Question

Find whether the following sequences converge or not


A) {2+(-1)^n}

B)(4n^3+n)/(2n^3+7n)

Answer 1

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Answer on Question #85212 – Math – Real Analysis

Question

Find whether the following sequences converge or not

A) $\{2 + (-1)^n\}$

B) $(4n^3 + n) / (2n^3 + 7n)$

Solution

A) $\lim_{n\to \infty}(2 + (-1)^n) = \left\{ \begin{array}{ll}2 + 1, & \text{if } n = 2k\\ 2 - 1, & n = 2k - 1 \end{array} \right. = \left\{ \begin{array}{ll}3, & \text{if } n = 2k\\ 1, & n = 2k - 1 \end{array} \right.$

$$\lim_{n\to \infty}(2 + (-1)^n) = 1 \neq 3 = \varlimsup_{n\to \infty}(2 + (-1)^n), \text{ hence}$$

hence the sequence $\{2 + (-1)^n : n \geq 1\}$ does not converge.

**Answer**: this sequence $\left(\{2 + (-1)^n : n \geq 1\}\right)$ does not converge.

B) $\lim_{n\to \infty}\frac{4n^3 + n}{2n^3 + 7n} = \lim_{n\to \infty}\frac{4n^2 + 1}{2n^2 + 7} = \lim_{n\to \infty}\frac{4 + \frac{1}{n^2}}{2 + \frac{7}{n^2}} = \frac{4 + 0}{2 + 0} = 2$, (in other words, there exists $\lim_{n\to \infty}\frac{4n^3 + n}{2n^3 + 7n}$),

hence the sequence $\left\{\frac{4n^3 + n}{2n^3 + 7n} : n \geq 1\right\}$ converges.

**Answer**: this sequence $\left(\left\{\frac{4n^3 + n}{2n^3 + 7n} : n \geq 1\right\}\right)$ converges.