Convergence of the Sequences (-1)^n and (-1)^n/n

The sequences (-1)ⁿ and (-1)ⁿ/n are special type of sequences. In this page, we will study the convergence of the sequences (-1)^n and (-1)^n/n.

Convergence of (-1)ⁿ

Question: Discuss the convergence of the sequence {(-1)ⁿ}.

Answer:

The sequence {(-1)ⁿ} can be written as follows:

{(-1)ⁿ} = {-1, 1, -1, 1, -1, 1, …}.

Therefore, the sequence {(-1)ⁿ} oscillates between -1 and 1. So it does not converge to a definite limit. Hence, {(-1)ⁿ} is not convergent.

Convergence of (-1)ⁿ/n

Question: Discuss the convergence of the sequence $\left\{ \dfrac{(-1)^n}{n} \right\}$.

Answer:

We have:

-1 ≤ (-1)ⁿ ≤ 1 for all n

Dividing both sides by n, we get that

$-\dfrac{1}{n} \leq \dfrac{(-1)^n}{n} \leq \dfrac{1}{n}$ for all n

Now, taking limit on sides, we obtain that

$- \lim\limits_{n \to \infty} \dfrac{1}{n} \leq \lim\limits_{n \to \infty} \dfrac{(-1)^n}{n} \leq \lim\limits_{n \to \infty} \dfrac{1}{n}$

⇒ $0 \leq \lim\limits_{n \to \infty} \dfrac{(-1)^n}{n} \leq 0$

This implies the limit of the sequence (-1)ⁿ/n lies between 0 and 0. Therefore, lim_n→∞ (-1)ⁿ/n = 0, by squeeze theorem/sandwich theorem on limits.

So {(-1)ⁿ/n} is a convergent sequence converging to 0.

Solved Problems

$\boxed{\color{blue}\textbf{Question}:}$ Test the convergence of the sequence $\left\{ 1+\dfrac{(-1)^n}{n} \right\}$.

$\boxed{\color{red}\textbf{Answer:}}$ We have:

-1 ≤ (-1)ⁿ ≤ 1 ∀ n

Dividing both sides by n,

$-\dfrac{1}{n} \leq 1+\dfrac{(-1)^n}{n} \leq \dfrac{1}{n}$ ∀ n

Adding 1 to both sides,

$1-\dfrac{1}{n} \leq 1+\dfrac{(-1)^n}{n} \leq 1+\dfrac{1}{n}$ ∀ n

Taking limit on sides,

$\lim\limits_{n \to \infty} \left(1-\dfrac{1}{n} \right) \leq$ $\lim\limits_{n \to \infty} \left( 1+\dfrac{(-1)^n}{n} \right) \leq$ $\lim\limits_{n \to \infty} \left(1+\dfrac{1}{n} \right)$

⇒ $1 \leq \lim\limits_{n \to \infty} \dfrac{(-1)^n}{n} \leq 1$

Thus by squeeze theorem on limits, it follows that lim_n→∞ 1+(-1)ⁿ/n = 1. Hence, the sequence {1+(-1)ⁿ/n} is convergent with limit 1.