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

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

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).

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

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 $$n((1+\frac{1}{n})^n-e)=-\frac{e}{2}+\mathcal{O}(\frac{1}{n})$$.
• @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. – J.G. Nov 3 '18 at 13:52
• @ЮрійЯрош Using $\exp\epsilon=1+\epsilon+\mathcal{O}(\epsilon^2)$ for $|\epsilon|\ll 1$. – J.G. Nov 3 '18 at 18:36
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 $$L=-e/2$$.