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.


Let $P_n = (1+1/n)^n$. Then

$$\log{P_n} = n \log{\left ( 1+\frac1{n} \right )} = n \left (\frac1{n} - \frac1{2 n^2} + \cdots \right) = 1-\frac1{2 n} + \cdots$$

$$P_n = e^{1-1/(2 n)+\cdots} = e \left (1-\frac1{2 n} + \cdots \right ) $$


$$\lim_{n \to \infty} n (e-P_n) = \frac{e}{2} $$

  • $\begingroup$ 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 $e/2$? $\endgroup$ – Jsevillamol Jun 1 '15 at 20:06
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    $\begingroup$ @Jsevillamol: use the Taylor series for the exponential: $e^{-x} = 1-x + \cdots$. $\endgroup$ – Ron Gordon Jun 1 '15 at 20:08
  • $\begingroup$ I don't see how it applies. Expanding $e^{(n log(1+\frac{1}{n}))}$ only gives me a terrible headache. $\endgroup$ – Jsevillamol Jun 1 '15 at 20:25
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    $\begingroup$ @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? $\endgroup$ – Ron Gordon Jun 1 '15 at 20:28
  • $\begingroup$ 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. $\endgroup$ – Jsevillamol Jun 1 '15 at 21:46

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