## University Calculus: Early Transcendentals (3rd Edition)

Let us make the 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 $a_{M+1}=-(1-\displaystyle \frac{1}{M+1})=\frac{1}{M+1}-1=\frac{1-(M+1)}{M+1}=-\frac{M}{M+1}$ $\left[\begin{array}{lll} |L-a_{M}|\lt 0.5 & & \\ -1/2\lt L-\frac{M-1}{M}\lt 1/2 & & \\ -M/2\lt ML-(M-1)\lt M/2 & & \\ -M/2+(M-1)\lt ML\lt M/2+(M-1) & & \\ M/2-1\lt ML\lt 3M/2-1 & & \\ 1/2-\frac{1}{M}\lt L\lt 3/2-\frac{1}{M} & & \end{array}\right]$ and, $\left[\begin{array}{l} |L-a_{M+1}|\lt 1/2\\ -1/2\lt L+\frac{M}{M+1}\lt 1/2\\ -(M+1)/2\lt L(M+1)+M\lt(M+1)/2\\ -(M+1)/2-M\lt L(M+1)\lt(M+1)/2-M\\ -\frac{1}{2}-\frac{M}{M+1}\lt L\lt\frac{1}{2}-\frac{M}{M+1} \end{array}\right]$ 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, so $\displaystyle \lim_{n\rightarrow\infty}\{a_{n}\}$ does not exist. $\{a_{n}\}$ diverges.