How to Show Convergence of Complex Integral to Zero as n Approaches Infinity

Question

Let $f = (f_0,f_1,\ldots,f_n,\ldots) \in \mathcal{P}(\mathbb N)$ be a probability distribution on $\mathbb N$ and denote by $$\hat{f}(z) = \sum_{n\geq 0} z^n f_n$$ for its probability generating function. It has been shown in this post that we have the following Bromwich's inversion formula (using complex integration over the positively oriented unit circle $\mathbb{S}^1$): $$\sum\limits_{i > n} f_i = \frac{1}{2 \pi i} \oint_{\mathbb{S}^1} \frac{1-\hat{f}(\xi)}{1-\xi}\cdot \frac{\mathrm{d} \xi}{\xi^{n+1}} \label{1}\tag{1}$$ for each $n\geq 0$. Note that the left hand side of equation \eqref{1} represents the probability $\mathbb{P}(X > n)$ for a $f$-distributed random variable $X$. Now my question is, since we have the obvious asymptotic limit $$\lim\limits_{n \to \infty} \mathbb{P}(X > n) = 0.$$ How can we show that the right hand side of equation \eqref{1} also tends to zero as $n \to \infty$ (without using the probabilistic interpretation of the contour integration)?

Answer 1

$\newcommand\R{\mathbb R}$Assume that the integral $$I_n:=\oint_{\mathbb{S}^1} \frac{1-\hat{f}(\xi)}{1-\xi}\,\frac{d\xi}{\xi^{n+1}} =\int_0^{2\pi} \frac{1-\hat{f}(e^{it})}{1-e^{it}}\,\frac{e^{it}\,i\,dt}{e^{i(n+1))t}} $$ exists in the Lebesgue sense -- that is, $$\int_0^{2\pi} \Big|\frac{1-\hat{f}(e^{it})}{1-e^{it}}\Big|\,dt<\infty. $$ Then $$I_n=\int_0^{2\pi} \frac{1-\hat{f}(e^{it})}{1-e^{it}}\,\frac{\,i\,dt}{e^{int}}\to0$$ (as $n\to\infty$) by the Riemann–Lebesgue lemma.

Detail: The Riemann–Lebesgue lemma states that $\int_\R g(t)e^{-itx}\,dt\to0$ as $|x|\to\infty$ if $\int_\R |g(t)|\,dt<\infty$. Here the Riemann–Lebesgue lemma is used with $g(t):=\frac{1-\hat{f}(e^{it})}{1-e^{it}}\,i\,1(0<t<2\pi)$.