Calculate limit of $x_n=\sum_{i=1}^nf(\frac{2i-1}{n^2}a)$

Question

Suppose when $x\rightarrow 0$,$f(x) \sim x$, $x_n=\sum_{i=1}^nf(\frac{2i-1}{n^2}a)$ Want to prove: $\lim_{n \to \infty}{x_n}=a(a>0).$

Also, I don't understand what $\sim$ mean?

Answer 1

Hint: Suppose $f(x)=x$, then we see that \begin{align} x_n =&\ \frac{a}{n^2}\sum^n_{i=1}(2i-1)=\frac{2a}{n^2}\sum^n_{i=1}i-\frac{a}{n}\\ =&\ \frac{2a}{n^2}\frac{n(n+1)}{2}-\frac{a}{n}\rightarrow a \end{align} as $n\rightarrow \infty$.

Also, $f(x)\sim x$ as $x \rightarrow 0$ means \begin{align} \lim_{x\rightarrow 0}\frac{f(x)}{x}=1 \end{align} or more usefully \begin{align} f(x)=x+\text{ something small and goes to 0} \end{align} when $x$ is close to $0$.

Answer 2

$f(x)\sim x$ means that $f(x)=x+o(x)$ where $o(.)$ is little-o notation. Therefore by substituting$$x_n=a+\sum_{i=1}^{n} g(a{2i-1\over n^2})$$where $$\lim_{x\to 0}{g(x)\over x}=0$$then you can write $$|g(a{2i-1\over n^2})|\le a{2i-1\over n^2}\epsilon$$for large enough $n$ which means that $$\left|\sum_{i=1}^{n} g(a{2i-1\over n^2})\right|\le \sum_{i=1}^{n} |g(a{2i-1\over n^2})|\le \sum a{2i-1\over n^2}\epsilon=a\epsilon$$therefore$$\lim_{n\to \infty}\sum g(a{2i-1\over n^2})=0$$and $$\lim x_n=a$$