Uniform convergence of a specific sequence of function

Question

i have trouble proving that $f_n(x)= sin(\frac{nx}{n+1})$ is not uniformly convergent on [1,$\infty)$. $x \in \mathbb{C} $. I know that it converges pointwise to sin(x) but i don't know how to proceed. I tried maximizing but that seems to difficult. I also have not found an n to offer a counterexample yet. There are threads in similar looking functions but i can't seem to apply those.

Thanks in advance

Answer 1

Note that $$\left|\sin\left(\frac{nx}{n+1}\right) - \sin x\right| = 2\left|\cos\left(\frac{(2n+1)x}{(2n+2)}\right)\sin\left(\frac{x}{2n+2}\right)\right| = 2|\cos((2n+1)y)||\sin y|$$

where $y = \frac x{2n+2}$. It is clear to see that for each $x$ the difference goes to zero, since the sine term goes to zero with increasing $n$.

However, for example fixing $n$, it can be easily checked with $y= \frac{\pi}{4}$ that the absolute difference is just $1$, since both the cosine and sine terms are just $\frac 1{\sqrt 2}$. This gives $x_n = \frac{(n+1)\pi}2$.

At this $x_n$, the difference is $1$. Therefore, you can see that for any $\epsilon <1$, the conditions for uniform convergence do not hold. However, note that since the "counterexample" is going to infinity as $n$ increases, we get an understanding of at least why ordinary convergence is not hindered by the above fact.

Answer 2

Consider $x=\frac{n+1}{2}\pi$. Note that $$\sin(\frac{n(n+1)\pi}{n+1})=\sin(\frac{n}{2}\pi)=\sin(x-\frac{1}{2}\pi).$$ So for all $N\geq 0$ there exists an $n>N$ and an $x\in[1,\infty)$ such that $$|\sin(x)-\sin(\frac{nx}{n+1})|=1.$$ Clearly $f_{n}$ does not converge uniformly to $\sin(x)$.