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](https://math.stackexchange.com/questions/220705/how-to-obtain-probability-function-from-its-probability-generating-function) 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**)?



Show that $\frac{1}{2 \pi i} \oint_{\mathbb{S}^1} \frac{1-\hat{f}(\xi)}{1-\xi}\cdot \frac{\mathrm{d} \xi}{\xi^{n+1}} \to 0$ as $n \to \infty$
$\newcommand\R{\mathbb R}$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](https://en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_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 $g(t):=\frac{1-\hat{f}(e^{it})}{1-e^{it}}\,i\,1(0<t<2\pi)$.


