I have to find if the following two function series are simply and uniformly convergent or not:
a) $f_n:R\rightarrow R,f_n(x)=\frac{1}{n}\arctan(x^n), \forall n\in N^*$
b) $f_n:R\rightarrow R,f_n(x)=\sqrt{x^2+\frac{1}{n}}$
I've come to the conclusion that both are simply convergent, yet I don't know how to search for the uniform convergence.
A sequence of function converges uniformly iff it converges w.r.t to the supremums norm…
I guess you found the following (pointwise) limits:
a) $f(x) = 0$
b) $f(x) = |x| = \sqrt{x^2}$
Then it holds for a)
$$||f_n - f||_\infty = \frac{1}{n}||\arctan||_\infty = \frac{2\pi}{n}$$ and for b) $$\begin{align*}||f_n - f||_\infty &= \sup_{x\in R} \left|\sqrt{x^2 + \frac{1}{n}} - \sqrt{x^2}\right| \\ \\ &= \frac{1}{n} \sup_{x\in R} \frac{1}{\sqrt{x^2 + \frac{1}{n}} + \sqrt{x^2}} \\ &\le \frac{1}{n} \sqrt{n} \\ &= \frac{1}{\sqrt{n}}\end{align*}$$
Because both tend to $0$ for $n\to\infty$ both sequences converge uniformly…