9
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $a_n\to 1$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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