Im trying to compute this limit $$\lim_{n\to\infty} e^{-itn^{1/4}}\left(1-\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}e^{i\frac{t}{n^{1/4}}} \right)^{n} $$ Note: ($t$ is real)
I have tried to expand $$e^{i\frac{t}{n^{1/4}}}=\sum_{k=0}^\infty \left(\frac{it}{n^{1/4}}\right)^{k}\frac{1}{k!}=1+\frac{it}{n^{1/4}}-\frac{t^{2}}{2n^{1/2}}+O\left(\frac{t^{3}}{n^{3/4}}\right) $$
Which gives me $$\lim_{n\to\infty} e^{-itn^{1/4}}\left(1-\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}\left(1+\frac{it}{n^{1/4}}-\frac{t^{2}}{2n^{1/2}}+O(\frac{t^{3}}{n^{3/4}})\right)\right)^{n}=\lim_{n\to\infty} e^{-itn^{1/4}} \left(1+\frac{it}{n^{3/4}}-\frac{t^{2}}{2n}+O\left(\frac{t^{3}}{n^{5/4}} \right) \right)^{n} $$ Im stuck here, and also not even sure if Im using the big O notation correctly (or maybe if its even more practical to use little O, instead ? )
Anyhow the answer is $e^{-\frac{t^{2}}{2}}$, I appreciate help.
Note that
$$e^{i\frac{t}{n^{1/4}}}=1+i\frac{t}{n^{1/4}}-\frac{t^2}{2n^{1/2}}+o(n^{-1/2})$$
[Note that the previous error was here, indeed we need an order 2 since later we need to multiply by n obtaining a term o(1)]
then
$$1-\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}e^{i\frac{t}{n^{1/4}}}= 1+i\frac{t}{n^{3/4}}-\frac{t^2}{2n}+o(1/n)$$
and
$$\left(1-\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}e^{i\frac{t}{n^{1/4}}} \right)^{n}=e^{n\log\left(1-\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}e^{i\frac{t}{n^{1/4}}} \right)}= e^{n\log\left(1+i\frac{t}{n^{3/4}}-\frac{t^2}{2n}+o(1/n)\right)}=e^{n\left(i\frac{t}{n^{3/4}}-\frac{t^2}{2n}+o(1/n)\right)}= e^{itn^{1/4}-\frac{t^2}{2}+\color{red}{o(1)}}$$
and
$$e^{-itn^{1/4}}\left(1-\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}e^{i\frac{t}{n^{1/4}}} \right)^{n}= e^{-itn^{1/4}}e^{itn^{1/4}-\frac{t^2}{2}+o(1)}=e^{-\frac{t^2}{2}+o(1)}\to e^{-\frac{t^2}{2}}$$