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Consider the real sequence $\left\{x_{n}\right\}$ given by $$x_{n}=\frac{1}{\ln n}$$ for $n \geq 2$. Provide an $\varepsilon-N$ proof that $x_{n} \rightarrow 0$.
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Given $\varepsilon>0$, let $N=e^{1 / \varepsilon}$. Then, for all $n>N=e^{1 / \varepsilon}$, we have that
$$e^{1 / \varepsilon}<n$$
Taking the natural logarithm of both sides, we have that
$$\frac{1}{\varepsilon}<\ln n$$
Since $n \geq 2, \ln n>0$ and $\varepsilon>0 .$ So we can cross divide to get that
$$\frac{1}{\ln }<\varepsilon$$
This is equivalent to
$$\begin{array}{l} \left|\frac{1}{\ln n}-0\right|<\varepsilon \\ \text { So, } 1 / \ln n \rightarrow 0 \end{array}$$
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