# On sequences and series of functions

I started studying sequences an series of functions and I found a problem which I don't quite understand. The problem is:

$f_n:(0,1)\rightarrow \mathbb{R}, f_n(x)=\frac{1}{nx+1}$, $n \geq 0$. Study the pointwise and uniform convergence of $f_n(x)$.

I showed that $f$ is pointwise convergent to $f(x)=0$ but I don't understand why is not uniform convergent.The solution in the book is:

$\underset{n \to \infty}{\lim}\hspace{1 mm} \underset{x\in (0,1)}{\sup} \Big | \frac{1 }{nx+1} \Big | =1$. Why? In another example I saw that he took $x=\frac{1}{n}$ to show that a sequence is not uniform convergent.

On series of functions: Can you give me some tips on how to show that a series is uniform convergent? I am confused because I don't know what theorem should I use or what result from the theory. Thank you.

• You can compute the derivative, and find the maximum. – Itay4 Aug 9 '17 at 14:13

For $n>1$,

$$M_n=\sup_{x\in (0,1)}|f_n (x)-0|$$ $$=\sup_{x\in (0,1)}|f_n (x)|\ge |f_n (\frac {1}{n}) |$$

because $\frac {1}{n}\in (0,1)$.

but $$f_n (\frac {1}{n})=\frac {1}{2}$$ thus $$M_n\ge \frac {1}{2}$$ $$\implies \lim_{n\to+\infty}M_n\ge \frac {1}{2}$$ $$\implies \lim_{n\to+\infty}M_n \ne 0$$ hence, the convergence is not uniform at $(0,1)$.

Done!

When $f,g$ are bounded functions from $(0,1)$ to $\mathbb R,$ let $\|f-g\|=\sup \{|f(x)-g(x)|: x\in (0,1)\}.$

Uniform convergence of a sequence $(f_n)_n$ to $f$ is equivalent to $\lim_{n\to \infty}\|f-f_n\|=0.$

If $f(x)=0$ for all $x\in (0,1)$ then $\|f-f_n\|=\|f_n\|.$ So uniform convergence of $(f_n)_n$ to $0$ is equivalent to $\lim_{n\to \infty}\|f_n\|=0.$ But $\|f_n\|\geq |f_n(1/n)|=1/2,$ so convergence is not uniform.

Footnote: In fact, $\|f_n\|=1.$