Question

Find the limit superior and the limit inferior of the following sequences

a) {(1 +

1

𝑛

)

𝑛+1

}

b) {

(−1)

𝑛

𝑛2

}

Answer 1

Solution:

Assume the sequences are as follows:

(a) $\{a_n\}=\{\frac{1}{n}+(-1)^n\}$$|a_n|=|\frac{1}{n}+(-1)^n|\leq2$ for all integers $n\geq1$. Hence $\{a_n\}$ is bounded.

The first few terms are $\{0,\frac{3}{2},\frac{-2}{3},\frac{5}{4},\frac{-4}{5},\frac{7}{6},\cdots\}$. The subsequence $\{a_{2n}\}=\{\frac{3}{2},\frac{5}{4},\frac{7}{6},\cdots\}$ converges to 1 and the subsequence $\{a_{2n-1}\}=\{0,\frac{-2}{3},\frac{-4}{3},\cdots\}$ converges to -1. Hence $S=\{-1,1\}$

$\liminf a_n=\inf(S)=-1\\ \limsup a_n=\sup(S)=1$

(b)

$x_{n}=(-1)^{n} \frac{1+n}{n}$

Since $\frac{1+n}{n}=1+\frac{1}{n} \geq 0$ for any $n \in \mathbb{N}$ we have

$\lim _{n \rightarrow \infty} \sup x_{n}=\lim _{k \rightarrow \infty} x_{2 k}=1$

$\limsup _{n \rightarrow \infty} x_{n}=\lim _{k \rightarrow \infty} x_{2 k}=1$

and

$\liminf _{n \rightarrow \infty} x_{n}=\lim _{k \rightarrow \infty} x_{2 k+1}=-1$