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这样证明是可以的。下面給出一另种证法:
固定\(d>0,\,\)取正整数\(k\)使\(d>\frac{1}{k}.\) 于是\(\,0<\small\dfrac{1}{n^d}\le \dfrac{1}{n^{1/k}}\)
对\(\,\varepsilon>0,\,\)取\(\,N_{\varepsilon}=\lceil\varepsilon^{-k}\rceil,\,\)则\(\,n>N_{\varepsilon}\implies \varepsilon>\small\dfrac{1}{n^{1/k}}\ge\dfrac{1}{n^d}>0.\)
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