Respuesta :

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Answer:

The limit of the infinite series is equal to zero.

The nth term test is inconclusive ∵ the limit is equal to 0.

By the Comparison Test, this sum diverges.

General Formulas and Concepts:
Calculus

Limits

  • Limit Rule [Variable Direct Substitution]:                                                 [tex]\displaystyle \lim_{x \to c} x = c[/tex]

Series Comparison Tests

  • nth Term Test
  • Direct Comparison Test (DCT)

Step-by-step explanation:

Step 1: Define

[tex]\displaystyle \sum^{\infty}_{n = 1} \frac{3n^5}{6n^6 + 1}[/tex]

Step 2: Find Convergence

  1. [Series] Define:                                                                                            [tex]\displaystyle a_n = \frac{3n^5}{6n^6 + 1}[/tex]
  2. [Series] Set up [nth Term Test]:                                                                   [tex]\displaystyle \sum^{\infty}_{n = 1} \frac{3n^5}{6n^6 + 1} \rightarrow \lim_{n \to \infty} \frac{3n^5}{6n^6 + 1}[/tex]
  3. [nth Term Test] Evaluate limit [Limit Rule - VDS]:                                       [tex]\displaystyle \lim_{n \to \infty} \frac{3n^5}{6n^6 + 1} = 0[/tex]
  4. [nth Term Test] Determine Conclusiveness:                                              [tex]\displaystyle 0 ,\ \sum^{\infty}_{n = 1} \frac{3n^5}{6n^6 + 1} \ \text{can't be concluded that it converges}[/tex]

Therefore, the nth term test is inconclusive and another test must be done.

Step 3: Find Convergence Pt. 2

  1. [DCT] Condition 1 [Define comparing series]:                                            [tex]\displaystyle \frac{1}{2} \sum^{\infty}_{n = 1} \frac{1}{n}[/tex]
  2. [DCT] Condition 1 [Test convergence of comparing series]:                   [tex]\displaystyle p = 1 ,\ \frac{1}{2} \sum^{\infty}_{n = 1} \frac{1}{n} \ \text{divergent by p-series (harmonic series)}[/tex]

∴ since the comparison series is divergent, then our original series is also divergent according to the Direct Comparison Test.

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Learn more about limits: https://brainly.com/question/26091024

Learn more about Taylor Series: https://brainly.com/question/23558817

Topic: AP Calculus BC (Calculus I + II)

Unit: Taylor Series

semsee45

Answer:

zero

Step-by-step explanation:

L’Hospital’s rule states that for two functions f(x) and g(x), if either:

[tex]\ \lim_{x \to \infty} f(x)= \lim_{x \to \infty} g(x)=0, \textsf{or}[/tex]

[tex]\ \lim_{x \to \infty} f(x)=\pm \infty \ \textsf{and} \ \lim_{x \to \infty} g(x)\pm \infty[/tex]

[tex]\textsf{Then provided that} \ \lim_{x \to \infty} \dfrac{f(x)}{g(x)} \textsf{ exists, }\lim_{x \to \infty} \dfrac{f(x)}{g(x)}=\lim_{x \to \infty} \dfrac{f'(x)}{g'(x)}[/tex]  

[tex]\lim_{n \to \infty} \dfrac{3n^5}{6n^6+1} \rightarrow \dfrac{\infty}{\infty}[/tex]

[tex]\textsf{Let } f(n)=3n^5 \textsf{ and let } g(n)=6n^6+1[/tex]

[tex]\implies f^{'}(n)=15n^4 \textsf{ and } g^{'} (n)=36n^5[/tex]

By L’Hopital’s rule:

[tex]\lim_{n \to \infty} \dfrac{3n^5}{6n^6+1}=\lim_{n \to \infty} \dfrac{15n^4}{36n^5}= \lim_{n \to \infty} \dfrac{5}{12n}=0[/tex]

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