Uniform convergence method?

Question

I am self-studying analysis and have reached the topic of uniform convergence. When it comes to series, it seems that Weierstrass' M-test is a powerful tool to determine uniform convergence. However, do similar tests exist in the case of sequences?

In my book (not by anybody famous), there is a single example in which uniform convergence is shown: as per the definition, we wish to show that $d(f_n,f) \rightarrow 0$. Coincidentally, this turned out to be the supremum of a neat function defined on a closed, bounded interval: so calculate the point of maxima, insert it, and see if it goes to zero. Fairly straight-forward.

What else can one do? Are all problems at my level (first course analysis, early undergrad) expected to be solved in this way?. What other techniques are common? When do they work, when don't they?

Here's one I picked from the exercises: $$f_n(x) = \frac{\ln x^n }{1 + x^n}, \ \ x \in [\alpha,\infty), \ \alpha > 1$$

Answer 1

For a series of positive functions, the Weierstrass M-test is the only test you can use.

For a series where the sign of the function value changes with change in the argument, you can use Abel's test or Dirichlet test.

Answer 2

For a sequence $\{f_n\}$ that converges pointless to a function $f$, the usual way to prove uniform convergence is to find a number sequence $M_n$ such that $$ |f_n(x)-f(x)|\le M_n\quad\forall x,\quad\text{and}\quad\lim_{n\to\infty}M_n=0. $$ Sometimes this can be achieved using calculus t0 find the maximum of $|f_n(x)-f(x)|$.

In your example it is easy to see that $f_n$ converges pointwise to $0$. If $x\ge\alpha>1$ and $n\ge2$ then $$ 0<f_n(x)=\frac{n\log x}{1+x^n}\le n\,\frac{\log x}{x}\frac{1}{x^{n-1}}\le\frac{n}{\alpha^{n-1}}. $$ Since $\alpha>1$, $\lim_{n\to\infty}n/\alpha^{n-1}=0$ and the convergence is uniform.

To disprove uniform convergence, a useful technique is to find $x_n$ and $\delta>0$ such that $|f_n(x_n)-f(x_n)|\ge\delta$ or all $n$.