What is $ \lim_{n\to\infty}\frac{1}{e^n}\Bigl(1+\frac1n\Bigr)^{n^2}$?

Solution 1:

Pass to the logarithm inside the exponential: $$e^{-n}(1+1/n)^{n^2}=\exp(-n+n^2\ln(1+1/n))$$ Since $\ln(1+x)=x-x^2/2+o(x^2)$ at $0$, you get $-n+n^2\ln(1+1/n)=-1/2+o(1)$ so the sequence $e^{-n}(1+1/n)^{n^2}$ converges and its limit is $e^{-1/2}$.

Solution 2:

Why don't you want to use the fact that: $$\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = e$$ So it will be $$\lim_{n \to \infty}\frac{1}{e^n}\bigg(\left(1 + \frac{1}{n}\right)^n\bigg)^n $$

Solution 3:

\begin{eqnarray*} \displaystyle &&\lim_{n\rightarrow\infty}(1+\frac{1}{n})^{n^2}\frac{1}{e^n}\\ &=&\displaystyle e^{n^2ln(1+\frac{1}{n})-n}\\ &=&e^{\displaystyle \lim_{n\rightarrow\infty}\left(n^2ln(1+\frac{1}{n})-n\right)}\\ &=&e^{\displaystyle \lim_{m\rightarrow0^+}\left(\left(\frac{1}{m}\right)^2ln(1+m)-\frac{1}{m}\right)}\\ &=&e^{\displaystyle \lim_{m\rightarrow0^+}\left(\frac{ln(1+m)-m}{m^2}\right)}\\ &=&e^{\displaystyle \lim_{m\rightarrow0^+}\frac{\frac{1}{1+m}-1}{2m}}\\ &=&e^{\displaystyle \lim_{m\rightarrow0^+}\frac{-m}{2m(1+m)}}\\ &=&e^{\displaystyle \lim_{m\rightarrow0^+}\frac{-1}{2(1+m)}}\\ &=&e^{-\frac{1}{2}} \end{eqnarray*}