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Let $ a_n = \left ( 1 + \frac {1}{n} \right)^n. $(a) Show that if $ 0 \le a < b, $ then $ \frac {b^{n + 1} - a^{n + 1}}{b -a } < (n + 1)b^n $(b) Deduce that $ b^n[(n + 1)a - nb] < a^{n + 1}. $(c) Use $ a = 1 + 1/(n + 1) $ and $ b = 1 + 1/n $ in part (b) to show that $ \left \{ a_n \right \} $ is increasing.(d) Use $ a = 1 $ and $ b = 1 + 1/(2n) $ in part (b) to show that $ a_{2n} < 4. $(e) Use parts (c) and (d) to show that $ a_n < 4 $ for all $ n. $(f) Use Theorem 12 to show that $ \lim_{n \to\infty} (1 + 1/n)^n $ exists.(The limit is $ e. $ See Equation 3.6.6.)

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Calculus 2 / BC

Chapter 11

Infinite Sequences and Series

Section 1

Sequences

Series

Oregon State University

University of Michigan - Ann Arbor

University of Nottingham

Boston College

Lectures

01:59

In mathematics, a series is, informally speaking, the sum of the terms of an infinite sequence. The sum of a finite sequence of real numbers is called a finite series. The sum of an infinite sequence of real numbers may or may not have a well-defined sum, and may or may not be equal to the limit of the sequence, if it exists. The study of the sums of infinite sequences is a major area in mathematics known as analysis.

02:28

In mathematics, a sequence is an enumerated collection of objects in which repetitions are allowed. Like a set, it contains members (also called elements, or terms). The number of elements (possibly infinite) is called the length of the sequence. Unlike a set, order matters, and exactly the same elements can appear multiple times at different positions in the sequence. Formally, a sequence can be defined as a function whose domain is either the set of the natural numbers (for infinite sequences) or the set of the first "n" natural numbers (for a finite sequence). A sequence can be thought of as a list of elements with a particular order. Sequences are useful in a number of mathematical disciplines for studying functions, spaces, and other mathematical structures using the convergence properties of sequences. In particular, sequences are the basis for series, which are important in differential equations and analysis. Sequences are also of interest in their own right and can be studied as patterns or puzzles, such as in the study of prime numbers.

01:13

Let $$a_{n}=\left(1+\frac{…

07:46

Let $a_{n}=\left(1+\frac{1…

We were given a sequence a end by the given formula here and is one plus one over n to the end power and in part, A. We like to show that if this inequality is true, then dis implies the following inequality. So this is for party, eh? So let's give a solution or proof of this for party. So recall the following We can factor out a B minus a here and then we're left with the following. If you notice here and any of these terms, if you add the exponents, you get end and this can be verified by just go ahead and distributing the being the way through the presidency's. So now let's go ahead and divide both sides of this equation by B minus A. So this implies. Now we're slip with the second term here on the right hand side, all the way up to just eight of the n. And now we'LL use this fact here that is less than be so each time we see a in the right hand side, we'LL just go ahead and use this fact here to replace it with a B. So have a B and then be the end minus one. That's B then and so on. So now it's just a matter of noting how many beat of the ends we have here. Well, this has beat of zero. This is B to the one all the way to be to the end. So that's a total of N plus one terms. And that's exactly what we wanted here for the apology. So this resolves part A. Let's go on to the next page for part B. So here we wanted deduce the following from party. All right, so what's going on? Give a solution to this. So from part, eh? Well, we just did on the previous page. Let's just recite that inequality here. So let's go ahead and just multiply this B minus eight of both sides. Okay, so then at this point, it's just a matter of simplification. So let's go ahead and move this to the right hand side and then bring this to the left hand side and then using a little algebra here, a factoring here, you could factor out, be too then, and after simplifying, you can write the inside as such, so one can go ahead and verify the step here by just using some algebra. And this is the inequality that we wanted for a part B So let's remember this fact because we'LL need this later. Now we're on to parties. So this is where we'LL go ahead and use and let's be equals one plus one over n and B. So let's use these to show that the sequence and this increasing and remember what an was by definition. Okay, so now let's go ahead and get started there. So we're just going to go straight to the expression in from part B. So let me recall that here in a different color. So this is the previous fact that were being used here. So now plugging this inn for B and this in for a we have the following. So is plugging that be first? Then we have our n plus one and then now eh minus and and then times B and then less than a to the n plus one. And unless you simplify So inside the larger apprentices here we'LL just get n plus one plus one minus and minus one. So this is leftover. Let's see here So I think I made a mistake here and plus one not best. Fine. So the end's cancel one's cancelling your leftover with just the one. So this now becomes and this is exactly what we wanted because now we can rewrite this as eight of the end less than eight of the n plus one. And that's what exactly what it means for the sequence to be increasing. So that resolves party. Let's go on to the next one for party suffer party. Now let's use this value of A and this value of b in the inequality from part B to show the following that eight of the two ends the two in term of this sequence or, in other words, every even term of this sequence is less than four. So once again using B. So there's our beat of the end. You have your n plus one and then you're a minus. And then times B. So again, this's coming from B. Pardon me, and now licious simplify. So inside the bracket here we'LL have n plus one minus and minus a house and then less than one. So this just becomes a half, so it's more supply that's at the other side. And then let's go ahead and square both sides. And this is what we want here because this is equivalent to saying a to end his lesson for Recall the definition of a n wass. And so this implies that these two are equal and that resolves part deep. Let's go on to the next one for party. This is where we'LL use part c and D so show that a N is less than four for all him. So let's fight and show this. So solution softened Parsi. We showed that Anne was increasing, which we can write as a N is less than and plus one and then from party. We showed all the even terms. We're less than four. So in particular d implies if chaos, even because even means that Kay's of the form to end for some him. So now we'd like to show that this is also true if an Izod So let's go buy contradiction. Suppose there exist and R K. Let's say K equals two and plus one every odd numbers of this form with a king bigger than or equal to four. So that would be the contradictions over supposing crewed by contradiction. Then we have the following. We have four less than or equal to K K. But since the sequence is increasing, we have this inequality. Here, however, I can now rewrite. This's a two one plus two using this, and that's a two and plus one. However, this is a even sub spirit. So from party, this has to be less than four. And here's the contradiction, because now we have four is less than or equal to four. And this contradicts the fact that we had a K and odd cave with this and therefore every K even or odd, we must have a K listen for and that resolves party. We have one more part. Let's go on to the next page. So here, let's go ahead and use theorems wealth to show that the limit exists the limit of an what? So let's give a solution for this final part. So from Parsi, we know that the sequence and increases from Part E. We know that the sequence is founded above because we showed and was listened for for all in so theorems. Whoa, which is monitoring sequence their own says that if you have an increasing sequence that's founded above or a decreasing sequence that's founded below, then it converges. So since ours is increasing inbounded above, the sequence also emerges, and that resolves part F and the entire problem.

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