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# Integral test

AP.CALC:
LIM‑7 (EU)
,
LIM‑7.A (LO)
,
LIM‑7.A.6 (EK)

## Video transcript

let's explore a bit the infinite series from N equals one to infinity of 1 over N squared which of course is equal to 1 plus 1/4 it's 1 over 2 squared plus 1 over 3 squared which is 1/9 plus 1/16 and it goes on and on and on forever so there's a couple of things that we know about it the first thing is all of the terms here are positive so all of the terms here are positive so they're all positive and that they're decreasing there when it looks like there D quickly they're decreasing quite quickly here from 1/2 1/4 to 1/9 1/16 and so they're quickly approaching 0 which makes us feel pretty good that this thing may you know has it has a chance of converging and because they're all positive we know that this sum right over here if it does converge is going to be greater than 0 so the reason the only reason why it wouldn't converge is if somehow it goes unbounded towards infinity which we know if this was 1 over and it would be unbounded towards infinity so this says that's that's a possibility here so if we could show that this is bounded then that'll be a pretty good argument for why this thing right over here converges because the only reason why you could diverge is if you went to either positive infinity or negative infinity we already know that this thing isn't going to go to negative infinity because it's all positive terms or you could diverge if this thing oscillates but it's not going to oscillate because all of these terms are just adding to the sum none of them are taking away none of because none of these terms are negative so let's see if we can make up make a good argument for why this sum right over here is bounded and especially if we can come up with the bound then that's a pretty good argument for this that that this infinite series should converge and the way that we're going to do that is we're going to explore a related function so what I want to do is I want to explore f of X is equal to 1 over x squared you could really view this right over here 1 over N squared as f of n if I if I if I were to write it this way so why is this interest well let's graph it let's graph it so that's the graph of y is equal to f of X y is equal to f of X and notice this is a continuous positive decreasing function especially over the interval that I've or over over the interval that I care about right over here I guess we could say over the 4 for positive values of X it is a continuous but it is a continuous positive decreasing function and what's interesting is we can use this as really an under estimate for this area right over here what do I mean by that well one this first term right over here you could view that as the area of this block right over here that is f of F of n or excuse day F of one high and one wide so it's going to be 1 it's going to be 1 times 1 over 1 squared or 1 this term this let me make sure I'm using different colors this term right over here that could represent the area of this block which is 1/4 high and 1 wide so it is going to be it's going to have an area of 1/4 what could this one represent well the area of the next block if we're trying to estimate the area under the curve and this might look familiar from when we first got exposed to the integral or even before we got exposed to the integral we're taking Riemann sums so that right over here that area is going to be equal to 1/9 so what's intriguing about this is we know how to find the exact area especially or the exact area from 1 to infinity from x equals 1 to infinity so maybe we can use that somehow we know what this area is right over here from which we can denote as the improper integral from 1 to infinity of f of X DX we know what that is and I'll figure out it a little bit and if we know what this is if we can figure out the value that's going to be an upper bound for 1/4 plus 1/9 plus 1/16 on and on and on and so I will allow us to essentially bound what this series evaluates to and as we said earlier that would be a very good argument for its convergence so the whole point here I'm not doing a rigorous proof but really getting you the the underlying conceptual understanding for a very popular test for convergence or divergence which is called the integral test let me just write that down just so you know what this is kind of the mental foundations for so what do I mean here so let me write this sum again let me write it a little bit different so our original series from N equals 1 to infinity of 1 over N squared it's going to be equal to this first block the area of this first block plus the area of all of the rest of the blocks the 1/4 plus 1/9 plus 1/16 which we could write which we could write as the reduce in a new color which we could write as the sum from N equals 2 to infinity of 1 over N squared so I just I just kind of expressed this as the sum of this plus all of that stuff now what's interesting is that this this would I just what I just wrote in this blue notation that's this block plus this block plus the next block which is going to be less than this definite integral right over here this definite integral it notice it's an underestimate it's always below the curve this it's going to be less than that definite integral so we can write that this thing is going to be less than 1 1 plus instead of writing this I'm going to write the definite integral 1 plus the definite integral from 1 to infinity of 1 over 1 over x squared DX now why is that useful well we know how to evaluate this and I encourage you to review our the section on him on Khan Academy on improper integrals if this looks unfamiliar but I'll evaluate this I'll evaluate this down here we know that this is the same thing is the limit as I'm going to introduce a variable here T approaches infinity of the definite integral from 1 to T of and I'll just write this is X to the negative 2 DX which is equal to the limit as T approaches infinity of the CD of negative x to the negative 1 or actually I could write that as negative 1 over X negative 1 over X so we're going to evaluate that as T and at 1 which is equal to the limit as T approaches infinity of negative 1 over T and then minus negative 1 over 1 so that would just be plus 1 and as T approaches infinity this term right over here is going to be 0 so this is just going to simplify to 1 so this whole thing evaluates to 1 so just like that we were able to place an upper bound on this series we're able to say that the series under question or in question so the infinite sum from N equals 1 to infinity of 1 over N squared is going to be less than 1 plus 1 or it's going to be less than it's going to be let me do this in a new color it's going to be less than it's going to be less than 2 or another way to think about it's going to be the 2 is this area that's 1 right over there Plus this area right over here so we're saying that this sum is going to be less than 2 so we have bounded it above so we know it cannot go to positive infinity because all the terms are positive it's definitely not going to go to negative infinity and because all the terms are positive we also know that this isn't going to oscillate between two different values so this gives us a pretty good sense that this series is or that this series converges and the logic we just used here the logic we just use here to argue for why this converges once again not a rigorous proof but this is the underlying logic of the integral test
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