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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 $x_{n} \rightarrow 0$.
in Mathematics by Bronze Status (8,688 points) | 9 views

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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
\left|\frac{1}{\ln n}-0\right|<\varepsilon \\
\text { So, } 1 / \ln n \rightarrow 0
by Bronze Status (8,688 points)

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