Answer:
The sum [tex]\displaystyle \sum^{\infty}_{n = 1} \frac{2^n}{(n + 1)!}[/tex] converges absolutely.
General Formulas and Concepts:
Calculus
Limits
- Limit Rule [Variable Direct Substitution]: [tex]\displaystyle \lim_{x \to a} x = a[/tex]
Series Convergence Tests
- Ratio Test: [tex]\displaystyle \lim_{n \to \infty} \bigg| \frac{a_{n + 1}}{a_n} \bigg|[/tex]
Step-by-step explanation:
Step 1: Define
Identify.
[tex]\displaystyle \sum^{\infty}_{n = 1} \frac{2^n}{(n + 1)!}[/tex]
Step 2: Find Convergence
- [Series] Define: [tex]\displaystyle a_n = \frac{2^n}{(n + 1)!}[/tex]
- [Series] Set up [Ratio Test]: [tex]\displaystyle \sum^{\infty}_{n = 1} \frac{2^n}{(n + 1)!} \rightarrow \lim_{n \to \infty} \bigg| \frac{2^{n + 1}}{(n + 2)!} \cdot \frac{(n + 1)!}{2^{n}} \bigg|[/tex]
- [Ratio Test] Simplify: [tex]\displaystyle \lim_{n \to \infty} \bigg| \frac{2^{n + 1}}{(n + 2)!} \cdot \frac{(n + 1)!}{2^{n}} \bigg| = \lim_{n \to \infty} \bigg| \frac{2}{n + 2} \bigg|[/tex]
- [Ratio Test] Evaluate Limit [Limit Rule - Variable Direct Substitution]: [tex]\displaystyle \lim_{n \to \infty} \bigg| \frac{2^{n + 1}}{(n + 2)!} \cdot \frac{(n + 1)!}{2^{n}} \bigg| = 0[/tex]
- [Ratio Test] Check Conclusiveness: [tex]\displaystyle 0 < 1 \ \checkmark[/tex]
∴ since 0 is less than 1, the Ratio Test defines the sum [tex]\displaystyle \sum^{\infty}_{n = 1} \frac{2^n}{(n + 1)!}[/tex]absolutely convergent.
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Learn more about the Ratio Test: https://brainly.com/question/16654521
Learn more about Taylor Series: https://brainly.com/question/23558817
Topic: AP Calculus BC (Calculus I + II)
Unit: Taylor Series
