Respuesta :

rokettojanpu

Answer:

The statement in the question is wrong. The series actually diverges.

Step-by-step explanation:

We compute

[tex]\lim_{n\to\infty}\frac{n^2}{(n+1)^2}=\lim_{n\to\infty}\left(\frac{n^2}{n^2+2n+1}\cdot\frac{1/n^2}{1/n^2}\right)=\lim_{n\to\infty}\frac1{1+2/n+1/n^2}=\frac1{1+0+0}=1\ne0[/tex]

Therefore, by the series divergence test, the series [tex]\sum_{n=1}^\infty\frac{n^2}{(n+1)^2}[/tex] diverges.

EDIT: To VectorFundament120, if [tex](x_n)_{n\in\mathbb N}[/tex] is a sequence, both [tex]\lim x_n[/tex] and [tex]\lim_{n\to\infty}x_n[/tex] are common notation for its limit. The former is not wrong but I have switched to the latter if that helps.

Space

Answer:

[tex]\displaystyle \sum^{\infty}_{n = 1} \frac{n^2}{(n + 1)^2} = \text{div}[/tex]

General Formulas and Concepts:

Calculus

Limits

  • Special Limit Rule [Coefficient Power Method]:                                         [tex]\displaystyle \lim_{x \to \pm \infty} \frac{ax^n}{bx^n} = \frac{a}{b}[/tex]

Series Convergence Tests

  • nth Term Test:                                                                                               [tex]\displaystyle \sum^{\infty}_{n = 1} a_n \rightarrow \lim_{n \to \infty} a_n[/tex]
  • Integral Test:                                                                                                 [tex]\displaystyle \sum^{\infty}_{n = a} f(n) \rightarrow \int\limits^{\infty}_a {f(x)} \, dx[/tex]
  • P-Series:                                                                                                         [tex]\displaystyle \sum^{\infty}_{n = 1} \frac{1}{n^p}[/tex]
  • Direct Comparison Test (DCT)
  • Limit Comparison Test (LCT)
  • Alternating Series Test (AST)
  • Ratio Test:                                                                                                     [tex]\displaystyle \sum^{\infty}_{n = 0} a_n \rightarrow \lim_{n \to \infty} \bigg| \frac{a_{n + 1}}{a_n} \bigg|[/tex]

Step-by-step explanation:

*Note:

Always apply the nth Term Test as the first test to use for convergence.

Rules:

  1. If  [tex]\displaystyle \lim_{n \to \infty} S_n = 0[/tex], then the nth Term Test is inconclusive.
  2. If  [tex]\displaystyle \lim_{n \to \infty} S_n = l[/tex]  (some number l), then the series is divergent by the nth Term Test.

Step 1: Define

Identify

[tex]\displaystyle \sum^{\infty}_{n = 1} \frac{n^2}{(n + 1)^2}[/tex]

Step 2: Find Convergence

  1. Substitute in variables [nth Term Test]:                                                       [tex]\displaystyle \sum^{\infty}_{n = 1} \frac{n^2}{(n + 1)^2} \rightarrow \lim_{n \to \infty} \frac{n^2}{(n + 1)^2}[/tex]
  2. Expand:                                                                                                         [tex]\displaystyle \lim_{n \to \infty} \frac{n^2}{(n + 1)^2}= \lim_{n \to \infty} \frac{n^2}{n^2 + 2n + 1}[/tex]
  3. Evaluate limit [Special Limit Rule - Coefficient Power Method]:                 [tex]\displaystyle \lim_{n \to \infty} \frac{n^2}{(n + 1)^2} = 1[/tex]
  4. Compute [nth Term Test]:                                                                             [tex]\displaystyle \sum^{\infty}_{n = 1} \frac{n^2}{(n + 1)^2} = \text{div}[/tex]

∴ by the nth Term Test, the series diverges.

Topic: AP Calculus BC (Calculus I + II)

Unit: Convergence Tests

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