# Limit at Infinity: $\lim\limits_{n\rightarrow\infty}n\left(1-\frac{1}{\ln(n)}\right)^n$

Maple says that this limit is zero but I can't prove it. Any help or tips would be appreciated.

$\displaystyle\lim_{n\rightarrow\infty}n\left(1-\frac{1}{\ln(n)}\right)^n$

• Hint: take logs, rearrange, L'H. – vadim123 Jan 28 '15 at 19:38
• See this. – Mhenni Benghorbal Jan 28 '15 at 19:41

$$\lim_{n\to\infty}n\bigg(1-\frac{1}{\ln(n)}\bigg)^n=\lim_{n\to\infty}n\bigg[\bigg(1-\frac{1}{\ln(n)}\bigg)^{\ln(n)}\bigg]^{\frac{n}{\ln(n)}}=\lim_{n\to\infty}\frac{n}{e^{\frac{n}{\ln(n)}}}=0$$ Where I used the fact that $$\lim_{a_n\to\infty}\left(1-\frac{1}{a_n}\right)^{a_n}=e^{-1}$$ and for $\ n\to\infty$ $$\ n>>\ln(n)$$
$$n\left(1-\frac{1}{\ln(n)}\right)^n=n\exp\left(n\ln\left(1-\frac{1}{\ln(n)}\right)\right)\sim_\infty n\exp\left(-\frac{n}{\ln n}\right)\xrightarrow{n\to\infty}0$$ and the last limit can be proved using the L'Hôpital's rule.
• Thanks. I see that $\displaystyle\lim_{n\rightarrow\infty}-\ln\left(1-\frac{1}{\ln(n)}\right)\ln(n)=1$ by L'Hospital's rule so this does indeed work. – Lucas Jan 28 '15 at 19:55