How to say limit of this expression is finite I have to show that
$(1-\frac{t^2}{2r}+O(r^{-\frac{3}{2}}))^{-r}\rightarrow e^{\frac{t^2}{2}}$ as $r\rightarrow\infty$
I have expanded the given expression like below:
$(1-\frac{t^2}{2r}+O(r^{-\frac{3}{2}}))^{-r}$
$= (1-\frac{t^2}{2r})^{-r} - rO(r^{-\frac{3}{2}})(1-\frac{t^2}{2r})^{-r-1} + \binom{r+1}{2}(O(r^{-\frac{3}{2}}))^2(1-\frac{t^2}{2r})^{-r-2}-...$
As $r\rightarrow\infty$,
$(1-\frac{t^2}{2r})^{-r}\rightarrow e^{\frac{t^2}{2}}$
$rO(r^{-\frac{3}{2}})\rightarrow 0$, $\binom{r+1}{2}(O(r^{-\frac{3}{2}}))^2\rightarrow 0$ etc.
But what about the terms $(1-\frac{t^2}{2r})^k, k=-r-1, -r-2,....$
How can I say that they tend to a finite value? Can I say they are finite because $(1-\frac{t^2}{2r})^{-r}$ tends to a finite value(i.e. $e^{\frac{t^2}{2}}$)?
 A: Using Taylor Theorem, it is easy to show that $\log(1-x)=-x+O(x^2)$, from which we have
$$\begin{align}
\left(1-\frac{t^2}{2r}+O\left(r^{-3/2}\right)\right)^{-r}&=e^{-r\log\left(1-\frac{t^2}{2r}+O\left(r^{-3/2}\right)\right)}\\\\
&=e^{t^2/2}e^{-O(r^{-1/2})}
\end{align}$$
Therefore, we see that
$$\lim_{r\to \infty }\left(1-\frac{t^2}{2r}+O\left(r^{-3/2}\right)\right)^{-r}=e^{t^2/2}$$
as was to be shown!


EDIT:
The OP was interested in proceeding with a binomial expansion.  To that end, we now use a generalized binomial expansion of the term if interest to write
$$\begin{align}
\left(1-\frac{t^2}{2r}+O\left(r^{-3/2}\right)\right)^{-r}&=\sum_{k=0}^\infty \binom{-r}{k}\left(1-\frac{t^2}{2r}\right)^{-r-k}O\left(r^{-3/2}\right)^k\tag1
\end{align}$$
where the generalized binomial coefficient, $\binom{-r}{k}$, is given by
$$\binom{-r}{k}=\frac{-r(-r-1)\cdots (-r-k+1)}{k!}$$
Owing to the uniform convergence of the series on the right-hand side of $(1)$, we can interchange the order of the limit and the summation to obtain
$$\begin{align}
\lim_{r\to\infty}\sum_{k=0}^\infty \binom{-r}{k}\left(1-\frac{t^2}{2r}\right)^{-r-k}O\left(r^{-3/2}\right)^k&=\sum_{k=0}^\infty \lim_{r\to \infty}\left(\binom{-r}{k}\left(1-\frac{t^2}{2r}\right)^{-r-k}O\left(r^{-3/2}\right)^k\right)\\\\
&=e^{t^2/2}
\end{align}$$
since for $k\ne 0$,
$$\binom{-r}{k}O\left(r^{-3/2}\right)^k=O\left(r^{-1/2}\right)^k\to 0$$
as $r\to \infty$ and
$$\lim_{r\to \infty}\left(1-\frac{t^2}{2r}\right)^{-r-k}=e^{t^2/2}$$
