A nice limit with a trick.

here we're gonna evaluate a nice little limit using a result that is tied to the ratio test and the route test from testing series for convergence so we want to look at the limit as n goes to infinity of the nth root of n factorial over N we're gonna use two main tools the first is like a standard limit representation of the natural base for an exponent E and that's the limit as n goes to infinity of 1 plus 1 over N to the N equals e so we'll show this just for completeness and then next is this nice result regarding this ratio of members of a sequence versus the root of members of a sequence so we have if a n is bigger than 0 for all N and the limit as n goes to infinity of a n plus 1 over a n and the limit as n goes to infinity of the nth root of a n both exist and are finite then they must be equal okay so let's go ahead and prove this first and then we'll move on to the second one so how I want to do this is we'll start by setting the limit of this equal to L so we have L equals the limit as n goes to infinity of 1 plus 1 over N to the N and notice that as n goes to infinity this guy inside of the parenthesis goes to 1 and then the exponent goes to infinity so this is an indeterminate form of type 1 to the infinity and so generally to evaluate these types of indeterminate forms you pass to a logarithm so let's do that so now we have the natural log of the limit is equal to the limit as n goes to infinity of now we have n times the natural log of 1 plus 1 over N now here we just used the logarithm fact that it will bring an exponent down as a multiplier now if we zoom in on this notice that as n goes to infinity n goes to infinity and then this thing goes to the natural log of 1 which is 0 so now we've transformed this in you an indeterminate form of type infinity times zero and that's really close to the type that we can apply l'hopital's rule to so here we have this is the limit as n approaches infinity of the natural log of 1 plus 1 over n over 1 over n so that's like kind of our standard trick and now notice this is of type zero over zero so we've got it all set for l'hopital's rule now I'm gonna do something kind of subtle but it's not super necessary as long as you know what's going on and that is I'm gonna pass this from a discreet variable to a continuous variable so I'm gonna write this is the limit as X goes to infinity of the natural log of 1 plus 1 by x over 1 over X notice that's still an indeterminate form but now we have a continuous limit which means we can apply l'hopital's rule by taking the derivative of the numerator and the denominator you can't necessarily do that until you have a continuous variable okay so let's go ahead and do that so using l'hopital's rule we need to take the derivative of the numerator so that's gonna be 1 over 1 plus 1 over x times negative 1 over x squared great and let's sneak a lemon in here so we have the limit as X goes to infinity and then the denominator the limit of the derivative of that will be minus 1 over x squared now notice that these minus 1 over x squared terms cancel and we're taking the limit as X goes to infinity of 1 over 1 plus 1 over X and so that's gonna give us 1 because notice it gives us 1 over 1 plus 0 but that's the natural log of our goal limit so in other words we have the natural log of our goal limit is equal to 1 which means our goal limit is equal to e the natural base for the exponential function okay so we've got this first tool taken care of and now we're ready to move on to the second and we'll do this using the epsilon in definition of a limit as well as the squeeze theorem so let's go ahead and suppose that each of these limits exists and let's named to one of the limits will give a name to this limit of the quotient here so let's go ahead and set the limit as n approaches infinity of a sub n plus 1 over a sub n equal to L so what that means is for any epsilon bigger than 0 there is some natural number n so we're thinking about epsilon being super small and this end being pretty sizable such that if little n is bigger than or equal to capital n we have the nth term of this sequence is as close to L as epsilon so in other words we have a n plus 1 over a n is between L minus epsilon and L plus Epsilon so sometimes this is written slightly differently sometimes this is written as the absolute value of a n plus 1 over a and minus L is less than Epsilon but this is like an equivalent kind of compound inequality instead of an absolute value inequality which will be helpful for our purposes here now what we want to do is notice that this is true for all n bigger than or equal to n so that means that I can scale this down all the way until work at this capital n so let's go ahead and do that so the next one underneath this will be a n over a n minus 1 is between L minus epsilon and L plus Epsilon and then the one below that will be a n minus 1 over a n minus 2 is between L minus epsilon and L plus Epsilon and then all the way down and so we want to stop at a sub capital n so we have a sub capital n plus 1 over a sub capital n so that's again between L minus epsilon and L plus Epsilon fantastic but now what we want to do is take the product of all of these inequalities so let's go ahead and do that we're going to do the product of all of these inequalities and notice since all of the parts here are positive we maintain the direction of the inequalities and you might say well how many of these do I have well I have exactly little n plus 1 minus big n and you can count that from kind of this starting point to that ending point right there so let's see what that's going to give us that's going to give us L minus epsilon to the little n plus 1 minus capital and so that's going to be on the left hand side of the inequality then in the middle of the inequality notice a bunch of stuff is going to cancel notice that this a n is going to cancel with this a and when we take that product this a n minus 1 with this a n minus 1 this a n minus 2 with something before or after I should say and this a n plus 1 something before so all we're left with is this a sub little n plus 1 and a sub capital n so that means we can put here a sub little n plus 1 over a sub capital N and then we have something similar to the left hand side on the right hand side in particular we have L plus epsilon to the n plus 1 minus n ok great so now what I want to do is I'll clean up this kind of stuff that we have and I'll bring this over and we'll continue the next step so the last board we ended with the following inequality so we have L minus epsilon to the quantity n plus 1 minus n is less than a sub little n plus 1 over a sub capital n which is less than L plus epsilon to that same exponent n plus 1 minus n so now what we want to do is isolate that a sub n plus 1 and see what we get from there so here we're going to have L minus epsilon to the n plus 1 over L plus epsilon to the capital N and in times a n so notice I rewrote this a little bit and then multiplied by the capital a in and then here we have that's less than a and plus one which is less than now I'm going to do the same kind of thing here so we've got L plus epsilon to the n plus 1 over L minus epsilon to the capital N and then we have a sub n so that that should be L plus Epsilon fantastic now what we want to do is take the n plus first root of all parts of the inequality and we're motivated to do that because that's what we're trying to get at is the limit of this nth root of the a n term but if we do the n plus first root of a a n plus one term that's essentially the same thing so let's go ahead and do that and see what we get so taking the n plus first root of this notice we have an L minus Epsilon so that's nice kind of on its own and then we have the n plus first root of a sub n over L plus epsilon to the N great and then here we have the n plus first root of a n plus one and now this is going to be L plus Epsilon and then the n plus first root of a n over L plus epsilon to the capital n ok great now let's maybe put this in parentheses now we're ready to do the squeeze theorem so now let's go ahead and take the limit of all parts of this as little in approaches infinity so notice as little n approaches infinity the n plus first root of anything will go to one so we won't prove that but that's kind of a result that would occur earlier in the class in this so if we let n go to infinity that's going to go to 1 if we let n go to infinity that's going to go to 1 as well which tells us we have L minus epsilon is less than the limit as n goes to infinity of the n plus first root of a sub n plus 1 which is less than L plus Epsilon but now a statement like this is true for all Epsilon but whenever a statement like this is true for all epsilon bigger than zero that gives you the exact value of your Lim so in other words this tells us that the limit as n approaches infinity of the n plus first root of a sub n plus 1 equals L which is exactly what we wanted to show because we wanted that to be equal to this ratio on it okay so now we've got the second tool taken care out and now we're ready for our final goal which is this limit as n goes to infinity of the nth root of n factorial over n so I guess in order to be very careful you should probably show that this limit exists first but notice all of the terms are positive which means that it's bounded below by zero and furthermore you can show that this is decreasing by induction pretty easily and then by the monotone sequence there this thing converges so since this thing converges if we can write it either as a quotient or a root we can use this result that we just proved and this root right here tells us we probably want to write the whole thing as a root and then rewrite it as this quotient type thing so let's go ahead and do that so notice I can rewrite this as the limit as n tends towards infinity and now I have the nth root of n factorial in the numerator and then the nth root of n to the N in the denominator okay great well so obviously the nth root of n to the N is just in so I've helped myself out there so that allows me to rewrite this as the limit as n approaches infinity of the nth root of n factorial over N to the N and so here we're playing our game from the second tool where this n factorial over N to the N is our a sub n term so now this tool says that our limit in question should be the same as the limit as n goes to infinity of a sub n plus 1 over a sub n so let's go ahead and write that down so this should be equal to the limit as n tends towards infinity so let's look at a sub n plus 1 that's going to be n plus 1 factorial over n plus 1 to the n plus 1 all over so that big numerator there is a sub n plus 1 and then this is going to be all over in factorial over N to the N now I'm going to rewrite this so white terms are on top of each other so like factorials I consider as like terms and then the exponents are the other type of white terms so let's see what that gives us that's going to give us this limit as n tends towards infinity of so we'll have n plus 1 factorial over N factorial and then next we'll have n to the N over N plus 1 to the n plus 1 again that's putting the like type terms on top of each other so let's see how we can simplify here so here I want to use the fact that if we have n plus 1 factorial we can decompose that into n plus 1 times n factorial so that's a standard trick and now we can cancel this n factorial with this n factorial and furthermore we can cancel this n plus 1 with the plus 1 part of this right here so that leaves us with N to the N over n plus 1 to the N so that's kind of nice so let's see what we have we have the limit as n approaches infinity of n over n plus 1 all to the N here I just brought that exponent out of the whole thing ok great now this almost looks like our tool right here and it will be exactly this tool if we use an algebraic property of limits involving a quotient of a limit and we can do that notice that's going to be the same thing as 1 over the limit as n goes to infinity of the reciprocal of this thing which is n plus 1 over N all to the nth power again this is an algebraic property of limits which is pretty standard but now that gives us 1 over we have the limit as n tends towards infinity of 1 plus 1 by N to the end after simplifying that guy right there but let's see what we've got that is our first tool which we showed to be equal to e so our final answer is 1 over e ok great so while this part of the calculation didn't use any like fancy analysis type stuff it was just like symbolic manipulation and our first calculation was also like a calculus type calculation notice we had to use something that looked like real analysis from this second tool so if you're interested in learning more about that I have a whole playlist that is in the process of being built about real analysis for my upcoming class that I'm teaching this fall ok that's a good place to stop

Channel: Michael Penn

Views: 105,799

Keywords: math, mathematics, number theory, abstract algebra, calculus, differential equations, Randolph College, randolph, Michael Penn

Id: 8fI0S-HeYrQ

Length: 15min 55sec (955 seconds)

Published: Mon Jul 06 2020