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 $\left\{{\displaystyle \frac{(-1{)}^n}{n}}\right\}$.

Answer:

We have:

-1 ≤ (-1)ⁿ ≤ 1 for all n

Dividing both sides by n, we get that

$-{\displaystyle \frac{1}{n}}\le {\displaystyle \frac{(-1{)}^n}{n}}\le {\displaystyle \frac{1}{n}}$ for all n

Now, taking limit on sides, we obtain that

$-\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{n}}\le \underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{(-1{)}^n}{n}}\le \underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{n}}$

⇒ $0\le \underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{(-1{)}^n}{n}}\le 0$

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

$\boxed{{\displaystyle \mathbf{\text{Question}}:}}$ Test the convergence of the sequence $\{1+{\displaystyle \frac{(-1{)}^n}{n}}\}$.

$\boxed{{\displaystyle \mathbf{\text{Answer:}}}}$ We have:

-1 ≤ (-1)ⁿ ≤ 1 ∀ n

Dividing both sides by n,

$-{\displaystyle \frac{1}{n}}\le {\displaystyle \frac{(-1{)}^n}{n}}\le {\displaystyle \frac{1}{n}}$ ∀ n

Adding 1 to both sides,

$1-{\displaystyle \frac{1}{n}}\le 1+{\displaystyle \frac{(-1{)}^n}{n}}\le 1+{\displaystyle \frac{1}{n}}$ ∀ n

Taking limit on sides,

$\underset{n\to \mathrm{\infty }}{lim}(1-{\displaystyle \frac{1}{n}})\le$ $\underset{n\to \mathrm{\infty }}{lim}(1+{\displaystyle \frac{(-1{)}^n}{n}})\le$ $\underset{n\to \mathrm{\infty }}{lim}(1+{\displaystyle \frac{1}{n}})$

⇒ $1\le \underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{(-1{)}^n}{n}}\le 1$

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.