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

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

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 $$L=-e/2$$.