I came across that limit:
$\lim_{s\to\infty} s\bigg(\big(1+\frac{1}{s}\big)^{s} - e\bigg)$ I tried to solve it using l'Hospital's rule:
$\lim_{s\to\infty} s\bigg(\big(1+\frac{1}{s}\big)^{s} - e\bigg) = \lim_{t\to0} \frac{1}{t} \bigg(\big(1+t\big)^{\frac{1}{t}} - e\bigg) = \lim_{t\to0} e^{\frac{1}{t}\log(1+t)}\big(\frac{1}{t(t+1)} - \frac{\log(1+t)}{t^{2}}\big)$
I used l'Hospital's rule after the second equality.
However I got nothing really valuable. Do you have any ideas how to solve that problem?

  • $\begingroup$ Use $\log(1+x)=x-\frac12x^2+o(x^2)$ when $x\to0$ and conclude. Warning: This yields an awfully quick solution. $\endgroup$
    – Did
    Jan 30, 2017 at 15:06
  • $\begingroup$ Likewise, $$\lim_{s\to+\infty}\left(\left(1+\frac1s\right)^s-e\right)s^2+\frac{e}2s=\frac{11e}{24}$$ Curious to see how the various approaches suggested below, allow to attack this one (the limited expansion $\log(1+x)=x-\frac12x^2+\frac13x^3+o(x^3)$ when $x\to0$ yields it instantly). $\endgroup$
    – Did
    Jan 30, 2017 at 18:17

4 Answers 4


Note that we can write

$$\begin{align} \left(1+\frac1s\right)^s&=e^{s\log\left(1+\frac1s\right)}\\\\ &=e^{\left(1-\frac{1}{2s}+O\left(\frac1{s^2}\right)\right)}\\\\ &=e\left(1-\frac{1}{2s}+O\left(\frac1{s^2}\right)\right) \end{align}$$


$$\begin{align} \lim_{s\to \infty}\left(s\left(\left(1+\frac1s\right)^s-e\right)\right)&=\lim_{s\to \infty}\left(-\frac12 e+O\left(\frac1s\right)\right)\\\\ &= -\frac12 e \end{align}$$

  • $\begingroup$ Same idea at the same time ! $\endgroup$ Jan 30, 2017 at 15:07
  • $\begingroup$ @ClaudeLeibovici Indeed. $\endgroup$
    – Mark Viola
    Jan 30, 2017 at 15:24
  • $\begingroup$ @Abhinavchakraborty Please avoid qualifying as nonrigorous, fully rigorous arguments that you fail to get. $\endgroup$
    – Did
    Jan 30, 2017 at 16:03
  • $\begingroup$ @Did Didier, I must have missed a comment that has since been deleted. But much appreciative of yours. -Mark $\endgroup$
    – Mark Viola
    Jan 30, 2017 at 16:30
  • 2
    $\begingroup$ @user144921 $e^{1+x}=e\,e^x$ and $e^x=(1+x+O(x^2))$. $\endgroup$
    – Mark Viola
    Feb 5, 2017 at 17:41

If you dont want to write all those ugly oh's all over the place you could write your expression as

$$es\frac{e^{s\ln(1+\frac{1}{s})-1}-1}{ s\ln(1+\frac{1}{s})-1} (s\ln(1+\frac{1}{s})-1)$$

and then

$$s(s\ln(1+\frac{1}{s})-1)\to -\frac{1}{2}$$

is the well known

$$\lim\limits_{x\to 0}\frac{\ln (1+x)-x}{x^2}=-\frac{1}{2}$$

  • $\begingroup$ "Ugly" as in, leading to the solution with no sweat and with no bizarre ingeniosity required? Why are we supposed to know that $\frac1{x^2}(\log(1+x)-x)\to-\frac12$ but not that $\log(1+x)=x-\frac12x^2+o(x^2)$? $\endgroup$
    – Did
    Jan 30, 2017 at 15:49
  • $\begingroup$ @did as in lacking in imagination and elegance. O dear, looks like I have touched a nerve. $\endgroup$ Jan 30, 2017 at 15:52
  • $\begingroup$ No nerve involved (sorry). What about answering the second question in my comment? $\endgroup$
    – Did
    Jan 30, 2017 at 15:53
  • $\begingroup$ This is the most rigorous ans to the question till now. People find ohs understanding and intuitive but it doesn't give us a rigorous proof. $\endgroup$ Jan 30, 2017 at 15:55
  • $\begingroup$ @did the limit at the end follow from l'Hospital and occurs quite often in these questions. $\endgroup$ Jan 30, 2017 at 15:58

Let us consider $$A= \frac{1}{t} \bigg(\big(1+t\big)^{\frac{1}{t}} - e\bigg) $$ where $t$ is small compared to $1$.

First, let us look at $$B=\big(1+t\big)^{\frac{1}{t}}\implies \log(B)=\frac{1}{t}\, \log(1+t)$$ Now, using Taylor expansions $$\log(1+t)=t-\frac{t^2}{2}+\frac{t^3}{3}+O\left(t^4\right)$$ $$\log(B)=1-\frac{t}{2}+\frac{t^2}{3}+O\left(t^3\right)$$ $$B=e^{\log(B)}=e-\frac{e t}{2}+\frac{11 e t^2}{24}+O\left(t^3\right)$$

I am sure that you can take it from here and find not only the limit but also how it is approached.

  • $\begingroup$ The line where u wrote B = e^(logB)= ?something?. But, e^x = 1 + x + x^2/2! +.... Then how did your e^(logB) evaluated to ?something?. $\endgroup$
    – Shobhit
    Jan 30, 2017 at 15:18
  • $\begingroup$ @Claude Leibovici How do you get $$B=e^{\log(B)}=e-\frac{e t}{2}+\frac{11 e t^2}{24}+O\left(t^3\right)$$ ? $\endgroup$
    – user144921
    Feb 5, 2017 at 17:15
  • 1
    $\begingroup$ @user144921. In more steps : $\log(B)=1-\frac{t}{2}+\frac{t^2}{3}+\cdots=\log(e)-(\frac{t}{2}-\frac{t^2}{3}+\cdots)$. So, $B=e\times\exp(-(\frac{t}{2}-\frac{t^2}{3}+\cdots))=e \times e^Y=e(1+Y+\frac{Y^2}2+\cdots)$. Replace $Y=-(\frac{t}{2}-\frac{t^2}{3}+\cdots)$ and expand again using binomial theorem. $\endgroup$ Feb 6, 2017 at 6:31

To sum up the topic and post the full solution:
$\lim\limits_{x\to \infty}s\bigg(\big(1+\frac{1}{s}\big)^{s}-e\bigg) = \lim\limits_{x\to \infty}\frac{\big(1+\frac{1}{s}\big)^{s}-e}{\frac{1}{s}} = \lim\limits_{x\to \infty}\frac{\big(1+\frac{1}{s}\big)^{s}\bigg(\log(1+\frac{1}{s})-s\frac{\frac{1}{s^{2}}}{1+\frac{1}{s}}\bigg)}{\frac{-1}{s^{2}}} = \lim\limits_{x\to \infty}\frac{\big(1+\frac{1}{s}\big)^{s}\big(\frac{1}{s+1}-\log(1+\frac{1}{s})\big)}{\frac{1}{s^{2}}} = \lim\limits_{x\to \infty}\big(1+\frac{1}{s}\big)^{s}\cdot\lim\limits_{x\to \infty}\bigg(\frac{\frac{1}{s+1}-\log(1+\frac{1}{s})}{\frac{1}{s^{2}}}\bigg)$
Now let's calculate only the second limit using l'Hospital's rule:
$\lim\limits_{x\to \infty}\bigg(\frac{\frac{1}{s+1}-\log(1+\frac{1}{s})}{\frac{1}{s^{2}}}\bigg)= \lim\limits_{x\to \infty}\frac{\frac{1}{(s+1)^{2}}-\frac{1}{s(s+1)}}{\frac{2}{s^{3}}}=\frac{-1}{2}\lim\limits_{x\to \infty}\frac{s^{2}}{s(s+1)^{2}}=\frac{-1}{2}$
$\lim\limits_{x\to \infty}s\bigg(\big(1+\frac{1}{s}\big)^{s}-e\bigg)=\frac{-e}{2}$
That's a pretty nice solution I think.


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