Solution:
\( \frac{n-1}{n} \leq \frac{n+(-1)^n}{n} \leq \frac{n+1}{n} \)
And
\( \lim \limits _{n \rightarrow \infty} \frac{n-1}{n} =\lim \limits _{n \rightarrow \infty} 1 - \frac{1}{n}=1 \)
Likewise
\( \lim \limits _{n \rightarrow \infty} \frac{n+1}{n} =\lim \limits _{n \rightarrow \infty} 1 + \frac{1}{n}=1 \)
Therefore,
\( \lim \limits _{n \rightarrow \infty} \frac{n+(-1)^n}{n} =1 \) by Sandwich Thm.
Therefore \( a_n= \frac{n+(-1)^n}{n} \) converges.
