Mathematics, 14.08.2021 14:00 caveman171

Answer the question with explanation;

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Answer from: jazzy2971

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 from: alvaradovanessa14

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

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:

If $\displaystyle \lim_{n \to \infty} S_n = 0$ , then the nth Term Test is inconclusive.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

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

Answer from: Quest

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Answer from: Quest

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