Without a calculator, so I cannot use the simple expansion of log(1+x) Using wolfram we can see that
$\log(\frac{\sqrt{n}}{2t}[\exp(\frac{2t}{\sqrt{n}})-1)])$
$=\frac{t}{\sqrt{n}}+\frac{t^{2}}{6n}+...$
But by hand,
$\exp(\frac{2t}{\sqrt{n}})=1+\frac{2t}{\sqrt{n}}+\frac{4t^{2}}{2!n}+...$
$\exp(\frac{2t}{\sqrt{n}})-1=\frac{2t}{\sqrt{n}}+\frac{4t^{2}}{2!n}+...$
and so plugging in and using the log taylor directly doesnt give the simplified result. so how can it be seen?
We have $$\exp\left( \frac{2t}{\sqrt{n}}\right)\approx1+\frac{2t}{\sqrt{n}}+\frac{4t^2}{2n}+\frac{8t^3}{6n\sqrt{n}}$$
\begin{align}\log(\frac{\sqrt{n}}{2t}[\exp(\frac{2t}{\sqrt{n}})-1)]) &\approx \log \left(\frac{\sqrt{n}}{2t}\left(\frac{2t}{\sqrt{n}} +\frac{4t^2}{2n}+\frac{8t^3}{6n\sqrt{n}}\right) \right)\\&= \log \left(1 +\frac{t}{\sqrt{n}}+\frac{2t^2}{3n} \right)\\ &=\sum_{n=1}^\infty (-1)^{n+1}\frac{\left(\frac{t}{\sqrt{n}}+\frac{2t^2}{3n}\right)^n}{n}\\ &\approx\frac{t}{\sqrt{n}}+\frac{2t^2}{3n} -\frac12\left(\frac{t}{\sqrt{n}} \right)^2\\ &= \frac{t}{\sqrt{n}}+\frac{t^2}{6n} \end{align}