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Problem 61. Use the definition to prove lim,-o n100 # 0. n+100

Problem 61. Use the definition to prove lim,-o n100 # 0. n+100

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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#61
**Convergence of Sequences and Series** We can negate this definition to prove that a particular sequence does not converge to zero. **Example 5.** Use the definition to prove that the sequence \((1 + (-1)^n)_{n=0}^\infty = (2, 0, 2, 0, 2, \ldots)\) does not converge to zero. Before we provide this proof, let’s analyze what it means for a sequence \((s_n)\) to not converge to zero. Converging to zero means that any time a distance \(\varepsilon > 0\) is given, we must be able to respond with a number \(N\) such that \(|s_n| < \varepsilon\) for every \(n > N\). To have this not happen, we must be able to find some \(\varepsilon > 0\) such that no choice of \(N\) will work. Of course, if we find such an \(\varepsilon\), then any smaller one will fail to have such an \(N\), but we only need one to mess us up. If you stare at the example long enough, you see that any \(\varepsilon\) with \(0 < \varepsilon \leq 2\) will cause problems. For our purposes, we will let \(\varepsilon = 2\). **Proof:** Let \(\varepsilon = 2\) and let \(N \in \mathbb{N}\) be any integer. If we let \(k\) be any non-negative integer with \(k > \frac{N}{2}\), then \(n = 2k > N\), but \(|1 + (-1)^n| = 2\). Thus no choice of \(N\) will satisfy the conditions of the definition for this \(\varepsilon\) (namely that \(|1 + (-1)^n| < 2\) for all \(n > N\)) and so \(\lim_{n \to \infty} (1 + (-1)^n) \neq 0\). **Problem 60.** Negate the definition of \(\lim_{n \to \infty} s_n = 0\) to provide a formal definition for \(\lim_{n \to \infty} s_n \neq
Transcribed Image Text:**Convergence of Sequences and Series** We can negate this definition to prove that a particular sequence does not converge to zero. **Example 5.** Use the definition to prove that the sequence \((1 + (-1)^n)_{n=0}^\infty = (2, 0, 2, 0, 2, \ldots)\) does not converge to zero. Before we provide this proof, let’s analyze what it means for a sequence \((s_n)\) to not converge to zero. Converging to zero means that any time a distance \(\varepsilon > 0\) is given, we must be able to respond with a number \(N\) such that \(|s_n| < \varepsilon\) for every \(n > N\). To have this not happen, we must be able to find some \(\varepsilon > 0\) such that no choice of \(N\) will work. Of course, if we find such an \(\varepsilon\), then any smaller one will fail to have such an \(N\), but we only need one to mess us up. If you stare at the example long enough, you see that any \(\varepsilon\) with \(0 < \varepsilon \leq 2\) will cause problems. For our purposes, we will let \(\varepsilon = 2\). **Proof:** Let \(\varepsilon = 2\) and let \(N \in \mathbb{N}\) be any integer. If we let \(k\) be any non-negative integer with \(k > \frac{N}{2}\), then \(n = 2k > N\), but \(|1 + (-1)^n| = 2\). Thus no choice of \(N\) will satisfy the conditions of the definition for this \(\varepsilon\) (namely that \(|1 + (-1)^n| < 2\) for all \(n > N\)) and so \(\lim_{n \to \infty} (1 + (-1)^n) \neq 0\). **Problem 60.** Negate the definition of \(\lim_{n \to \infty} s_n = 0\) to provide a formal definition for \(\lim_{n \to \infty} s_n \neq
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Step 1

Limits:

 

Let's pretend we have a function f(x). The value that a function achieves when the variable x approaches a specific value, say a, is referred to as its limit. 'a' denotes a value that has already been allocated. It's referred to as

limx→af(x)=l

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