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(a) an = (b) an = (c) an = 4(2n) + 3n4 +6 3(2n) + n²+2n³¹ n5 + 3n² + 2n 4n4 + 2n² - n' n = n = 1. n² + 3n² + 2(n!) n³ + 2n² + (−1)^(n!)'

(a) an = (b) an = (c) an = 4(2n) + 3n4 +6 3(2n) + n²+2n³¹ n5 + 3n² + 2n 4n4 + 2n² - n' n = n = 1. n² + 3n² + 2(n!) n³ + 2n² + (−1)^(n!)'

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Determine whether each of the following sequences (an) converges, and find the limit of each convergent sequence. Name any results or rules that you use. You may use the basic null sequences listed in Theorem D7 from Unit D2. (a) an = (b) an (c) an = 4(2n) + 3n4 +6 3(2n) +n4+2n³¹ n5 + 3n² + 2n 2 4n4 + 2n² - n n = 1, 2, ... n = 1, 2, ... n² + 3n² + 2(n!) n³ + 2n² + (−1)n (n!)' n = 1, 2, ...
Transcribed Image Text:Determine whether each of the following sequences (an) converges, and find the limit of each convergent sequence. Name any results or rules that you use. You may use the basic null sequences listed in Theorem D7 from Unit D2. (a) an = (b) an (c) an = 4(2n) + 3n4 +6 3(2n) +n4+2n³¹ n5 + 3n² + 2n 2 4n4 + 2n² - n n = 1, 2, ... n = 1, 2, ... n² + 3n² + 2(n!) n³ + 2n² + (−1)n (n!)' n = 1, 2, ...
We now show that there are various basic types of sequences that are null. By applying the rules from the previous subsection to these 'basic null sequences', we can deduce the existence of many different null sequences without having to use the definition. It is important that you are familiar with these types of basic null sequences and are able to use them. Reading the proof that they are null may help you with this, but skim read it if you are short of time and return to it when time permits. Theorem D7 Basic null sequences The following sequences are null. (a) (1/nº), for p > 0. (b) (c), for c| < 1. (c) (nºch), for p > 0, c < 1. (d) (cn/n!), for c E R. (e) (nº/n!), for p > 0. Proof (a) To prove that (1/nº) is null for p > 0, we apply the Power Rule to the sequence (1/n), which we know is null. (b) To prove that (c) is null for c < 1, first note that it is sufficient to consider only the case 0 ≤ c < 1, because any sequence (an) is null if and only if the sequence (an) is null. If c= 0, then the sequence is obviously null. Thus we can assume that 0 < c < 1, so we can write where a > 0. C = 1 1 + a
Transcribed Image Text:We now show that there are various basic types of sequences that are null. By applying the rules from the previous subsection to these 'basic null sequences', we can deduce the existence of many different null sequences without having to use the definition. It is important that you are familiar with these types of basic null sequences and are able to use them. Reading the proof that they are null may help you with this, but skim read it if you are short of time and return to it when time permits. Theorem D7 Basic null sequences The following sequences are null. (a) (1/nº), for p > 0. (b) (c), for c| < 1. (c) (nºch), for p > 0, c < 1. (d) (cn/n!), for c E R. (e) (nº/n!), for p > 0. Proof (a) To prove that (1/nº) is null for p > 0, we apply the Power Rule to the sequence (1/n), which we know is null. (b) To prove that (c) is null for c < 1, first note that it is sufficient to consider only the case 0 ≤ c < 1, because any sequence (an) is null if and only if the sequence (an) is null. If c= 0, then the sequence is obviously null. Thus we can assume that 0 < c < 1, so we can write where a > 0. C = 1 1 + a
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