Absolute convergence does not imply uniform convergence Let 
$$f_n(x) =
\begin{cases}
0,  & \text{if $x<\frac{1}{n+1}$} \\[2ex]
\sin^2\frac{\pi}{x}, & \text{if $\frac{1}{n+1}\leq{x}\leq\frac{1}{n}$}\\[2ex]
0, & \text{if $\frac{1}{n}<x$}
\end{cases}$$
Show that $(f_n)_n$ converges to a continuous function, but does not converge uniformly.
Use the series $\sum_nf_n$ to show that the absolute convergence does not imply the uniform convergence.
The information I have:


*

*The Weiestrass test which implies that if $f_n\in\mathcal{C^b}$ and $M_n=\sup|{f_n(x)}|$ and $\sum_{n=1}^\infty{M_n}$ converges then the serie converges uniformly. Recalling that $\mathcal{C^b}$ is the set of continuous bounded functions, which is the case of $f_n$. I know it isn't enough, but it's a beggining.


The questions I have:


*

*In the first step, i want to show that $(f_n)_n$ does not converges uniformly by absurd. I saw the graphic and its obvious to see it, but i'm not so sure how to start the proof. My difficulty is that the functions doesn't change by $n$, what changes is the interval where $x$ is.
See: 

*Then, the question asks to show that $(f_n)_n$ converges to a continuous function. That's right (by image), but how can i prove it? To me, the sequence is going to a non-function, a vertical line in $x\to\infty$ and y between 0 a 1 because the maximum of the map is not decreasing. Am i wrong?

*And about the $\sum_n{f_n}$? What does it means (specifically about this $f_n)$?


This question seems very hard to me, but i need to do it and learn it, Step by step.
 A: The functions $f_n$ are obviously continuous on $]\frac{1}{n+1},\frac1n[$ (because here $f_n(x)$ is $\sin(\pi/x)$, which is continuous everywhere on $]0,\infty[$), and they are continuous on $[0,\frac{1}{n+1}[$ and $]\frac1n,\infty[$, because $f_n(x)=0)$ there$. Since
$$\lim_{x\to(\frac1{n})^+}f_n(x)=\lim_{x\to(\frac1{n})^-}f(x)=0$$
and likewise at $\frac1{n+1}$, $f_n$ is continuous everywhere.

For a given $x>0$, you can prove that $f_n(x)\to0$, simply because all the values of $f_n(x)$ are zero, except possibly finitely many. And $f_n(0)=0$ for all $n$. Hence $f_n$ converges "pointwise" to the zero function.
So, we have a sequence of continuous funtctions, that converges pointwise to the zero function, which is also continuous.
However, you are right, the maximum of $f_n(x)$ is always $1$, because for $x_n=\dfrac1{n+1/2}$, which is in $[1/(n+1),1/n]$,
$$\sin^2(\pi/x_n)=\sin^2(n\pi+\pi/2)=1$$
However, $f_n(x)$ converges uniformly if convergence happens at each point "at once", that is, for all $\epsilon>0$, there is some $n_0$ such that for all $x$ and all $n>n_0$, $|f_n(x)|<\epsilon$. It has to happen for instance for $\epsilon=1/2$. But this can't be true, since, with $x_n$ as above, $|f_n(x_n)|=1>\epsilon$.
Hence the sequence $f_n$ does not converge uniformly.

Now for $\sum_n f_n(x)$. The successive intervals $[1/(n+1),1/n]$ are disjoint except at $1/n$ and $1/(n+1)$, where $f_n$ is zero anyway, and their union is $]0,1]$, hence the sum is simply the function $f(x)=\sin(\pi/x)$ for $x\in]0,1]$ and $f(x)=0$ elsewhere. Your plot shows a part of it.
We have a theorem (see here, that states that if a series of continuous functions converges uniformly to a function $f$, then $f$ has to be continuous.
Here we have a sum of continuous functions, which converge to a function that is certainly not continuous at $0$: the limit $\lim_{n\to0^+}f(x)$ does not exists, and $f$ oscillates in every interval $[0,\epsilon[$. Hence the convergence can't be uniform.
