Question

What is the limit of the infinite series? ∞∑n=1 (3n^5 / 6n^6 + 1)

Answer 1

Answer:

zero

Step-by-step explanation:

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

$\ \lim_{x \to \infty} f(x)= \lim_{x \to \infty} g(x)=0, \textsf{or}$

$\ \lim_{x \to \infty} f(x)=\pm \infty \ \textsf{and} \ \lim_{x \to \infty} g(x)\pm \infty$

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

$\lim_{n \to \infty} \dfrac{3n^5}{6n^6+1} \rightarrow \dfrac{\infty}{\infty}$

$\textsf{Let } f(n)=3n^5 \textsf{ and let } g(n)=6n^6+1$

$\implies f^{'}(n)=15n^4 \textsf{ and } g^{'} (n)=36n^5$

By L’Hopital’s rule:

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

Answer 2

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]:                                                 $\displaystyle \lim_{x \to c} x = c$

Series Comparison Tests

• nth Term Test
• Direct Comparison Test (DCT)

Step-by-step explanation:

Step 1: Define

$\displaystyle \sum^{\infty}_{n = 1} \frac{3n^5}{6n^6 + 1}$

Step 2: Find Convergence

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

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

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

---

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

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

Topic: AP Calculus BC (Calculus I + II)

Unit: Taylor Series