(a) Triangle Inequality Let a and b be real numbers. Then (b) Reverse Triangle Inequality Let a and b be real numbers. Then lal - IM S la -M CONVERGENCE OF SEQUENCES AND SERIES 80 Problem 65. (a) Prove Lemma 0/ Hint: For the Reverse Triangle Inequality, consider Jal - la -b+./

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### Convergence of Sequences and Series

**Problem 64.** Let \( c_n = c \) (i.e., \( c, c, c, \ldots \)) be a constant sequence. Show that \(\lim_{n \to \infty} c_n = c\).

In proving the familiar limit theorems, the following will prove to be a very useful tool.

**Lemma 1.**
(a) **Triangle Inequality**: Let \( a \) and \( b \) be real numbers. Then 

\[ |a + b| \leq |a| + |b|. \]

(b) **Reverse Triangle Inequality**: Let \( a \) and \( b \) be real numbers. Then 

\[ ||a| - |b|| \leq |a - b|. \]

---

**Problem 65.**
(a) *Prove Lemma 1.* [Hint: For the Reverse Triangle Inequality, consider \( |a| = |a - b + b|. \)]

(b) *Show* \( ||a| - |b|| \leq |a - b| \). [Hint: You want to show \( |a| - |b| \leq |a - b| \) and \(-(|a| - |b|) \leq |a - b|. \)]

**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 this 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
Transcribed Image Text:Certainly! Here's a transcription suitable for an educational website: --- ### Convergence of Sequences and Series **Problem 64.** Let \( c_n = c \) (i.e., \( c, c, c, \ldots \)) be a constant sequence. Show that \(\lim_{n \to \infty} c_n = c\). In proving the familiar limit theorems, the following will prove to be a very useful tool. **Lemma 1.** (a) **Triangle Inequality**: Let \( a \) and \( b \) be real numbers. Then \[ |a + b| \leq |a| + |b|. \] (b) **Reverse Triangle Inequality**: Let \( a \) and \( b \) be real numbers. Then \[ ||a| - |b|| \leq |a - b|. \] --- **Problem 65.** (a) *Prove Lemma 1.* [Hint: For the Reverse Triangle Inequality, consider \( |a| = |a - b + b|. \)] (b) *Show* \( ||a| - |b|| \leq |a - b| \). [Hint: You want to show \( |a| - |b| \leq |a - b| \) and \(-(|a| - |b|) \leq |a - b|. \)] **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 this 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
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