algebraic manipulation question

Question

$M_{z_n}(t)$ is a particular moment generating function, and it is given that $\lambda_n$ approaches $\infty$ as $n$ approaches $\infty$:

Could someone help me see how the above was derived?

Answer 1

The big idea here is that $$ e^x=1+x+\frac{x^2}{2}+O(x^3)\text{ as }x\rightarrow0. $$ So, since $t/\sqrt{\lambda_n}\rightarrow0$, we have $$ e^{t/\sqrt{\lambda_n}}=1+\frac{t}{\sqrt{\lambda_n}}+\frac{t^2}{2\lambda_n}+O\left(\frac{1}{\lambda_n^{3/2}}\right)\text{ as }n\rightarrow\infty, $$ so that $$ -t\sqrt{\lambda_n}+\lambda_n(e^{t/\sqrt{\lambda_n}}-1)=-t\sqrt{\lambda_n}+t\sqrt{\lambda_n}+\frac{t^2}{2}+O\left(\frac{1}{\lambda_n^{1/2}}\right)=\frac{t^2}{2}+O\left(\frac{1}{\lambda_n^{1/2}}\right). $$ Since $\lambda_n\rightarrow\infty$, this last expression converges to $\frac{t^2}{2}$ as $n\rightarrow\infty$.

(We assume here, of course, that $t$ does not change with $n$.)

Answer 2

$$ e^{t/\sqrt{\lambda_n}} = \sum_{k=0}^\infty \frac{(t/\sqrt{\lambda_n})^k}{n!} = 1 + \frac{t}{\sqrt{\lambda_n}} + \frac{t^2}{2\lambda_n} + o\left(\frac{t^3}{\lambda_n^{3/2}}\right) $$ and now substitute this in to get $$ \log M(t) = -t/\sqrt{\lambda_n} + \left[t/\sqrt{\lambda_n} + t^2/2 + o\left(\frac{t^3}{\lambda_n}\right)\right] = t^2/2 + o\left(\frac{t^3}{\lambda_n}\right) $$ and taking the limit as $n\to \infty$, we have $\lambda_n \to \infty$ so the lower order term vanishes.