How find this limit $$\displaystyle \lim_{n\to\infty}n\left(1-\dfrac{\ln n}{n}\right)^n$$


Let $$f(n) = n\left(1 - \frac{\log n}{n}\right)^{n}\tag{1}$$ and we need to calculate the limit of $f(n)$ as $n \to \infty$ through integer values. The best approach would be to analyze the behavior of $\log f(n)$. Clearly we have $$\log f(n) = \log n + n\log\left(1 - \frac{\log n}{n}\right)\tag{2}$$ and if $n > 1$ we know that $$0 < \frac{\log n}{n} < 1\tag{3}$$ We also know that the following inequality $$x < -\log(1 - x) < \frac{x}{1 - x}\tag{4}$$ holds for $0 < x < 1$. Replacing $x$ with $(\log n)/n$ in the above inequality we get $$\frac{\log n}{\log n - n} < \log\left(1 - \frac{\log n}{n}\right) < -\frac{\log n}{n}$$ Multiplying by $n$ we get $$\frac{n\log n}{\log n - n} < n\log\left(1 - \frac{\log n}{n}\right) < -\log n\tag{5}$$ Using $(2)$ we now have $$\frac{(\log n)^{2}}{\log n - n} < \log f(n) < 0\tag{6}$$ Now we can see that \begin{align} A &= \lim_{n \to \infty}\frac{(\log n)^{2}}{\log n - n}\notag\\ &= \lim_{n \to \infty}\dfrac{(\log n)^{2}}{n\left(\dfrac{\log n}{n} - 1\right)}\notag\\ &= \lim_{n \to \infty}\dfrac{(\log n)^{2}}{n}\cdot\dfrac{1}{\dfrac{\log n}{n} - 1}\notag\\ &= 0\cdot\frac{1}{0 - 1} = 0\tag{7} \end{align}

In the above derivation we have used the standard result that $$\lim_{n \to \infty}\frac{(\log n)^{a}}{n^{b}} = 0\tag{8}$$ for any positive numbers $a, b$. Using Squeeze theorem in equation $(6)$ and noting the equation $(7)$ we get that $\log f(n) \to 0$ as $n \to \infty$. Hence $f(n) \to 1$ as $n \to \infty$. The desired limit is therefore equal to $1$.

Update: Some other answers make use of the symbol $\sim$, but it is wrong unless provided with further justification. The definition of the symbol $\sim$ in the current context is like this. If $$\lim_{n \to \infty}\frac{a(n)}{b(n)} = 1$$ then we write $a(n) \sim b(n)$. And because of this definition we can replace $a(n)$ by $b(n)$ while calculating limits where $a(n)$ is used in the multiplicative context. To be more specific if we have $a(n) \sim b(n)$ then while calculating the limit of an expression like $a(n)c(n)$ we can replace $a(n)$ by $b(n)$ and just calculate the limit of $b(n)c(n)$ to get final answer. This is justified because we can write $$\lim_{n \to \infty}a(n)c(n) = \lim_{n \to \infty}\frac{a(n)}{b(n)}\cdot b(n)c(n) = \lim_{n \to \infty}1\cdot b(n)c(n)$$ Replacement of $a(n)$ by $b(n)$ in other contexts must be justified by further analysis and it may generate wrong answer also.

Further Update: In case you have access to powerful technique of series expansions then the limit can be calculated easily as follows: \begin{align} \log f(n) &= \log n + n\log\left(1 - \frac{\log n}{n}\right)\notag\\ &= \log n - n\left\{\frac{\log n}{n} + \frac{(\log n)^{2}}{2n^{2}} + o\left(\frac{(\log n)^{2}}{n^{2}}\right)\right\}\notag\\ &= -\frac{(\log n)^{2}}{2n} + o\left(\frac{(\log n)^{2}}{n}\right)\notag \end{align} Using the fact that $(\log n)^{2}/n \to 0$ as $n \to \infty$ we can see that $\log f(n) \to 0$ and hence $f(n) \to 1$ as $n \to \infty$. My preferred approach is to use simpler tools (theorems on algebra of limits, Squeeze theorem etc), but advanced tools like series expansions and L'Hospital give the answer very easily.


we have $$a_n=n\left(1-\frac{\ln(n)}{n}\right)^{n}=ne^{n\ln \left (1-\frac{\ln(n)}{n}\right)} $$ and because $\ln(1-\frac{\ln(n)}{n})\sim\frac{-\ln(n)}{n}$ hence : $$a_n\sim 1 $$ and from here it follows that $a_n\to 1$

  • $\begingroup$ How is $\ln\left(1-\dfrac{\ln(n)}{n}\right)\sim \dfrac{\ln(n)}{n}$ for $n\to\infty$ ? $\endgroup$ – Prasun Biswas Apr 5 '15 at 21:40
  • $\begingroup$ because $ln(1-x)\sim -x$ when $x\to 0$ $\endgroup$ – Elaqqad Apr 5 '15 at 21:41
  • $\begingroup$ Yes, now it's fixed. $\endgroup$ – Prasun Biswas Apr 5 '15 at 21:41
  • $\begingroup$ Please see the "Update" in my answer regarding the use of $\sim$ in limit calculation. $\endgroup$ – Paramanand Singh Apr 9 '15 at 3:53

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