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. $\endgroup$– vadim123Jan 28, 2015 at 19:38
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$\begingroup$ See this. $\endgroup$– Mhenni BenghorbalJan 28, 2015 at 19:41
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2 Answers
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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. $\endgroup$– LucasJan 28, 2015 at 19:55
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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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$\begingroup$ 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 :) $\endgroup$– ParaH2Feb 6, 2015 at 11:57
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$\begingroup$ 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? $\endgroup$– user7610Feb 6, 2015 at 13:08