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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?

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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. – 2017-01-30
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    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). – 2017-01-30

4 Answers 4

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.

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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,

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

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    Same idea at the same time ! – 2017-01-30
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    @ClaudeLeibovici Indeed. – 2017-01-30
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    @Abhinavchakraborty Please avoid qualifying as nonrigorous, fully rigorous arguments that you fail to get. – 2017-01-30
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    @Did Didier, I must have missed a comment that has since been deleted. But much appreciative of yours. -Mark – 2017-01-30
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    Indeed, Abhinav posted a comment (now deleted) to your answer, similar to the one on Rene's answer, only slightly more strident. (+1 to your answer.) – 2017-01-30
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    @Did Thank you! Much appreciative. – 2017-01-30
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    @Dr. MV Why is $$ e^{\left(1-\frac{1}{2s}+O\left(\frac1{s^2}\right)\right)} =e\left(1-\frac{1}{2s}+O\left(\frac1{s^2}\right)\right) $$ ? – 2017-02-05
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    @user144921 $e^{1+x}=e\,e^x$ and $e^x=(1+x+O(x^2))$. – 2017-02-05
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    Oh... thank you. Just look weird but now it is clear. – 2017-02-05
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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

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

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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 $\log(1+x)=x-\frac12x^2+o(x^2)$? – 2017-01-30
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    @did as in lacking in imagination and elegance. O dear, looks like I have touched a nerve. – 2017-01-30
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    No nerve involved (sorry). What about answering the second question in my comment? – 2017-01-30
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    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. – 2017-01-30
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    @did the limit at the end follow from l'Hospital and occurs quite often in these questions. – 2017-01-30
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    @ReneSchipperus You probably realize that the "occurs quite often" argument is entirely reversible. – 2017-01-30
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    @Abhinavchakraborty Proofs by little-ohs are not rigorous? This is a new one, I must say. – 2017-01-30
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    @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. – 2017-01-30
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    @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? – 2017-01-30
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    @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)? – 2017-01-30
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    @Hendrra Comment addressed at the OP? – 2017-01-30
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    @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. – 2017-01-30
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    @Dr.MV apologies if my comments have been in any sense disrespectful, will take care not to repeat in the future. – 2017-01-30
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    @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? – 2017-01-30
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    You have bracket error, look closely at the expression in my answer, the second $s$ only muliplies the log and not the one. – 2017-01-30
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    @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. – 2017-01-30
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    @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. – 2017-01-30
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    Its equal to $s^2\ln(1+\frac{1}{s})-s$ hope thats clearer. – 2017-01-30
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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.

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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?. – 2017-01-30
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    @Claude Leibovici How do you get $$B=e^{\log(B)}=e-\frac{e t}{2}+\frac{11 e t^2}{24}+O\left(t^3\right)$$ ? – 2017-02-05
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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. – 2017-02-06