Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

3 Answers

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.

share|improve this answer
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
add comment

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

share|improve this answer
add comment

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

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.