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How can I show that $$ \lim_{n\to\infty} \frac{1}{\ln(n)} = 0 \quad (n\geq 2) $$ using $\varepsilon$ definition of convergence?, that is given $\varepsilon > 0$ find $K(\varepsilon)\in\mathbb N$ so that for all $n\geq K(\varepsilon)$ we have that $|\frac{1}{\ln(n)}| < \varepsilon$. It is clear that the sequence is bounded and it is monotone decreasing so it necessarily converges. Thanks in advance.

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up vote 1 down vote accepted

You want $$ \frac{1}{\ln n}<\epsilon \quad\Leftrightarrow \quad \ln n>\frac{1}{\epsilon} \quad\Leftrightarrow\quad n>e^{1/\epsilon}. $$ Now if you want to be explicit for $K(\epsilon)$, use ceiling function.

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Rather $\lceil e^{1/\epsilon} \rceil + 1$, to get a strict inequality: $n\geq \lceil e^{1/\epsilon} \rceil + 1 \geq e^{1/\epsilon} + 1 > e^{1/\epsilon}$. – 1015 Feb 2 '13 at 16:25

Let $\varepsilon>0$ and $k \in \mathbb{R}$, $k \geq 2$, such that $\frac{1}{k} \leq \varepsilon$. Then

$$\ln(e^k) = k$$


$$\left| \frac{1}{\ln(n)} \right| \leq \left| \frac{1}{\ln (e^k)} \right| = \frac{1}{k} \leq \varepsilon$$

for all $n \geq \lfloor e^k \rfloor +1=:K$.

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Suppose that there is $\epsilon > 0$ such that for all $K > 0$ we have $|1/\log(K)| \geq \epsilon$. This implies that $|\log(K)| \leq 1/\epsilon$.

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