\(求證:lim_{n\to \infty }\frac{1}{n^\alpha}=0,\alpha>0\)
\(證明:\)
\(分為兩種情況考慮,情況1:\alpha>=1,情況2:\alpha<1\)
\(情況1:當\alpha\geq 1\)
\(|\frac{1}{n^\alpha}-0|=\frac{1}{n^\alpha}<\frac{1}{n}\)
\(若需\frac{1}{n}<\epsilon\)
\(即n>\frac{1}{\epsilon},即可\)
\(即,取N=[\frac{1}{\epsilon}]+1\)
\(則當\quad n>N時,即有\quad \frac{1}{n} < \frac{1}{N}<\epsilon\)
