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

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

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  • 2
    $\begingroup$ Hint: take logs, rearrange, L'H. $\endgroup$ – vadim123 Jan 28 '15 at 19:38
  • $\begingroup$ See this. $\endgroup$ – Mhenni Benghorbal Jan 28 '15 at 19:41
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$$\lim_{n\to\infty}n\bigg(1-\frac{1}{\ln(n)}\bigg)^n=\lim_{n\to\infty}n\bigg[\bigg(1-\frac{1}{\ln(n)}\bigg)^{\ln(n)}\bigg]^{\frac{n}{\ln(n)}}=\lim_{n\to\infty}\frac{n}{e^{\frac{n}{\ln(n)}}}=0$$ Where I used the fact that $$\lim_{a_n\to\infty}\left(1-\frac{1}{a_n}\right)^{a_n}=e^{-1}$$ and for $\ n\to\infty$ $$\ n>>\ln(n)$$

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  • $\begingroup$ happy to help, when I can $\endgroup$ – Mosk Jan 28 '15 at 21:15
  • $\begingroup$ I think it could be better to transform the expression to verify if the limit exists and if it exists then write "lim of the expression = 0" It's more speed and better I think :) $\endgroup$ – ParaH2 Feb 6 '15 at 11:57
  • $\begingroup$ I don't understand your second equality, it looks as if you took the limit of the inner sequence in the brackets, and then took the limit of the entire sequence again, which is generally a false thing to do. What is your justification in this case? $\endgroup$ – user7610 Feb 6 '15 at 13:08
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We have

$$n\left(1-\frac{1}{\ln(n)}\right)^n=n\exp\left(n\ln\left(1-\frac{1}{\ln(n)}\right)\right)\sim_\infty n\exp\left(-\frac{n}{\ln n}\right)\xrightarrow{n\to\infty}0$$ and the last limit can be proved using the L'Hôpital's rule.

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  • $\begingroup$ 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. $\endgroup$ – Lucas Jan 28 '15 at 19:55

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