For a series, we can use Weierstrass to show uniform convergence. What test does one have for a sequence?
Here's the one I am working on right now: $$\frac{ln(x^n)} {1 + x^n} \ , \ x \in [a, \infty)$$
For $a = 1$, I wish to show lack of uniform convergence, and for $a > 1$, I wish to show that it does convergence uniformly.
It converges pointwise to the zero function, and so I guess we are interested in d$(f_n,0)$ and whether it approaches zero? The calculations get quite messy appealing to the definition, I feel .... any help is appreciated.
Your conjecture is right.
I'm going to study first the behaviour of $f_n$. Some elementary differential calculus proves that $f_n$ is firstly increasing then decreasing, and that it achieves its maximum at $x_n$ such that ${x_n}^n(\ln({x_n}^n)-1)=1$
If you let $g:x\to x(\ln(x)-1)$, there exists a unique $x_0$ such that $g(x_0)=1$ (for the record $x_0 \approx 3.951$). Hence $x_n^n=x_0$ and $x_n=x_0^{1/n}$
Furthermore, $\displaystyle f_n(x_n)=\frac{x_0}{1+x_0}\approx 0.2785$
Let $a>1$, and let $||f_n||:= \sup_{x\geq a}f_n(x)$.
Since $x_n\to 1$, there exists some $N$ such that for every $n\geq N$, $||f_n||=f_n(a)$.
Since $f_n(a) \to 0$, $||f_n||\to 0$ which proves uniform convergence.
If $a=1$, since $\displaystyle ||f_n||=\frac{x_0}{1+x_0}$, $||f_n||$ does not go to $0$,and the convergence is not uniform.