Log in hjy6296 5 years agoPosted 5 years ago. Direct link to hjy6296's post “I'm at 2:50 looking at th...” I'm at • (18 votes) Alex 5 years agoPosted 5 years ago. Direct link to Alex's post “The integral expression o...” 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. (7 votes) G Y 7 years agoPosted 7 years ago. Direct link to G Y's post “At 1:45, what if you use ...” At • (3 votes) Adam Thai 6 years agoPosted 6 years ago. Direct link to Adam Thai's post “The step of n is 1 so the...” The step of n is 1 so the step on x-axis must be 1 in correspondence. (14 votes) {Rayeed}^3 4 years agoPosted 4 years ago. Direct link to {Rayeed}^3's post “I understand it fully in ...” 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? • (7 votes) CJ Coudriet a year agoPosted a year ago. Direct link to CJ Coudriet's post “My intuition for this is ...” My intuition for this is somewhat mentioned in the comment below. Basically, the terms in the sequence decrease towards 0 at a slower and slower rate, allowing the sum to approach infinity. Have you ever looked at the graph of ln(x)? If you look on desmos (https://www.desmos.com/calculator/iing9fbgoc) and slowly scroll to the right, it feels like the values keep growing by smaller and smaller values, but nevertheless always growing. It makes sense that the graph of ln(x) must approach infinity since you can plug any number into e^x (the "answer" for ln(x)). Then, zoom out in the graph and go as far right as you'd like. The numbers are much larger, and they never seem to stop like convergent series do, even though it grows so slowly. Even if that didn't make any sense, think about this: the difference between any two consecutive values doesn't change very fast, so the sum is able to continually grow. Reasoning from Ian Pulizzotto (thanks!): "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." This doesn't work with 1/x^2, for instance: I'm not sure though, bc not I'm more confused after writing this comment. EDIT: Read my comment below! It has more examples! (3 votes) Chris Offner 5 years agoPosted 5 years ago. Direct link to Chris Offner's post “At 2:10 don't quite under...” At But I'm not sure how we've shown that the p-series is between those two integral expressions. • (5 votes) Eric Bandera 4 years agoPosted 4 years ago. Direct link to Eric Bandera's post “So, the key thing is that...” 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 (7 votes) G. Tarun 5 years agoPosted 5 years ago. Direct link to G. Tarun's post “What does "*divergent*" m...” What does "divergent" mean in the context of a p-series? • (1 vote) Ian Pulizzotto 5 years agoPosted 5 years ago. Direct link to Ian Pulizzotto's post “Divergence of a series do...” 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. (10 votes) Michele Franzoni 4 years agoPosted 4 years ago. Direct link to Michele Franzoni's post “I've read all the answers...” 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? • (3 votes) JI YONG Ahn 3 years agoPosted 3 years ago. Direct link to JI YONG Ahn's post “1) Integration of P-serie...” 1) Integration of P-series from 1 to infinity is the white shade. 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) and 2) are different. (2 votes) Soldier King 5 years agoPosted 5 years ago. Direct link to Soldier King's post “at 4:48, how do you know ...” at • (1 vote) Alex 5 years agoPosted 5 years ago. Direct link to Alex's post “We define it to be. Any p...” We define it to be. Any p <= 0 would simply diverge (which can be shown through the nth term test). (5 votes) Charizard-Max 3 years agoPosted 3 years ago. Direct link to Charizard-Max's post “why is the sum of 1/n^p w...” why is the sum of 1/n^p with 0<p<1 diverging on why is that? • (3 votes) Wantedsolo a year agoPosted a year ago. Direct link to Wantedsolo's post “This is quite a late answ...” This is quite a late answer but ig it's for future reference.. I think you are a bit confused between divergence of series and divengence of nth term: The y values of the graph itself only show the individual values of 1/n^p at different points. But the sum of the series (and what we're trying to calculate) is the area of the rieman sums of the graph and thus bounded on both sides by the 2 graphs. These sums are not necessarily finite and the conditions that change the sums 'finiteness' is what was proved in this video. Hope this resolved your doubt. (1 vote) Hexuan Sun 9th grade 2 years agoPosted 2 years ago. Direct link to Hexuan Sun 9th grade's post “What will happen if the p...” What will happen if the p exponent is less than 0? • (2 votes) Art 3 months agoPosted 3 months ago. Direct link to Art's post “It still diverges as p is...” It still diverges as p is still less than or equal to 1. The idea is that since it is to the negative first, the denominator will be 1/whatever, which means that the numbers get greater with each increase in n. (1 vote) stspapon1 4 years agoPosted 4 years ago. Direct link to stspapon1's post “so, technically the integ...” so, technically the integral is the p-series if dx was 1? • (2 votes)Want to join the conversation?
2:50 1:45
1/1+1/4+1/9+1/16+1/25+...
> 1+(1/9+1/9)+(1/25+1/25)
which doesn't diverge to anything. It simply gives another sequence. How I think about it is that 1/x^n, where n > 1, has the denominator of 1/k skipping more and more values (1, 1/4, 1/9,1/16, 1/25, 1/36, etc.) while 1/x passes through every value which sums up to infinity. 2:10
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?
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.
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). 4:48 8:52
the final term of 1/n^p for any non negative value of p will always approach 0 as lim n -> infinity. But that says nothing about the sum of the series.
FAQs
How do you prove that the P-series converges? ›
p -series have the general form ∑ n = 1 ∞ 1 n p where p is any positive real number. They are convergent when p > 1 and divergent when 0 < p ≤ 1 .
What is the convergence rule of the P-series? ›With p-series, if p > 1, the series will converge, or in other words, the series will add up to a specific numerical value. If 0 < p ≤ 1, the series will diverge, which means that the series won't add up to a specific numerical value.
How to find where the AP series converges? ›If it's a p-series X 1 np , you know if it converges or not. It converges when p > 1. If the terms don't approach 0, you know it diverges. If you can dominate a known divergent series with the series, it diverges.
What are the values of P for which series converges? ›Therefore, the infinite series converges when p > 1, and diverges when p is in the interval (0,1).
Do all P-series converge absolutely? ›To summarize, the convergence properties of the alternating p-series are as follows. If p > 1, then the series converges absolutely. If 0 < p ≤ 1, then the series converges conditionally. If p ≤ 0, then the series diverges.
How do you know if the series converges? ›If a series is a p-series, with terms 1np, we know it converges if p>1 and diverges otherwise. If a series is a geometric series, with terms arn, we know it converges if |r|<1 and diverges otherwise.
What is the criteria for P-series? ›A test exists to describe the convergence of all p-series. That test is called the p-series test, which states simply that: If p > 1, then the series converges, If p ≤ 1, then the series diverges.
How do you prove a series converges to a value? ›Ratio test.
If r > 1, then the series diverges. If r = 1, the ratio test is inconclusive, and the series may converge or diverge. where "lim sup" denotes the limit superior (possibly ∞; if the limit exists it is the same value). If r < 1, then the series converges.
To determine if a given sequence is convergent, we use the following two steps:
- Find a formula for the nth term, or an, of the sequence.
- Find the limit of that formula as n approaches infinity. If the limit exists, the sequence is convergent. If not, the sequence is not convergent.
Given an alternating series , ∑ ( − 1 ) k a k , if the sequence of positive terms decreases to 0 as , k → ∞ , then the alternating series converges. Note that if the limit of the sequence is not 0, then the alternating series diverges.
For which values does p converge? ›
For x−p, if p<1, the integral converges because the function x−p does not become infinitely large as x approaches 0. Conversely, if p≥1, the function becomes infinitely large as x approaches 0, causing the integral to diverge.
How do you find the point of convergence of a series? ›In order for a series to converge the series terms must go to zero in the limit. If the series terms do not go to zero in the limit then there is no way the series can converge since this would violate the theorem.
How do you prove convergence of a complex series? ›To establish the fact that the sum of a series is a given number , it is often convenient to define the remainder after terms, using the partial sums: ρ N = S − S N Thus S = S N + ρ N . Now, since | S N − S | = | ρ N − 0 | , then a series converges to a number if and only if the sequence of remainders tends to zero.
How do you prove a power series converges? ›If L=limn→∞|cn|1/n exists, then:
- If L<1 then the series converges absolutely.
- If L>1 then the series diverges.
- If L=1 then the test gives no information.
May 2, 2023