Answer
$\{a_{n}\}$ diverges.
Work Step by Step
ASSUMPTION:
The sequence $ \displaystyle \{a_{n}\}=\{(-1)^{n}(1-\frac{1}{n})\}$ has a limit L, $\displaystyle \lim_{n\rightarrow\infty}\{a_{n}\}=L.$
Then, by definition, if we choose $\epsilon=0.5$, there must exist an index N such that
for $n\gt N $ (for sequence terms after the Nth), $|L-a_{n}|\lt 1.$
Let M be the first even number after N.
Then, $a_{M}=1-\displaystyle \frac{1}{M}=\frac{M-1}{M}$,
and $a_{M+1}=-(1-\displaystyle \frac{1}{M+1})=\frac{1}{M+1}-1=\frac{1-(M+1)}{M+1}=-\frac{M}{M+1}$
$\left.\begin{array}{lll}
|L-a_{M}|\lt 1/2 & and, & |L-a_{M+1}|\lt 1/2\\
-1/2\lt L-\frac{M-1}{M}\lt 1/2 & & -1/2\lt L+\frac{M}{M+1}\lt 1/2\\
-M/2\lt ML-(M-1)\lt M/2 & & -(M+1)/2\lt L(M+1)+M\lt(M+1)/2\\
-M/2+(M-1)\lt ML\lt M/2+(M-1) & & -(M+1)/2-M\lt L(M+1)\lt(M+1)/2-M\\
M/2-1\lt ML\lt 3M/2-1 & & -\frac{1}{2}-\frac{M}{M+1}\lt L\lt\frac{1}{2}-\frac{M}{M+1}\\
1/2-\frac{1}{M}\lt L\lt 3/2-\frac{1}{M} & &
\end{array}\right.$
Now, letting $ M\rightarrow\infty$, it follows that
$L\displaystyle \in(\frac{1}{2},\frac{3}{2})$ and $L\displaystyle \in(-\frac{1}{2},\frac{1}{2})$
which is impossible.
Our assumption was wrong, $\displaystyle \lim_{n\rightarrow\infty}\{a_{n}\}$ does not exist.
$\{a_{n}\}$ diverges.
