Asymptotic equation $e^{-1}(1 + t^{-1})^t = 1 - \dfrac{1}{2}t^{-1} + O(t^{-2})$

Question

I'm trying to prove the following

$$e^{-1}(1 + t^{-1})^t = 1 - \dfrac{t^{-1}}{2} + O(t^{-2})$$

When $t \geq 1$.

What I've tried:

$$(1 + t^{-1})^t = e^{t\log(1 + t^{-1})}$$

Expanding the series of $\log(1 + z)$ when $|z| < 1$ (Taking $z = t^{-1}$ and assuming that $t > 1$, and we can see $t = 1$ independently, but didn't check...)

$$ e^{t \sum_{v = 1}^{\infty} \dfrac{(-1)^{v - 1}}{v} (t^{-1})^v} $$

$$ e^{1 -t^{-1}/{2} + O(t^-2)} $$

Which is somewhat close to where I'm trying to get but I need to get rid of the $e^{\ldots}$ and also adding the $e^{-1}$ kills the leading $1$, so I'm guessing this is the wrong approach.

Any hints or tips would be welcome.

Answer 1

Looks to me like you are exactly right.

$\begin{array}\\ e^{1 -t^{-1}/{2} + O(t^-2)} &=e\ e^{-t^{-1}/{2} + O(t^-2)}\\ &=e\ (1-t^{-1}/{2} + O(t^-2))\\ \end{array} $

Answer 2

$$A=\frac 1 e \left(1+\frac{1}{t}\right)^t\qquad \implies \qquad \log(A)=-1+t \log \left(1+\frac{1}{t}\right)$$

Let $t=\frac 1 x$ $$\log(A)=-1+\frac 1 x \log(1+x)=-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+O\left(x^4\right)$$ Back to $t$ $$\log(A)=-\frac{1}{2 t}+\frac{1}{3 t^2}-\frac{1}{4 t^3}+O\left(\frac{1}{t^4}\right)$$

Using Taylor again $$A=e^{\log(A)}=1-\frac{1}{2 t}+\frac{11}{24 t^2}-\frac{7}{16 t^3}+O\left(\frac{1}{t^4}\right)=1-\frac{1}{2 t}+O\left(\frac{1}{t^2}\right)$$