.

Harmonic series: Both a type of p-series with p = 1, and a general harmonic series with a = 1 and b = 0. Written as Σ(1/x). It's useful to know that the harmonic series diverges.

Let me know if this helps.(2 votes)

- Would it still be considered a p-series if it was -(1/n) instead of 1/n?(1 vote)
- At1:23, why the series will be divergent when p=1?(1 vote)
- The sum of 1/n for all n > 0 (i.e. the harmonic series) is known to diverge. One way to prove this is with the integral test (a monotonically decreasing series converges if and only if the integral of the function converges). The integral of 1/n is ln(n) which diverges as n approaches infinity. Therefore, the harmonic series must also be divergent.(1 vote)

- How can you solve alternating series using p-series? Eg:(-1)^n*n(1 vote)
- If these series diverge in this case, will it have an asymptote with x axis?(1 vote)

## Video transcript

- [Instructor] So we have
an infinite series here, one plus one over two to the fifth, plus one over three to the fifth, and we just keep on going forever. We could write this as the sum from n equals one to infinity of one over n to the fifth power, one over n to the fifth power. And now you might recognize,
notice, when n is equal to one, this is one over one to the fifth, that's that over there,
and we could keep on going. Now you might immediately
recognize this as a p-series, and a p-series has the
general form of the sum, going from n equals one to infinity, of one over n to the p, where p is a positive value. So in this particular case, our p, for this p-series,
is equal to five. P is equal to five. Now you might already recognize,
under which conditions for a p-series does it
converge or diverge? It's going to converge. It's going to converge when
your p is greater than one, which is clearly the case in
this scenario right over here. Our p is clearly greater than one. We would diverge, we would diverge if our p is greater than zero and less than or equal, or less than or equal to one. This would be a divergent. So if this was like .9 here, or if this was a, you know, 3/4, then we would be diverging. So at least for this
one, we are convergent. Let's do another one of these. All right. So here, you might again
recognize this as a p-series. Let me rewrite this infinite sum. So this is the sum from n equals one to infinity of one over, let's see, we have a square root of two, a square root of three. So you could do this as two to the 1/2, three to the 1/2, four to the 1/2. So it's one over n to the 1/2. Notice, this is when n is equal to one. One over one to the 1/2 is one. One over two to the 1/2, well,
that's this right over here. And we keep on going on and on and on. Well, in this case, we
still have a p-series. We have one over n to some power, and that power is positive. But notice, in this
case, our p falls between zero and one. So 1/2 is our p. So p for our p-series is equal to 1/2, and that's between zero and one. Remember, we're divergent, divergent, when our p is greater than zero and less than or equal to one, which was clearly the
case right over here. So this is going to be divergent.