I want to find the value of $$\lim\limits_{n \to \infty}n\left(e-\left(1+\frac{1}{n}\right)^n\right)$$
I have already tried using L'Hôpital's rule, only to find a seemingly more daunting limit.
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$\begingroup$ You can find such limit through the Hermite-Hadamard inequality, look at math.stackexchange.com/questions/1307812/… $\endgroup$– Jack D'AurizioJun 1, 2015 at 20:30
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$\begingroup$ related: math.stackexchange.com/questions/73243/… $\endgroup$– HenryNov 19, 2020 at 15:43
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1 Answer
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Let $P_n = (1+1/n)^n$. Then
$$\log{P_n} = n \log{\left ( 1+\frac1{n} \right )} = n \left (\frac1{n} - \frac1{2 n^2} + \frac1{3 n^3} - \cdots \right) = 1-\frac1{2 n} + \frac1{3 n^2} - \cdots$$
$$\therefore P_n = e^1e^{-1/(2 n)+ 1/(3n^2)-\cdots} = e \left (1-\frac1{2 n} + \frac{11}{24 n^2} \cdots \right ) $$
Thus
$$\lim_{n \to \infty} n (e-P_n) = \lim_{n \to \infty} e \left(\frac12-\frac{11}{24n} \cdots\right)=\frac{e}{2} $$
answered Jun 1, 2015 at 19:52
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$\begingroup$ I don't fully understand the answer. In the first line I take that you are expanding $\log P_n$ as a Taylor series, but then in the second line, how do you bring down the exponent? And how does that imply that the limit equals to $e/2$? $\endgroup$ Jun 1, 2015 at 20:06
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1$\begingroup$ @Jsevillamol: use the Taylor series for the exponential: $e^{-x} = 1-x + \cdots$. $\endgroup$ Jun 1, 2015 at 20:08
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$\begingroup$ I don't see how it applies. Expanding $e^{(n log(1+\frac{1}{n}))}$ only gives me a terrible headache. $\endgroup$ Jun 1, 2015 at 20:25
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2$\begingroup$ @Jsevillamol: Take an aspirin and have another look. I tried to break everything down one step at a time. Are you familiar with the series for log and exp? $\endgroup$ Jun 1, 2015 at 20:28
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1$\begingroup$ Got lost on the fact that the exponent and the factor in line 2 are not the same, even though they share the first two terms. Thanks, Gordon. $\endgroup$ Jun 1, 2015 at 21:46