Find the value of $\alpha$ for which following limit exists and is equal to a non-zero constant $$\lim_{n \rightarrow \infty}\frac{e \left(1-\frac{1}{n}\right)^n -1}{n^\alpha}$$
I don't know how to begin.
I don't want a full solution, I would like only a hint on how to begin.
Thanks in advance.
On
Hint:
You have to find an equivalent of the numerator. First find the beginning of Taylor's expansion of the log of first term: \begin{align} \log\biggl(\mathrm e\Bigl(1-\frac1n\Bigr)^n\biggr)&=1+n\log\Bigl(1-\frac1n\Bigr)\\&=1+n\Bigl(-\frac1n-\frac1{2n^2}+o\Bigl(\frac1{n^2}\Bigr)\Bigr)= -\frac1{2n}+o\Bigl(\frac1{n}\Bigr), \end{align} so $$\mathrm e\Bigl(1-\frac1n\Bigr)^n-1=\mathrm e^{-\tfrac1{2n}+o\bigl(\tfrac1{n}\bigr)}-1=1-\frac1{2n}+o\Bigl(\frac1{n}\Bigr)-1\sim_\infty-\frac1{2n}$$ and the given expression is asymptotically equivalent to $$\frac{-\frac1{2n}}{n^\alpha}=-\frac1{2n^{\alpha+1}}.$$
On
By the binomial theorem, $$e\left(1-\frac1n\right)^n-1=e\left(1-1+\frac{n(n-1)}2\frac1{n^2}-\frac{n(n-1)(n-2)}{3!}\frac1{n^3}+\cdots\right)-1\\ =e\left(1-1+\frac12\left(1-\frac1n\right)-\frac1{3!}\left(1-\frac1n\right)\left(1-\frac2n\right)+\cdots\right)-1.$$
The constant terms yield
$$e\left(1-1+\frac12-\frac1{3!}+\cdots\right)-1=e\frac1e-1=0.$$
The remaining terms are
$$e\left(\frac12\left(-\frac1n\right)-\frac1{3!}\left(\left(1-\frac1n\right)\left(1-\frac2n\right)-1\right)+\cdots\right).$$
The coefficient of $1/n$ will be $e$ times
$$\frac12(-1)-\frac1{3!}(-1-2)+\frac1{4!}(-1-2-3)-\frac1{5!}(-1-2-3-4)+\cdots,$$ which is convergent.
More precisely,
$$-\sum_{k=2}^\infty\frac{(-1)^k}{k!}\frac{(k-1)k}2=-\frac12\sum_{k=2}^\infty\frac{(-1)^k}{(k-2)!}=-\frac1{2e}$$ so that the limit with $\alpha=-1$ is $-\dfrac12$.
On
$$\lim_{n \rightarrow \infty}\frac{e \left(1-\frac{1}{n}\right)^n -1}{n^\alpha}=\lim_{n \rightarrow \infty}\left(\frac{e \left(1-\frac{1}{n}\right)^n -1}{\ln\left(1+e \left(1-\frac{1}{n}\right)^n -1\right)}\cdot\frac{1+n\ln\left(1-\frac{1}{n}\right)}{n^\alpha}\right)=$$ $$=\lim_{n \rightarrow \infty}\frac{1+n\ln\left(1-\frac{1}{n}\right)}{n^\alpha}=-\frac{1}{2}$$ for $\alpha=-1$ because $1+n\ln\left(1-\frac{1}{n}\right)=1+n\left(-\frac{1}{n}-\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right)\right)$
$$e\left(1-\frac {1}{n}\right)^n=e^{n\left(\ln(1-\frac {1}{n})+\frac {1}{n}\right)} $$
$$\ln\left(1-\frac {1}{n}\right)=-\frac {1}{n}-\frac {1}{2n^2}(1+\epsilon (n) )$$
thus, we will compute
$$\lim_{n\to+\infty}\frac {-(1+\epsilon (n))}{2n^{1+\alpha}}$$
You should take $$\alpha=-1$$
the limit is then $=-\frac{1}{2}$.