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## Calculus, all content (2017 edition)

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

Lesson 8: Basic convergence tests

# Proof of p-series convergence criteria

A p-series converges for p>1 and diverges for 0.

## Want to join the conversation?

• The graphs are wrong! The right graph is integral from 1 to infinity of 1/x^p +1,kind of . But when Sal changed form left Reimann sum to right Reimann sum, the graph shrinks. If the left graph is the integral from 1 to infinity of 1/x^p, the right graph is not the left graph +1. Sal just transferred the left graph left 1 unit. That is confusing or least I don't understand. So I think we should make a new video or just explain to me......Please.
• It's a little confusing because there are only two graphs but there are THREE things in the inequality. The orange rectangles in both graphs are exactly the same, Σ[1, ∞] 1/n^p, and this is the infinite series we want to sandwich. The idea is that in the left graph, this set of orange rectangles is exactly a Left Endpoint Riemann Sum, and is clearly greater that the integral from 1 to ∞.

Now, when he shifts the rectangles left one unit, they almost form a Right Endpoint Riemann Sum (RRS), BUT the first rectangle (area = 1), which he colors in reddish-purple, is not part of the Right Riemann Sum for the integral from 1 to ∞. So the (integral from 1 to ∞) > RRS, but our orange rectangles are 1 + RRS, and we want something bigger than the orange rectangles, so the right end of the inequality is 1 + the integral.

To sum up:
Integral from 1 to ∞ of 1/x^p = INT
Infinite series = orange rectangles (OR) = Left RS = 1 + Right RS
FIRST
INT < Left RS
and thus
INT < OR

SECOND
Right RS < INT
and thus
1 + Right RS < 1 + INT
and thus
OR < 1 + INT

And so finally (since OR = orange rectangles = infinite series):
INT < infinite series < 1 + INT
• It doesn't effect the outcome, but around when Sal factors out the 1/(1-p) shouldn't it have made the rest [M^(1-p)]-1 ?
• At , he factored 1/(1-p) out from the 1st term only.
For the second term, he simplified 1^(1-p) to just 1 in the numerator because one to any power (any p-value) is still just 1. Because the simplified second term 1/(1-p) is just a constant, and will be the same regardless of the value of m, it does not affect whether the limit converges or diverges. So to determine convergence/divergence, he only uses the first term. I hope this helps.
• I don't understand p-series. is 1/(ln3)^n n = 1 to infinity a p series, ? does it converge or diverge.?
• Variable as the exponent => Geometric series
Variable as the base => p-series
• why does f(x) decrease if p > 0? If p is a fraction, couldn't it increase?
• Why are Riemann sums being used to visualise? Aren't definite integrals the exact area under the curve, not approximations?
(1 vote)
• The integral gives the exact area under the curve, but the p-series corresponds to the sum of the rectangles. So in this case it's not that Riemann sums are being used to approximate the area, but rather that the (exact) area is bounding the discrete sum. Interestingly, the integrals are giving the conditions under which the series converges, but they are not actually giving the value to which it converges. (There's no general formula for the sum in terms of `p`.)
(1 vote)
• why p have to be greater than 0?
(1 vote)
• Because we define p to be greater than 0.
We know that when p is zero, you're adding ∞ * 1.
We know that when p is negative, you're adding an infinite number of increasing positive terms, which obviously diverges.
(1 vote)
• Okay, but you said in the integral test videos, that if the a function failed the divergence integral test, it doesn't mean for sure it converges, can someone explain ?
(1 vote)
• What if the bounds were changed to 0 to infinity? How would the range of p change then? Thank you.
(1 vote)
• The integrals diverge because you have an improper integral, and the series diverges because you always have a 1/0, and thus no value of p makes the series converge.
(1 vote)