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What is the limit of

$$\lim_{n\to+\infty}\left(n\left(1+\frac{1}{n}\right)^n-ne\right)$$

using the squeeze theorem, without the use of differentiation, integrals and series?

I don't know how to get rid of that '$n$' that is multiplying the whole equation. I tried expanding it with the binomial formula and see if there are any connection but couldn't find it. The solution from Wolfram Alpha was $-\frac{e}{2}$.

Neil hawking
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Account_For_Education
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    Can we use the formula $e=\sum_{k=0}^{\infty}\frac{1}{k!}$? Otherwise, what facts about $e$ are allowed in your problem? – Sangchul Lee Jun 13 '21 at 23:07
  • I would guess this question follows a proof that $(1+1/n)^n \to e$. Maybe this new one can be proved based on the given proof of that. – GEdgar Jun 13 '21 at 23:12
  • @SangchulLee Lets say this is a homework problem we were given after learning about the squeeze and Stoltz theorem as well as the basics of sequences. The only formula we were given including e was $\lim\limits_{n \to \infty} ( 1 + (1/n))^n) = e$ . And as i saw many of the problems we faced required some sort of inequalities or the binomial formula. – Account_For_Education Jun 13 '21 at 23:13
  • If you can explain a bit about how the limit $(1+\frac{1}{n})^n\to e$ was dealt with in the class, it would help us better understand what is intended in your problem. (The answer obviously depends on the properties of $e$, so we need to know what is known about $e$ in your context.) Also, I am interested in what version of Stolz theorem you learned, because, it may be tackled using a version of Stolz theorem. – Sangchul Lee Jun 13 '21 at 23:46
  • @SangchulLee Sorry for the wait I was absent. We implemented the limit for e as a definition. – Account_For_Education Jun 13 '21 at 23:57
  • @SangchulLee With some transformations I managed to get this limit $e\cdot \lim_\limits{n\to\infty}(n\cdot e^{n ln(1+\frac{1}{n} ) - 1} -1)$ Which gives me $\frac{-e}{2}$. Maybe we shouldn't use the squeeze theorem, but I still don't know how to finish the limit to get the $\frac{-1}{2}$. – Account_For_Education Jun 14 '21 at 00:43
  • If the Taylor approximation is available, then finding the limit is not a hassle, since $$e^{n \log(1+\frac{1}{n})-1}=e^{-\frac{1}{2n}+\dots}=1-\frac{1}{2n}+\dots,$$ where $\dots$ represents degree-1 Taylor remainder terms. If this powerful machinery is not available to us, then fingers are crossed and we have to resort to very tricky approaches such as the inequalities proposed in Robert's answer (though the answer did not articulate how the inequalities themselves are derived). I have a couple of ideas for elementary proofs, but none of them are short or intuitive. – Sangchul Lee Jun 14 '21 at 00:48
  • @SangchulLee I was making some other transformations and got to the point of it looking like this I think im near the end just need to figure out what the next transformation should be. $e \lim_{n\to\infty} \frac{e^{nln(1+1/n) - 1} - 1}{nln(1+\frac{1}{n}) - 1}n(nln(1+\frac{1}{n}) - 1)$ – Account_For_Education Jun 14 '21 at 01:22
  • Maybe transform $n = \frac{1}{x}$, so we can use $\lim_{x\to 0}(1+x)^{\frac{1}{x}} = e$ and tidy up the logarithms, I'm tired and will go now. – Account_For_Education Jun 14 '21 at 01:24
  • Computing the limit of the sequence $$a_n=n\left(n\log\left(1+\frac{1}{n}\right)-1\right)$$ is essentially equivalent to resolving the asymptotic formula $$\log\left(1+\frac{1}{n}\right)=\frac{1}{n}+\frac{a_n}{n^2},$$ which is more or less figuring out the second-order term in the Taylor approximation of $\log(1+x)$ about $x=0$. I am not sure if this can be achieved by without any calculus machinery, especially thinking that defining logarithm already requires calculus. That is why I would like to know more about the context, so we can fathom what is the intention behind this problem. – Sangchul Lee Jun 14 '21 at 01:28

1 Answers1

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Hints:

$\Big(n+\dfrac{1}{2}\Big) \cdot \Big(1+ \dfrac{1}{n}\Big)^n-ne > 0$.

$\Big(n+1\Big)\cdot \Big(1+\dfrac{1}{n}\Big)^n - \Big(n+\dfrac{1}{2}\Big)e < 0$.

Robert
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    I can confirm that both inequalities are true and will lead to a solution. I wonder how these inequalities are proved. Is there a slick way of establishing these that meets OP's requirements? – Sangchul Lee Jun 13 '21 at 23:53