Asymptotic behavior of $f(x) = \sum_{k=1}^{\infty} \sqrt{ 1 - \frac{k^2}{x^2} }$ as $x\to\infty$

Let $g:\mathbb{R}\to\mathbb{R}$ be the function $$g(x) = \left\{ \begin{array}{cl}\sqrt{ 1 - \frac{1}{x^2} } &; x>1 \\ 0 &; x\leq 1 \\ \end{array}\right.$$ and $f:\mathbb{R}\to\mathbb{R}$ $$f(x) = \sum_{k=1}^{\infty} g\left(\frac{x}{k}\right).$$ What's the asymptotic behavior of $f(x)$ as $x\to\infty$?

Solution 1:

You have

$\begin{align} & g\left(\frac{x}{k}\right) = \begin{cases} \sqrt{1 - \frac{k^2}{x^2}} &;\; x > k \\ 0 &;\; x \le k . \end{cases} \end{align}$

Note that $ g\left(\frac{x}{k}\right) $ decreases with $ k $: for any fixed $ x $, $ g\left(\frac{x}{k}\right) \ge g\left(\frac{x}{k+1}\right) $ for all $ k $. On the other hand, $ g\left(\frac{x}{k}\right) $ increases with $ x $, i.e., $ x_2 > x_1 \implies g\left(\frac{x_2}{k}\right) \ge g\left(\frac{x_1}{k}\right) $.

Let $ \lceil x \rceil $ denote the smallest integer greater than or equal to $ x $. Then,

$\begin{align} & f(x) = \sum_{k=1}^\infty g\left(\frac{x}{k}\right) = \sum_{k=1}^{\lceil x \rceil - 1} g\left(\frac{x}{k}\right) + \sum_{k = \lceil x \rceil}^{\infty} g\left(\frac{x}{k}\right) = \sum_{k=1}^{\lceil x \rceil - 1} g\left(\frac{x}{k}\right) + 0 . \end{align}$

Given any $ M \ge 3 $, let $ 2 M_0 + 1 $ be the largest odd number less than or equal to $ M $. For any $ x \ge 2 M_0 + 1 $,

$\begin{align} f(x) &= \sum_{k=1}^{\lceil x \rceil - 1} g\left(\frac{x}{k}\right) \\ &\ge \sum_{k=1}^{2 M_0} g\left(\frac{x}{k}\right) \\ &> \sum_{k=1}^{M_0} g\left(\frac{x}{k}\right) &&\left( \text{because } g\left(\frac{x}{k}\right) > 0 \text{ for } x > k \right) \\ &\ge M_0 \sqrt{1 - \frac{M_0^2}{x^2}} &&\left( \text{because } g\left(\frac{x}{k}\right) \ge g\left(\frac{x}{M_0}\right) \text{ for } k \le M_0 \right) \\ &> M_0 \sqrt{1 - \frac{M_0^2}{(2 M_0)^2}} &&\left( \text{because } x > 2 M_0 \right) \\ &= M_0 \frac{\sqrt{3}}{2} . \end{align}$

Given any large $ K $, one can choose $ M $ such that $ M_0 \frac{\sqrt{3}}{2} > K $. Then, for all $ x \ge 2 M_0 + 1 $, $ f(x) > K $. Hence, $ \lim_{x\to \infty} f(x) = \infty $.