$\lim_{n \rightarrow \infty} \int_0^{\infty}f_n(x)dx = \int_0^{\infty} \frac{x}{e^x-1}dx$, where $f_n(x):=\frac{n}{e^x-1}\sin\frac{x}{n}$
Hi I'm working on some practice questions and having a bit of trouble with this one. I have been able to show that $f_n(x) \rightarrow f(x) = \frac{x}{e^x-1}$.
But I am having some trouble trying to find a 'dominant' function $g$, such that $|f_n(x)| \le g(x) $ for all $x > 0$, in which case I could then apply the dominated convergence theorem directly.
Looking at $f_n(x)$ I thought that I could just use $\frac{n}{e^x-1}$, but I realised that $g(x)$ can't depend on $n$. So how do I go about finding such a function?
Any help would be greatly appreciated!
Observe that : $\left|\sin\left(\dfrac{x}{n}\right)\right| < \dfrac{x}{n} < x, x > 0$ and you can prove this fact. Thus it follows that : $|f_n(x)| < g(x) = \dfrac{x}{e^x-1}$. Next consider $f(x) = \sin\left(\dfrac{x}{n}\right) - \dfrac{x}{n} \to f'(x) = \dfrac{\cos\left(\dfrac{x}{n}\right)}{n} - \dfrac{1}{n} < 0\to f(x) < f(0) = 0 \to \sin\left(\dfrac{x}{n}\right) < \dfrac{x}{n}$, and similarly you can prove $-\dfrac{x}{n} < \sin\left(\dfrac{x}{n}\right) \to |\sin\left(\dfrac{x}{n}\right)| < \dfrac{x}{n}$