Question

Answer the question with explanation;​

Answer 1

Answer:

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

Step-by-step explanation:

We compute

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

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

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

Answer 2

Answer:

$\displaystyle \sum^{\infty}_{n = 1} \frac{n^2}{(n + 1)^2} = \text{div}$

General Formulas and Concepts:

Calculus

Limits

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

Series Convergence Tests

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

Step-by-step explanation:

*Note:

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

Rules:

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

Step 1: Define

Identify

$\displaystyle \sum^{\infty}_{n = 1} \frac{n^2}{(n + 1)^2}$

Step 2: Find Convergence

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

∴ by the nth Term Test, the series diverges.

Topic: AP Calculus BC (Calculus I + II)

Unit: Convergence Tests