Math for Non-Geeks/The squeeze theorem

oben The squeeze theorem is a powerful tool to determine the limit of a complicated sequence. It is based on comparison to simpler sequences, for which the limit is easily determinable. Motivation thumb|Convergence proof for a root sequence (video in German) The intuition behind this theorem is quite simple: We are given a complicated sequence and want to know whether it converges. Often, one can leave out terms in the complicated sequence and get some simpler sequences and . If is a lower bound and an upper bound, then is "caught" in the space between both sequences. If both sequences converge to the same limit , they "squeeze" together this space to this single point and has no other option than to converge towards , as well. center|image about the principle of the sandwich lemma thumb|class=noprint|A ham & cheese sandwich You may also visualize this theorem by a "ham & cheese sandwich" (see image on the right). The upper and lower bounds and act as two "slices of toast", which confine , i.e. the "filling". If you squeeze the toast slices together, the filling in between will also squeeze to this point. The squeeze theorem thumb|Squeeze theorem (video in German, YouTube video from the YouTube channel Maths CA). The theorem reads as follows: Squeeze theorem Let be any sequence. If two sequences and exist with for all and for some common limit , then will also converge towards this limit . Let and be such that and for some . We need to prove , i.e. for each there is an , such that for all . Let be arbitrary. By assumption, the convergences and hold. Hence, there are two thresholds and with for all (lower bounding sequence) and for all (upper bounding sequence). Now, there is for . For each term we distinguish the two cases (above the limit) and (below the limit). In case there is The other case is treated analogously: So if the maximum of both thresholds is reached or exceeded by ( i.e. and hold), then for , and for , So in either case for all , which establishes convergence. The inequality does not need to be satisfied for all . Assume a finite number of sequence elements are not "caught between and ". Then, after some finite all of these finitely many elements will have been passed and the further sequence elements with will indeed be caught as . We illustrate the squeeze theorem using the root sequence , whose limit is for an arbitrary but fixed constant . Two cases need to be distinguished: Let us choose some , such that . Then, for all with there is , so , i.e. we have an upper bound. A lower bound is given by for any . In the previous chapter, we have established the limit . So the squeeze theorem yields for any fixed standing inside the root. In this case, , so we are led to the above case, replacing by . Hence, also and the limit theorems (quotient) yield: A useful special case We often encounter 0 as a sequence limit (null sequence). Since the squeeze theorem can be used to prove any to be a limit of a sequence , it can also be used for . In particular, for any converging to , the sequence must converge to 0: Special case of the squeeze theorem Let be a real sequence, , and a null sequence with for all . Then . We set . For the constant sequence there is (absolute values are always non-negative). Since the squeeze theorem also implies . Therefore, is a null sequence, which is equivalent to the statement that converges to . This special case of the squeeze theorem is also referred to as direct comparison argument for sequences . Let us take a look at . Not a simple case: both and tend to . But we can show that their difference converges to 0, since it is bounded by: Let us prove this inequality. There is so and the difference is positive, i.e. . Now, we remove the root by squaring it: The last inequality holds for all and implies But now, is a null sequence, so the squeeze theorem proves our assertion. Squeeze theorem: examples and problems Squeeze theorem: example & exercise 1 Squeeze theorem 1 Let us consider the sequence with thumb|The first 6 elements of the sequence , and We take a look at the first sequence elements . They seem to be alternately positive and negative, and they tend to zero. We use as lower or upper bound and . Then, so we caught for all . In addition, . The squeeze theorem hence implies . Question Which further sequences and can be used for catching and squeezing ? We could also use and . Or and for any real (not necessarily or ). Squeeze theorem 1 Investigate whether the sequence converges, using the squeeze theorem, where with For there is a with . By monotonicity of the power function, Hence The limit theorems yield and the root rule gives . Hence, by means of the squeeze theorem . Squeeze theorem: example & exercise 2 Squeeze theorem 2 Next, we prove that the sequence with converges to 0. thumb|The first elements of , and This sequence is bounded from below as all elements are positive for all . An upper bound can be found by multiplying out and : Obviously, both bounding sequences converge to zero: . So the squeeze theorem implies . Squeeze theorem 2 Use the squeeze theorem to show that and with both converge to 0. Part 1: The trick is multiplying out again. This time, for and : Since , the squeeze theorem implies . Part 2: Again, the elements are positive, i.e. . Bounding from above can be done using for all . This inequality can be shown by induction: Induction basis: . Induction step: . So for all , Since , the squeeze theorem implies . Squeeze theorem: example & exercise 3 Squeeze theorem 3 Now, let us consider the root sequences and with and First sequence: For , monotonicity of the -th root implies as well as For those two bounding sequences, the limit theorems imply and The squeeze theorem hence implies . Second sequence: For the root function is monotone, as well and We use the limit theorems for the upper and lower bounding sequences and So the squeeze theorem implies . Squeeze theorem 3 Use the squeeze theorem to determine the limits of the sequences and with Part 1: For the -th root is monotone, so further, for Hence, there is The limit theorems for the bounding sequences yield and So the squeeze theorem implies . Part 2: For we again have monotonicity of the -th root Further, The limit theorems for the bounding sequences render And by means of the squeeze theorem . Squeeze theorem: example & exercise 4 Squeeze theorem 4 Let be the sequence of finite sums Then, for all we bound: and hence For the lower bound, For the upper bound, So the squeeze theorem implies . Squeeze theorem 4 Investigate the limits of the two sequences of finite sums and defined by using the squeeze theorem. Part 1: For all we have: so we have some bounds For the lower bound, For the upper bound, So the squeeze theorem implies . Part 2: For all and there is: so we have the bounds The lower bound is just 0 and for the upper bound, So the squeeze theorem implies . Squeeze theorem: example & exercise 5 Squeeze theorem 5 Does the sequence converge? If yes, what is its limit? Provide proofs for your claims. Perhaps, you have already encountered the sequence which converges to Euler's number . This is an upper bound for , so with that knowledge, we may assert that it is bounded above. But what is the limit? Maybe, we should start by computing some initial sequence elements: It looks as if converges to . This gets particularly clear, if we plot the sequence elements in a diagram: center|500px|The first 20 elements of the sequence (1-n^(-2))^n What makes determining the limit hard is the exponent of . Otherwise, we could use some limit theorems or an epsilon proof and would be done quite fast. Luckily, there is a trick to get rid of the exponent: Bernoulli's inequality: The sequence is hence a lower bound on and it clearly converges to . We hence set as a lower bound for the squeeze theorem. How to get an upper bound converging to 1? This is quite easy, since is always smaller than : The expression lies in and therefore, also every positive integer power of it is smaller than . This includes being smaller than . So for the squeeze theorem, we have for all the inequality: Both the lower and the upper bounding sequences converge to the identical limit . So by means of the squeeze theorem, there must also be . Bernoulli's inequality implies And there is So we get the bounds Both bounding sequences converge to , so by means of the squeeze theorem, which was to be shown. If you are allowed to use and , then you can also obtain the result directly, using the third binomial formula: Squeeze theorem 5 Does the sequence converge? If yes, what is its limit? Prove your assertions. Again, Bernoulli's inequality can be used for getting rid of the exponent . There is so we have a bound from below converging to . But is already a lower bound converging to . So Bernoulli's inequality doesn't help us in this form. What we need is an upper bound converging to . This can be done by a little trick: we put expressions from the numerator in the denominator and apply Bernoulli's inequality afterwards: Hence, we have an upper bound . Dividing the numerator and denominator of the bound by gives a quotient of convergent sequences, both with limit , and the quotient rule gives . Therefore, we have two bounding sequences converging to and by the squeeze theorem . Let and . There is : The bounding sequences converge to and (by means of the limit theorem): So there is and the squeeze theorem implies . If the exponential series is known, one can also use the binomial theorem for a bound. A lower bound is given by . For the upper bound, we use the binomial theorem : This yields a suitable upper bound, and we get . The squeeze theorem then implies that converges to . We may use for a direct proof. The sequence is a subsequence of and converges to the same limit: Hence, for some , there is for all . The sequence can be recovered from this by taking the root . So for all : With , the squeeze theorem again implies that converges to . Examples & exercises: squeeze theorem for null sequences Squeeze theorem for null sequences 1 First, an easy example: we prove . We estimate by the null sequence as follows: Since , is a null sequence. We can use as an upper and as a lower bound for . The squeeze theorem then implies . Squeeze theorem for null sequences 2 Now, it gets a bit more complicated: We fix any and prove that . This can be done by bounding from above using a null sequence . The trick is to use Bernoulli's inequality, which will yield the upper bound : This implies So we have The squeeze theorem for null sequences directly implies . Squeeze theorem for null sequences Prove that . In order to use the squeeze theorem, we need to bound from above by a null sequence. The binomial theorem can be used for this bounding. In case : This inequality implies So the desired bound reads And by means of the squeeze theorem , which was to be shown. unten