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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?

Hendrra
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  • Use $\log(1+x)=x-\frac12x^2+o(x^2)$ when $x\to0$ and conclude. Warning: This yields an awfully quick solution. – Did Jan 30 '17 at 15:06
  • 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). – Did Jan 30 '17 at 18:17

4 Answers4

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Note that we can write

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

Therefore,

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

Mark Viola
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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

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

Rene Schipperus
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  • "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 $\log(1+x)=x-\frac12x^2+o(x^2)$? – Did Jan 30 '17 at 15:49
  • @did as in lacking in imagination and elegance. O dear, looks like I have touched a nerve. – Rene Schipperus Jan 30 '17 at 15:52
  • No nerve involved (sorry). What about answering the second question in my comment? – Did Jan 30 '17 at 15:53
  • 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. – Abhinav chakraborty Jan 30 '17 at 15:55
  • @did the limit at the end follow from l'Hospital and occurs quite often in these questions. – Rene Schipperus Jan 30 '17 at 15:58
  • @ReneSchipperus You probably realize that the "occurs quite often" argument is entirely reversible. – Did Jan 30 '17 at 16:01
  • @Abhinavchakraborty Proofs by little-ohs are not rigorous? This is a new one, I must say. – Did Jan 30 '17 at 16:01
  • @Did I mean that much justification is needed in each and every step of the argument involving oh notation, so if we start writing all the steps needed to justify our steps it would be cumbersome. Whereas this answer uses limited number of theorems to achieve the goal. – Abhinav chakraborty Jan 30 '17 at 16:14
  • @Abhinavchakraborty Sorry but the opposite holds. To begin with, which hypothesis should one check before applying L'Hopital? Note also your start(l)ing position that proofs by little-ohs are non rigorous (by essence?). Do you still hold it? – Did Jan 30 '17 at 16:33
  • @did That's a very nice solution but I have to ask something. You wrote that: \lim\limits_{x\to 0}\frac{\ln (1+x)-x}{x^2}=-\frac{1}{2} but: \lim\limits_{s\to \infty}s\big(s(\log(1+\frac{1}{s})-1)\big) = \lim\limits_{s\to \infty}\big(s^{2}(\log(1+\frac{1}{s})-s)\big) = \lim\limits_{x\to \0}\big(\frac{1}{x^{2}}(\log(1+x)-\frac{1}{x})\big)? – Hendrra Jan 30 '17 at 16:33
  • @Hendrra Comment addressed at the OP? – Did Jan 30 '17 at 16:35
  • @Hendrra Please use dollar signs also in the comments, as your comment is difficult to read, but I think the answer is to set $s=\frac{1}{x}$ and then the two expressions become equal. – Rene Schipperus Jan 30 '17 at 16:38
  • @Dr.MV apologies if my comments have been in any sense disrespectful, will take care not to repeat in the future. – Abhinav chakraborty Jan 30 '17 at 16:53
  • @ReneSchipperus I'm really sorry about that! I meant: $\lim\limits_{x\to 0}\frac{\ln (1+x)-x}{x^2}=-\frac{1}{2}$ but: $\lim\limits_{s\to \infty}s\big(s(\log(1+\frac{1}{s})-1)\big) = \lim\limits_{s\to \infty}\big(s^{2}(\log(1+\frac{1}{s})-s)\big) = \lim\limits_{x\to 0}\big(\frac{1}{x^{2}}(\log(1+x)-\frac{1}{x})\big)$ So I think that there should be $\frac{1}{x}$ instead of $x$ Am I right? – Hendrra Jan 30 '17 at 17:04
  • You have bracket error, look closely at the expression in my answer, the second $s$ only muliplies the log and not the one. – Rene Schipperus Jan 30 '17 at 17:10
  • @Abhinavchakraborty OK. Note that the tone of comments is one thing, but, at least to me, cogency of the mathematical arguments is even more important. – Did Jan 30 '17 at 18:08
  • @ReneSchipperus In your expression (I mean the second one: $s(s\ln(1+\frac{1}{s})-1)\to -\frac{1}{2}$) the second $s$ multiplies both log and 1. – Hendrra Jan 30 '17 at 21:22
  • Its equal to $s^2\ln(1+\frac{1}{s})-s$ hope thats clearer. – Rene Schipperus Jan 30 '17 at 21:27
2

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.

Claude Leibovici
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  • 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?. – Shobhit Jan 30 '17 at 15:18
  • @Claude Leibovici How do you get $$B=e^{\log(B)}=e-\frac{e t}{2}+\frac{11 e t^2}{24}+O\left(t^3\right)$$ ? – user144921 Feb 05 '17 at 17:15
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    @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. – Claude Leibovici Feb 06 '17 at 06:31
2

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:
$\lim\limits_{x\to \infty}\bigg(\frac{\frac{1}{s+1}-\log(1+\frac{1}{s})}{\frac{1}{s^{2}}}\bigg)= \lim\limits_{x\to \infty}\frac{\frac{1}{(s+1)^{2}}-\frac{1}{s(s+1)}}{\frac{2}{s^{3}}}=\frac{-1}{2}\lim\limits_{x\to \infty}\frac{s^{2}}{s(s+1)^{2}}=\frac{-1}{2}$
Thus:
$\lim\limits_{x\to \infty}s\bigg(\big(1+\frac{1}{s}\big)^{s}-e\bigg)=\frac{-e}{2}$
That's a pretty nice solution I think.

Hendrra
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