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This question already has an answer here:

$$\lim_{n\to\infty} \left(n\left(1+\frac{1}{n}\right)^{n}-ne\right) = -\frac{e}{2}$$

So I watched this video https://www.youtube.com/watch?v=FPHHv1UcrMA and it shows one way of proving it, but it basicaly transforms this limit into limit of a function at infinity and uses L'Hopital multiple times, but is there some sequence-like way of doing this limit ? I tried to apply Stolz theorem but it doesn't help: $$\lim_{n\to\infty} n((1+\frac{1}{n})^{n}-e) = \lim_{n\to\infty} \frac{n}{\frac{1}{(1+\frac{1}{n})^{n}-e}} = \lim_{n\to\infty} \frac{n+1-n}{\frac{1}{(1+\frac{1}{n})^{n}-e}-\frac{1}{(1+\frac{1}{n+1})^{n+1}-e}} = \lim_{n\to\infty} \frac{1}{\frac{1}{(1+\frac{1}{n})^{n}-e}-\frac{1}{(1+\frac{1}{n+1})^{n+1}-e}}$$ It doesn't seem to go well to me. Maybe you can suggest some way of doing this one (not nessesarily with Stolz theorem).

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marked as duplicate by YuiTo Cheng, Lee David Chung Lin, max_zorn, StubbornAtom, Cesareo Jul 4 at 7:59

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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For large $n$, $n\ln (1+\frac{1}{n})=1-\frac{1}{2n}+\mathcal{O}(\frac{1}{n^2})$, so $(1+\frac{1}{n})^n=e(1-\frac{1}{2n}+\mathcal{O}(\frac{1}{n^2}))$ and $n((1+\frac{1}{n})^n-e)=-\frac{e}{2}+\mathcal{O}(\frac{1}{n})$.

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  • $\begingroup$ The problem is, I am not quite familiar with O notation. $\endgroup$ – Юрій Ярош Nov 3 '18 at 13:42
  • $\begingroup$ @ЮрійЯрош We write $f=g+O(h)$ if $\lim_{n\to\infty}\frac{f-g}{h}$ exists and is finite and nonzero. Well, technically the definition is the weaker condition that $\frac{f-g}{h}$ is bounded for sufficiently large $n$, but either way you should be able to rewrite the proof in terms of limits. $\endgroup$ – J.G. Nov 3 '18 at 13:52
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    $\begingroup$ How did you get from first to second equation? $\endgroup$ – Юрій Ярош Nov 3 '18 at 18:35
  • $\begingroup$ @ЮрійЯрош Using $\exp\epsilon=1+\epsilon+\mathcal{O}(\epsilon^2)$ for $|\epsilon|\ll 1$. $\endgroup$ – J.G. Nov 3 '18 at 18:36
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    $\begingroup$ OK, then how can I derive all this formulas I mean first one in your post and one you mentioned in the comment? $\endgroup$ – Юрій Ярош Nov 3 '18 at 19:06
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Let $t=1/n$ then $$L=\lim_{t\rightarrow 0} \frac{(1+t)^{1/t}-e}{t}$$ Use McLaurin expansion $$(1+t)^{1/t}=e-et/2+11et^2/24$$ THen $L=-e/2$.

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