Prove that the series $\sum f_n$ has normal convergence only on an interval $[a, + \infty[$. 
We have the sequence of functions: $f_n(x) = \frac{1}{(1 + nx)(n + x)} \space $ on $I = ]0, + \infty[$.
We note: $f(x) = \sum_{n = 0}^{+ \infty} f_n (x)$.
  
  
*
  
*Examine the convergence and uniform convergence of $f_n$ on $ I$. 
  
*Prove that the series $\sum f_n$ converges on $I$.
  
*Prove that the series $\sum f_n$ has normal convergence on the interval $[a, + \infty[$ with $ a > 0$ and not on $I$. 
  
*Prove that $f$ is continuous on $I$.
  
*Prove that $f$ is integrable on $I$ and: 
  $$ \int_{0}^{+ \infty} f(x)dx = 1 + 2 \sum_{n = 2}^{+ \infty} \frac{ln(n)}{n^2- 1}$$




*We have: $|f_n(x)| \leq \frac{1}{n^2 x}$
And: $ \sum |f_n(x)| \leq \sum \frac{1}{n^2 x} \leq \frac{1}{n^2} $ only if $ x \in [1, + \infty [$
since $ \sum \frac{1}{n^2}$ is convergent, so is $\sum ||f_n(x)||_{\infty}$ for $x \geq 1$. I am not sure about this. Is it correct? 
From question 3, the series $\sum f_n$ converges normally on the interval $I = [a, + \infty[$, which means we have uniform convergence on the same interval. I can switch the sum and the integral.
$\begin{align}
\int_{0}^{+ \infty} f(x)dx = & \int_{0}^{+ \infty} \sum_{n = 0}^{+ \infty} f_n(x)dx \\
& =  \sum_{n = 0}^{+ \infty} \int_{0}^{+ \infty} f_n(x)dx \\
& = \int_{0}^{+ \infty} f_0(x)dx + \int_{0}^{+ \infty} f_1(x)dx + \sum_{n = 2}^{+ \infty} \int_{0}^{+ \infty} f_n(x)dx
\end{align}$ 
I have problem with: $ \int_{0}^{+ \infty} f_0(x)dx = \int_{0}^{+ \infty} \frac{1}{x}dx$ 
I don't know how to compute this improper integral. If it is $0$, I will have the wanted result. 
Although, if the series $\sum f_n$ has uniform convergence on the interval $[a, 
+ \infty[$. Why do we take the integral from $0$? Why not $a$ ?
Thank you. 
 A: Actually, if you sum from $0,$ part 5. is false, as $\int_0^{\infty} \frac{1}{x} \, dx = +\infty.$ A suitable correction would be to redefine $f(x) = \sum_{n=1}^{\infty} f_n(x).$ 
(1.), (2.),(3.) and (4.) follow by the OP's considerations: (1.) can be shown relatively easy by $f_n(x) \le \frac{1}{n} \to 0$ uniformly. (2.) then follows by the following passage:

since $ \sum \frac{1}{n^2}$ is convergent, so is $\sum ||f_n(x)||_{\infty}$ for $x \geq 1$. I am not sure about this. Is it correct? 

Yes, it is. Indeed, it holds all the way to any $a>0,$ as each summand will be $|f_n(x)| \le \frac{1}{n^2 a},$ which will converge on that range. This is enough to prove that the convergence is normal on $(a,+\infty)$ and that the limit function is continuous in $I$ (as the limit of a normal family of continuous functions is continuous). 
Also, one notices right away that $\lim \sum_{n=1}^N f_n(0) = 1 + \frac{1}{2} + \cdots + \frac{1}{n} + \cdots = + \infty.$ It is also not terribly difficult to prove that $\lim_{x \to 0} f(x) = +\infty$ - indeed, we see that, for $x$ close to $0$,
$$ f(x) > \frac{1}{2} \left(1 + \frac{1}{2} + \cdots + \frac{1}{n}\right),$$ 
for each $n >0.$ Here, how close $x$ is to zero depends on the $n$ we take. But we fix any $n$ such that $1 + \frac{1}{2} + \cdots + \frac{1}{n} > 2N,$ for $N$ big. This shows that the convergence is not normal over the whole $I$, otherwise we would have $f$ bounded on $(0,+\infty).$ 
Thus, we have answered (1.),(2.),(3.) and (4.) indeed.
We calculate the integral in number 5: 
First, we write $f_n(x)  = \frac{1}{n^2-1} \left(\frac{n}{1+nx} - \frac{1}{n+x}\right),$ for $n\ge2.$ As mentioned, we are allowed - by, for instance, Fubini - to exchange summation and integration. But 
$$ \int_0^{\infty} f_n(x) \, dx = \lim_{T \to \infty} \int_0^T f_n(x) \,dx = \lim_{T\to \infty}\frac{1}{n^2-1}\left(\int_0^T\left(\frac{n}{1+nx} - \frac{1}{n+x}\right) \, dx\right)  = $$
$$ = \lim_{T \to \infty} \frac{1}{n^2-1}\left(\log(nT+1)-\log(T+1) + \log(n)\right) = 2 \frac{\log(n)}{n^2-1}. $$
(5) follows directly from that and the calculation of $\int_0^{\infty}f_1(x) \, dx,$ which can be seen, e.g. by seeing that $F_1(x) = -\frac{1}{(1+x)}$ is a primitive to $f_1,$ to be $1$. 
Also, by redoing the calculations, we might see that integrating from $a$ only changes the calculation to 
$$ \int_a^{\infty} f_n(x) \, dx = \frac{\log(n) + \log(a+n) - \log(na+1)}{n^2-1}.$$
So, the answer is: one is allowed to take integration from $(0,+\infty)$ because of Fubini's theorem (all functions involved are positive, so if we integrate each $f_n$ over $(0,\infty)$, sum those integrals up and this is finite, so is the original integral). However, nothing deters us from doing the same thing from $(a,+\infty)$ and taking limits - both methods yield the same result. 
