Problem 66. Prove TheoremZ
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Author:Erwin Kreyszig
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![**Convergence of Sequences and Series**
**Problem 65.**
(a) *Prove Lemma 7.* [Hint: For the Reverse Triangle Inequality, consider |a| = |a - b + b|.]
(b) Show ||a| - |b|| ≤ |a - b|. [Hint: You want to show |a| - |b| ≤ |a - b| and -(|a| - |b|) ≤ |a - b|.]
**Theorem 7.** If \(\lim_{n \to \infty} a_n = a\) and \(\lim_{n \to \infty} b_n = b\), then \(\lim_{n \to \infty} (a_n + b_n) = a + b\).
We will often informally state this theorem as “the limit of a sum is the sum of the limits.” However, to be absolutely precise, what it says is that if we already know that two sequences converge, then the sequence formed by summing the corresponding terms of those two sequences will converge and, in fact, converge to the sum of those individual limits. We’ll provide the scrapwork for the proof of this and leave the formal write-up as an exercise. Note the use of the triangle inequality in the proof.
**SCRAPWORK:** If we let \(\varepsilon > 0\), then we want \(N\) so that if \(n > N\), then \(|(a_n + b_n) - (a + b)| < \varepsilon\). We know that \(\lim_{n \to \infty} a_n = a\) and \(\lim_{n \to \infty} b_n = b\), so we can make \(|a_n - a|\) and \(|b_n - b|\) as small as we wish, provided we make \(n\) large enough. Let’s go back to what we want, to see if we can close the gap between what we know and what we want. We have
\[
|(a_n + b_n) - (a + b)| = |(a_n - a) + (b_n - b)| \leq |a_n - a| + |b_n - b|
\]
by the triangle inequality. If we make this whole thing less than \(\varepsilon\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc074fc7e-4e48-46f1-bfc4-aee6ba4e6262%2F068379dc-3b86-4511-9b7b-7b45da64cae6%2Fhx662x4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Convergence of Sequences and Series**
**Problem 65.**
(a) *Prove Lemma 7.* [Hint: For the Reverse Triangle Inequality, consider |a| = |a - b + b|.]
(b) Show ||a| - |b|| ≤ |a - b|. [Hint: You want to show |a| - |b| ≤ |a - b| and -(|a| - |b|) ≤ |a - b|.]
**Theorem 7.** If \(\lim_{n \to \infty} a_n = a\) and \(\lim_{n \to \infty} b_n = b\), then \(\lim_{n \to \infty} (a_n + b_n) = a + b\).
We will often informally state this theorem as “the limit of a sum is the sum of the limits.” However, to be absolutely precise, what it says is that if we already know that two sequences converge, then the sequence formed by summing the corresponding terms of those two sequences will converge and, in fact, converge to the sum of those individual limits. We’ll provide the scrapwork for the proof of this and leave the formal write-up as an exercise. Note the use of the triangle inequality in the proof.
**SCRAPWORK:** If we let \(\varepsilon > 0\), then we want \(N\) so that if \(n > N\), then \(|(a_n + b_n) - (a + b)| < \varepsilon\). We know that \(\lim_{n \to \infty} a_n = a\) and \(\lim_{n \to \infty} b_n = b\), so we can make \(|a_n - a|\) and \(|b_n - b|\) as small as we wish, provided we make \(n\) large enough. Let’s go back to what we want, to see if we can close the gap between what we know and what we want. We have
\[
|(a_n + b_n) - (a + b)| = |(a_n - a) + (b_n - b)| \leq |a_n - a| + |b_n - b|
\]
by the triangle inequality. If we make this whole thing less than \(\varepsilon\
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