Math for Non-Geeks/Exercises: Subsequences, Accumulation points, and Cauchy sequences

oben Exercises: Subsequences Subsequences of the e-sequence Calculate the limit Calculate the limit Calculate the limit using 1) Subtask 1: As we know that the sequence has the limit , for the subsequence holds that: Subtask 2: With and being a subsequence of the sequence from 1, we get . Using the computation rules for limits, we get Subtask 3: For holds that with the sequence from subtask 1. As , is bounded from above, there exists an (actually ) with for all . Therefore, is an upper bound of . We even get As , using the squeeze theorem we get Exercises: Accumulation points Accumulation points Determine all accumulation points of the sequences with and . Regarding : As converges to , all of its subsequences converge to . The sequence alternates between and . Hence, its accumulation points are and . Regarding : For even -s, is a subsequence of the -sequence . For odd -s, is a subsequence of the sequence , which converges to . As , one accumulation point of is . Furthermore, we have . Hence, another accumulation point of is . These are the only accumulation points of because every other subsequence of either diverges or converges to or . We have , as . Thus, an accumulation point of is . Furthermore, we have , as . Therefore, another accumulation point of is . These are the only accumulation points of because every other subsequence of either diverges or converges to or . Exercises: Lim inf and Lim sup Aufgabenliminflimsup Lim inf and Lim sup Determine Lim inf and Lim sup of the sequences with and . We start off investigating whether is bounded from above/below. This sequence is bounded from below, but not from above. Hence, Lim inf equals the smallest accumulation point of . We can determine it looking at the subsequence, including the odd sequence elements. As the sequence is not bounded from above, Lim sup equals . As all of its subsequences are bounded, is bounded from above and below. In order to determine Lim sup and Lim inf of , we need to find the limits of its subsequences. As , we have . Hence, is bounded from below. then equals the smallest accumulation point of the sequence . For the subsequence holds that Therefore, is an accumulation point of . There can't be other accumulation points, as all other subsequences of our sequence either converge to or diverge. Hence, we have . Alternative solution: We have , as the odd sequence elements of monotonically decrease to . It follows that With , we get The Archimedean property then states that there exists a with for all . Therefore, is not bounded from above and . Regarding : We have Hence, only has the accumulation points and . Furthermore, is bounded, as all subsequences are bounded as well. We get and . Lim inf and Lim sup 2 ordered Let be a sequence of positive real numbers. Prove the following inequalities: Let be a sequence with Compute , , and . Derive the following statement from 1): Let be a sequence of positive real numbers with . Then we have . Prove the following limits: ordered Let be a positive sequence with We need to show the following inequalities: W.l.o.g. let and . By definition, is the smallest and the largest accumulation point of the sequence . Hence, for all () there exists an with Applying the inequality times setting , for an arbitrary with holds that: The expression in the middle is a telescoping product. Except for the numerator and the denominator , everything cancels out. Multiplying and taking the -th root leads to With , the term on the left side converges to and the one on the right side converges to for . Therefore, we get and As can be chosen arbitrarily small, we get     and     As the inequality holds by definition, the claim follows. For all we have , as the numerator and denominator are always positive. Hence, 0 is a lower bound of the quotient sequence . With being odd and even, it follows that for Hence, there exists a subsequence of the quotient sequence converging to zero. It follows that With being even and being odd, we get for Hence, the quotient series doesn't have an upper bound and With being odd, we have and being even leads to Hence, has the accumulation points and . It follows that and As , we get Using 1) and the squeeze theorem, we have It follows that . 1st limit: Let . It follows that Using part 3, we have . 2nd limit: Let . It follows that Again using 3. , we get . Take a look at the inequalities proven in subtask 1. In subtask 2 even holds that Taking a look at the limits from subtask 4 and using reciprocals, we get: Exercises: Cauchy sequences Cauchy criterion for sequences Let and be a sequence of real numbers with for all . Show that the sequence converges. Let and . Show that the sequence converges and compute its limit. Subtask 1: We will show that is a Cauchy sequence. The Cauchy criterion then states that converges. Using the assumptions made, we have: If and , there exists a with . It follows that From the inequalities follows that From the Archimedean axiom we can derive that for all there exists an such that for all holds Hence, is a Cauchy sequence and therefore convergent. Subtask 2: For the sequence holds true that With , the criterion from subtask 1 is met. Hence, converges. The limit can be computed as follows: Analogous to subtask 1 - but without taking the absolute values -, we have Applying the formula times leads to Hence, we get Using the computation rules for limits, we have unten