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## Integral Calculus

### Unit 5: Lesson 5

Harmonic series and p-series# Proof of p-series convergence criteria

AP.CALC:

LIM‑7 (EU)

, LIM‑7.A (LO)

, LIM‑7.A.7 (EK)

p-series have the general form sum, start subscript, n, equals, 1, end subscript, start superscript, infinity, end superscript, start fraction, 1, divided by, n, start superscript, start superscript, p, end superscript, end superscript, end fraction where p is any positive real number. They are convergent when p, is greater than, 1 and divergent when 0, is less than, p, is less than or equal to, 1.

The AP Calculus course doesn't require knowing the proof of this fact, but we believe that as long as a proof is accessible, there's always something to learn from it. In general, it's always good to require some kind of proof or justification for the theorems you learn.

## Want to join the conversation?

- I'm at2:50looking at the two graphs and I'm a little confused about the graph on the right side. If you add 1 to the integral, shouldn't the graph be something like a graph on the left side but the red bar on the right side graph added to it?(11 votes)
- The integral expression on the left includes the white area under the curve. The expression on the right includes the white area under the curve plus the red bar. If it's the orange series that's confusing you, it's simply because the indexes are shifted over by 1 in the graph on the right, making the red bar also belong to the orange series.(4 votes)

- At1:45, what if you use a rectangle width other than 1?(3 votes)
- The step of n is 1 so the step on x-axis must be 1 in correspondence.(13 votes)

- At2:10don't quite understand how we see from this visualisation that 1 + Integral(...) is larger than the p-series formula in the middle. It seems like we're merely comparing it to the left Integral(...) without the +1 added to it. It makes sense to me that if we add 1 (i.e. the area of the red bar) to the area under the curve, that's bigger than the area under the curve alone, i.e. Integral(1/x^p) < 1 + Integral (1/x^p).

But I'm not sure how we've shown that the p-series is**between**those two integral expressions.(4 votes)- So, the key thing is that the P series, shown in Orange are essentially left and right Riemann sums. And since the function is descending, it can be concluded that a left Riemann sum will be greater than a right Riemann sum. The integral that we are working with is from 1 to ∞. That isn't changing. So, first, the left Riemann sum from 1 to ∞. You can probably see why that is greater than the integral - there are the left over corners on the top. But, the right Riemann sum...well...we can shift the P series over to the left by one. Keep in mind the P series is equivalent to itself! We're just shifting the placement of it. But the problem is that it's not really a fair comparison after the shift since the P-series is now starting at 0 and going to ∞ instead of starting at 1 like the integral. We do know that the first term of the P series will always be 1. So...if we took the P series minus 1 would be the right Riemann series from 1 to ∞ (smaller than ∫) and the standard P series would be the left Riemann series from 1 to ∞ (bigger than ∫) This statement is: (P-series -1)< the integral < (P-series). Now keep all those inequalities in mind, and add 1 to each to get (P-series) < the integral +1 < (P-series +1). Combine them so that the integral is in the middle, and you get: the integral < (P-Series < the integral +1. This is a little convoluted, the way I explained it, but I feel after reading this the concepts will become clearer(5 votes)

- I understand it fully in mathematical terms but I don't have a nice intuition about it. If we look at the graphs of 1/x and 1/x^2, they both look almost the same. The fact that one of them has a finite area and the other has an infinite area seems counterintuitive. Do both of them 'touch' the x axis as the reach infinity? How does all this actually happen?(5 votes)
- What does "
**divergent**" mean in the context of a**p-series**?

Does it mean that as the series goes on—endlessly, toward infinity—the terms get*larger and larger*?

If=**p****1**, then the the p-series is divergent by definition, as a divergent p-series has a value of p greater than zero but lesser than or equal to 1 (as given in this article and the*Harmonic series and p-series*video in this lesson). But then, in a harmonic p-series whose p value is 1, don't the terms get smaller and smaller as the series goes on?(1 vote)- Divergence of a series does not always mean that the terms get larger or stay the same size. Divergence of a series means that the limit of the
**partial sums**of the terms does not exist (that is, the partial sums grow without bound, positively or negatively, or the partial sums oscillate without converging to a limit).

In the case of the harmonic series with p=1, yes the terms do become smaller and converge to zero, but the terms converge so slowly to zero that the partial sums still grow without bound! We can see this by observing that

1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+1/9+1/10+1/11+1/12+1/13+1/14+1/15+1/16+...

= 1+1/2+(1/3+1/4)+(1/5+1/6+1/7+1/8)+(1/9+1/10+1/11+1/12+1/13+1/14+1/15+1/16)+...

>= 1+1/2+(1/4+1/4)+(1/8+1/8+1/8+1/8)+(1/16+1/16+1/16+1/16+1/16+1/16+1/16+1/16)+...

= 1+1/2+2/4+4/8+8/16+...

= 1+1/2+1/2+1/2+1/2+..., which clearly diverges to infinity since the sequence 1,1.5,2,2.5,3,... clearly grows without bound.

So the harmonic series with p=1 diverges to infinity!

It is important the distinguish the behavior of the sequence of terms from the behavior of the partial sums of the terms, since these behaviors are not always the same.(7 votes)

- at4:48, how do you know p is greater than 0?(1 vote)
- We define it to be. Any p <= 0 would simply diverge (which can be shown through the nth term test).(5 votes)

- I've read all the answers and comments in this section but i can't really grasp the reason why the series is bounded the way it is.

If the series is greater than the improper integral but lesser than the improper integral plus one it means that such integral overestimates the series by less than one (when looking to the let graph). And the reason why this should be true is very far from clear. Could someone help me out?(2 votes)- 1) Integration of P-series from 1 to infinity is the white shade.

2) The summation of the P-series from 1 to infinity is the bars. In comparison to clause 1), it has larger surface area.

3) Integration of P-series from 1 to infinity + 1 is the red and white area.

4) now, see the left graph and right graph. They both have the bars. The bars are the summation of the P-series in both cases.

5) the right side graph is the left side bars moved to the left by 1. This is done by doing +1.

Therefore, "such integral overestimates" as you mentioned, because it is, by looking at it. It is the surface areas of the bars + the gap between the function's curve and the bars.

As a side note:

1) the area of overestimation on the left graph is, (the summation of P series) - (the integration of the P series). This is the overestimated area of the Summation of P-series compared to the area by integration.

2) the area of underestimation on the right graph, for the Summation of P-series is, (1 + integration of P series) - (summation of P series).

1) and 2) are different.(2 votes)

- why is the sum of 1/n^p with 0<p<1 diverging on8:52using m->infinity(using lim) but on the graph literally shows that as n-> infinity 1/n^p->0 (0<p<1)(even if it's slow but infinity is a really big value, in the end it still converge to zero for 1/n^p)?

why is that?(2 votes) - How is it about p=0?

Does it simplify to infinit sum of 1+1+..+1 which diverges?(1 vote) - What will happen if the p exponent is less than 0?(1 vote)